Interpolation of Polynomials and Singular Curves: Segre and Veronese Varieties

Abstract

We study an interpolation problem (objects singular at a prescribed finite set) for curves instead of hypersurfaces. We study singular curves in projective and multiprojective spaces. We construct curves that are singular (or with maximal dimension Zariski tangent space) at each point of a prescribed finite set, while the curves have low degree or low “complexity” (e.g., they are complete intersections of hypersurfaces of low degree). We discuss six open problems on the existence and structure of the base locus of the set of all hypersurfaces of a given degree and singular at a prescribed number of general points. The tools come from algebraic geometry, and some of the results are only existence ones or only asymptotic ones (but with as explicit as possible bounds). Some of the existence results are almost constructive, i.e., in our framework, random parameters should give a solution, or otherwise, take other random parameters.

1. Introduction

Key problems in algebraic geometry and linear algebra are related to secant varieties, which, in turn, are related to the following interpolation problem: finding (or proving the non-existence of) hypersurfaces of a prescribed degree that are singular at a prescribed number of general points [1,2]. In this paper, we consider a similar problem for curves instead of hypersurfaces, i.e., curves of prescribed “degree” (or with the lowest degree) that are singular at a prescribed number of “general” points. However, there are at least three “solutions” to this problem, and the solutions give different degrees for “minimal curves”. Hence, we obtain at least three problems with different answers. One problem is finding reduced and irreducible (also called integral) curves of minimal degree with a prescribed singular locus, the same problem dropping the irreducibility assumption and the same problem dropping both assumptions, allowing reducible curves with “multiple” components (see below for their definition). Degree 2 plane curves show that these three problems are different. Plane curves are hypersurfaces of the ambient variety, the plane. These problems are much more challenging for curves in higher dimensional varieties, where curves are not hypersurfaces. We focus on curves in projective and multiprojective spaces.

Our tools come from algebraic geometry. One of our tools (residual exact sequences) was heavily used not only in [1,2] for secant varieties but also in the study of singularities [3,4], but the queries we answer or ask are, as far as we know, quite different.

For any variety M and any positive integer s, let $S(M,s)$ denote the set of all $S\subset M$ such that $\#S=s$. The number, s, of the prescribed singularities is obviously an important parameter for the difficulty of the interpolation problem “find a curve C singular at the prescribed set $S\in S(M,s)$”. Another parameter is the dimension, n, of the projective space (or the multiprojective space) M in which we need to find the curve. Not all curves are the same. We may look at irreducible curves or reduced curves or reduced and connected curves or (the more general case) local Cohen–Macaulay curves. We recall that a projective scheme $(X,{\mathcal{O}}_X)$ is called a local Cohen–Macaulay curve if all local rings ${\mathcal{O}}_{X,p}$, $p\in X$, are one-dimensional local Cohen–Macaulay rings, i.e., they have dimension 1 and depth 1. We give many examples in which the type of curves we require gives different minimal degrees.

We use the Zariski topology even when the base field is the field of complex numbers. The main point of the Zariski topology is that non-empty open subsets U of an irreducible variety, M, are huge: they are dense and their complement has lower dimension. Hence, a “random point” of M should be in U. If not, try another random point. In all cases, the set parametrizing the “possible curves” is a non-empty Zariski open subset $\mathcal{V}$ of a large finite-dimensional vector space. The projective space is sometimes an open subset of a Grassmannian (Theorem 1), once a linear system on a toric surface (part (ii) of Example 1 and Proposition 3), often a finite product of vector spaces of multivariate homogeneous polynomials with a prescribed number of variables, $n+1$, and a prescribed degree. For the multiprojective spaces (Section 5, part (i) of Example 1 and Proposition 2), one uses multihomogeneous polynomials instead of homogeneous polynomials.

Not all singularities are the same. We also consider curves that are fully singular at prescribed points in the following sense.

Let W be a smooth and connected variety. Assume $n:=dimW>1$. For any $p\in M$, let $2p$ or $(2p,M)$ denote the closed subscheme of W with ${\left({\mathcal{I}}_p\right)}^2$ as its ideal sheaf. The scheme $2p$ is zero-dimensional; $\text{deg}\left(2p\right)=n+1$, and p is the reduction of $2p$. Fix $p\in W$, and let $C\subset W$ be a local Cohen–Macaulay one-dimensional scheme. We say that C is fully singular at p if $2p\subset C$. For any $S\in S(W,s)$, we say that C is fully singular along S or at all points of S if it is fully singular at each point of S.

For $n=2$, being singular is equivalent to being fully singular, but if $n>2$, it is weaker (Remark 9).

There are other criteria on the complexity of a curve, not only the degree. A reasonable criterion is to count (in some way) the minimal “weight” of the hypersurfaces needed to describe the curve. A reasonable “weight” may be the sum of the degrees of any set of hypersurfaces needed to describe the curve. For a general S, we have very strong (sometimes optimal) answers for projective spaces and for Segre–Veronese embeddings in multiprojective spaces.

Several geometric properties of projective varieties are translated first in a problem on secant varieties and then (using the Terracini Lemma) in an interpolation problem about double points. This is not just a nice abstract theory because several practical problems on tensors are translated in terms of secant varieties of multiprojective spaces [1,2,5]. The case of the additive decomposition of forms was solved by J. Alexander and A. Hirschowitz [6,7,8] in a series of fundamental papers that showed the power of their techniques. Their result was reproved by several authors, some of them using different techniques [9]. In the Alexander–Hirschowitz paper, it was shown that the interpolation is as good as possible, except for four exceptional cases, each of them previously known. Since the tensor case (and partially symmetric tensor case) is so important, many mathematicians gave positive results (and a few negative ones) [10,11,12,13,14,15,16,17,18].

Question 1: Fix integers $n\ge 2$, $d\ge 5$ and $x\le \left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-n-1$. Take a general $S\subset {\mathbb{P}}^n$ such that $\#S=x$. Is the base locus of ${\mathcal{I}}_{2S}\left(d\right)$ outside S empty or at least of very low dimension?

A Yes answer to Question 1 for very high integers x does not follow from vanishing theorems like the ones used to prove the Alexander–Hirschowitz theorem. Of course, for low x, scheme $2S$ is the scheme-theoretic base locus of ${\mathcal{I}}_{2S}\left(d\right)$ (Remark 1). As far as we know no, vanishing theorem gives the result even for, say, $3n+1\le \left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-(n+1)x\le n+1$, and the non-existence of non-imposed base points is a delicate problem even for curves and a single general double point [19,20,21]. In Section 4, we discuss Question 1 for (partially symmetric) tensors, i.e., for general finite subsets of a multiprojective space. As far as we know, Question 1 is not a consequence (or may be proven with similar techniques) as the generic identifiability problem proven in [22].

Question 2: Fix integers n, d such that $n\ge 3$, $d\ge 3$ and $(n,d)\notin \left\{\right(3,4),(4,3),(4,4\left)\right\}$. Compute the maximal integer x such that $2S$ is the scheme-theoretic intersection of the elements of $|{\mathcal{I}}_{2S}\left(d\right)|$, where S is a general subset of ${\mathbb{P}}^n$ with $\#S=x$.

Question 3: Fix integers n, d and t such that $n\ge 3$, $0\le t\le n-1$, $d\ge 3$ and $(n,d)\notin \left\{\right(3,4),(4,3),(4,4\left)\right\}$. Compute the maximal integer $\alpha (n,d,t)$ such that the base locus of ${\mathcal{I}}_{2S}\left(d\right)$ has dimension $\le t$, where S is a general subset of ${\mathbb{P}}^n$ with $\#S=\alpha (n,d,t)$.

By the theorem of Alexander–Hirschowitz, we have $\alpha (n,d,n-1)=\lfloor (\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-1)/(n+1)\rfloor$[6,7,8]. The same theorem gives $\alpha (n,d,t)\le \lfloor (\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-n+t)/(n+1)\rfloor$ (Remark 6).

We work over an algebraically closed field of characteristic 0 and obtain the following main results.

• We have $\alpha (n,d,n-2)=\lfloor (\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-2)/(n+1)\rfloor$ if $d\ge 2n+1$ (Theorem 6).
• We construct low degree rational curves that are singular or fully singular at a prescribed finite set (Theorems 1 and 4).
• We give a detailed description of two cases in which the type and geometry of the curves matter (Theorems 2 and 7).
• We prove two asymptotic results for the existence of integral curves singular (resp. fully singular) at a prescribed set and with arbitrarily large degrees (Theorems 3 and 8). The bounds are effective and explicit, but the construction of the integral singular curves is not “effective”.
• We construct low degree complete intersection space curves that are fully singular at a general subset with a prescribed cardinality (Corollary 1).
• We prove two asymptotic results on the existence of integral curves with prescribed singular (or fully singular) points in a multiprojective space (Theorems 9 and 11).

Question 4: Fix $n\ge 3$ and $s\ge n+1$. Take a general $S\in S({\mathbb{P}}^n,s)$. Compute the minimal degree of an irreducible curve (resp. a reduced curve, resp. a reduced and connected curve, resp. a local Cohen–Macaulay one-dimensional scheme) C such that $S\subseteq \mathrm{Sing}\left(C\right)$. For any such reduced curve, C is $S=\mathrm{Sing}\left(C\right)$? Now, we are restricted to integral curves. Are all degrees larger than the minimal degree of an integral curve D such that $S\subseteq \mathrm{Sing}\left(D\right)$?

In Section 5, we consider Question 4 with the appropriate modifications, e.g., multidegree instead of degree, in the case of multiprojective spaces (see Question 6).

In Section 6, we discuss Questions 1, 2 and 3.

We conclude with a section (discussion) reiterating the main question and explaining which parts of the paper may be made “explicit”.

2. Preliminaries

We work over an algebraically closed field of characteristic 0.

For any algebraic set or zero-dimensional scheme A of a projective space ${\mathbb{P}}^m$, let $\langle A\rangle$ denote the linear span of A, i.e., the intersection of all hyperplanes of ${\mathbb{P}}^m$ containing A with the convention $\langle A\rangle ={\mathbb{P}}^m$ if there is no such hyperplane. Note that $\langle A\rangle =\varnothing$ if and only if $A=\varnothing$. We write $dim\varnothing =-1$ when we see ∅ as a linear subspace of a projective space.

Remark 1.

Let $Z\subseteq W$ be zero-dimensional schemes contained in a projective space ${\mathbb{P}}^m$. We have $\langle Z\rangle \subseteq \langle W\rangle$, $dim\langle Z\rangle \ge max\{-1,m+1-\text{deg}(Z\left)\right\}$ and $dim\langle W\rangle \le dim\langle Z\rangle +\text{deg}\left(W\right)-\text{deg}\left(Z\right)$.

Remark 2.

Take a smooth quasi-projective variety $W\subseteq {\mathbb{P}}^r$ of dimension $n\ge 2$ and a complete and connected smooth curve $L\subset W$. Since L is closed in ${\mathbb{P}}^r$, it has a degree. Set $c:=\text{deg}\left(L\right)$. We want to construct a local Cohen–Macaulay structure C on L with $C_{\mathrm{red}}=L$ and $\text{deg}\left(C\right)=2c$. Note that $\text{deg}\left(C\right)$ is well-defined, because C is closed in ${\mathbb{P}}^r$. First, assume $n=2$. In this case, L is a Cartier divisor of W, and $2L$ (as a Cartier divisor of W) is the only possible solution. Now, assume $n>2$. We fix an integer $d\gg c$ and use the case $n=2$ applied to the intersection of W and $n-2$ general degree d hypersurfaces containing L.

For any zero-dimensional scheme Z of a variety M and any effective Cartier divisor T of M, let ${\mathrm{Res}}_T\left(Z\right)$ denote the residual scheme of Z with respect to T, i.e., the closed subscheme of M with ${\mathcal{I}}_Z:{\mathcal{I}}_T$ as its ideal sheaf.

For any smooth variety M and any finite set $S\subset M$ set $(2S,M):={\cup }_{p\in S}(2p,S)$. We often write $2S$ instead of $(2S,M)$.

3. Singular Curves in Projective Spaces

Theorem 1.

Fix integers $n\ge 2$ and $s>0$. Take $S\in S({\mathbb{P}}^n,s)$ and any integer $d\ge 2s+n$. Then there is an integral and rational degree d curve singular at all points of S and spanning ${\mathbb{P}}^n$.

Proof. 

Let $D\subset {\mathbb{P}}^d$ be a degree d rational normal curve. For any linear subspace $V⊊{\mathbb{P}}^d$, let ${\ell }_V:{\mathbb{P}}^n\setminus V\to {\mathbb{P}}^{n-dimV-1}$ denote the linear projection from V. We fix an ordering $o_1,\dots ,o_s$ of the points of S. Fix $2s$ points $p_{i,j}\in D$ with $i=1,\dots ,s$ and $j=1,2$. Set $E_i:=\langle \{p_{i,1},p_{i,2}\}\rangle$. Since D is a rational normal curve and $d\ge 2s-1$, the linear space $W:=\langle E_1\cup \cdots \cup E_s\rangle$ has dimension $2s-1$. We take a general n-dimensional linear subspace $M\subset {\mathbb{P}}^d$. We identify M with ${\mathbb{P}}^n$, and hence, we write $S\subset M$ and $(2S,M)$ instead of $2S$. If $dimV=d-n-1$ and $V\cap M=\varnothing$, we take M as the target of ${\ell }_V$, and hence, we obtain a morphism ${\ell }_V:{\mathbb{P}}^n\setminus V\to M$ whose restriction to M is the identity map. Since we took M general after fixing the points $p_{i,j}$, hence, after fixing W, $M\cap W=\varnothing$. For any positive integer x, let ${\sigma }_x\left(D\right)$ denote the x-secant variety of D[2,5]. Since D is an irreducible and non-degenerate curve and $d\ge 2s-1$, $dim{\sigma }_s\left(D\right)=2s-1$ ([5], Remark 1.6). Since $d\ge n+2s$ and M is general, $M\cap {\sigma }_s\left(D\right)=\varnothing$. Set $V_i:=\langle E_i\cup \left\{o_i\right\}\rangle$, $i=1,\dots ,s$, and take general lines $L_i\subset V_i$, $i=1,\dots ,s$. Set $W_1:=\langle L_1\cup \cdots \cup L_s\rangle$. Since $L_i$ is general in $V_i$ and $dimW=2s-1$, $dim{W}_1=2s-1$. Since $M\cap W=\varnothing$ and each $L_i$ is general in $V_i$, $W_1\cap M=\varnothing$. Since D is a rational normal curve, the scheme-theoretic intersection $W\cap D$ is the union of all $p_{i,j}$. Since $V_i$ is the linear span of $\langle \{p_{i,1},p_{i,2}\}\rangle$ and a point of M, $W_1\cap D=\varnothing$. Let $V\subset {\mathbb{P}}^d$ be a general $(d-n-1)$-linear subspace containing $W_1$. Since $W_1\cap M=W_1\cap D=\varnothing$ and V is general, $V\cap D=\varnothing$; hence, ${\ell }_V$ induces a morphism $\mu :D\to M$. Set $C:=\mu \left(D\right)$. Obviously, C is a rational curve. If $\text{deg}\left(\mu \right)=1$, i.e., if $\mu :D\to C$ is isomorphic to the normalization map, then $\text{deg}\left(C\right)=d$, and C is singular at each point of S. Hence, to conclude the proof, it is sufficient to prove that $\text{deg}\left(\mu \right)=1$. Hence, it is sufficient to prove that V only meets a finite number of secant lines. Since D is a rational normal curve and $d>2$, no point of ${\mathbb{P}}^n$ is contained in an infinite number of secant lines of D. Hence, it is sufficient to prove that $V\cap {\sigma }_2\left(D\right)$ is finite. Since D spans ${\mathbb{P}}^n$, the curve C spans ${\mathbb{P}}^n$. Since $W_1\cap {\sigma }_2\left(D\right)$ is finite, V is a general linear subspace containing $W_1$ and $d-dimV>3=dim{\sigma }_2\left(D\right)$, the set $V\cap {\sigma }_2\left(D\right)$ is finite, concluding the proof. □

Proposition 1.

Fix integers $n\ge 2$, $s>0$ and $g\ge 0$. Take any $S\in S({\mathbb{P}}^n,s)$ and any smooth genus g curve X. Then, for all integers $d\ge g+n+2s$, there is an integral degree d curve $C\subset {\mathbb{P}}^n$ such that $S\subseteq \mathrm{Sing}\left(C\right)$ and X is the normalization of C.

Proof. 

Since $d\ge g+n+2s$, there is a non-special embedding of X as a linearly normal curve $D\subset {\mathbb{P}}^{d-g}$. We take $2s$ general points of D and then make the linear projections as in the proof of Theorem 1. Alternatively (for the application to Proposition 10), we first take $2s$ general points of X and then take a general non-special embedding of degree d. We need to find an embedding $D\subset {\mathbb{P}}^{d-g}$ without Segre points in the sense of [23] and then preserve this condition in the linear projection. Since $d\ge g+n+2s$, it is sufficient to take as D the image of the embedding induced by the general degree d line bundle of C. □

Easy examples show that (for some S) Theorem 1 does not give the minimal degree of an integral curve singular at all points of S (Example 1). In the same example, we discuss lower degrees reducible or multiple curves singular at all points of S. The main advantage is that it is almost a constructive proof and the rational parametrizations are very cheap ways to describe curves (if they have a rational parametrization, i.e., only for very specific curves).

Remark 3.

Fix a finite set $S\subset {\mathbb{P}}^n$, $n\ge 2$, and let e (resp. f) be the minimal integer such that $|{\mathcal{I}}_S\left(e\right)|\ne \varnothing$ (resp. $|{\mathcal{I}}_S\left(f\right)|$ contains an irreducible hypersurface). Since S is a finite set and e is minimal, every $T\in |{\mathcal{I}}_S\left(e\right)|$ has no multiple components, and no proper union of the irreducible components of T contains S. Using the theorem of Bertini ([24] Th. 6.3; [25]), it is easy to check that f is the minimal degree of a hypersurface containing an integral curve C with $S\subseteq \mathrm{Sing}\left(C\right)$ and that e is both the minimal degree of a local Cohen–Macaulay curve $C_1$ with $S\subseteq \mathrm{Sing}\left(C_1\right)$ and of a reduced curve $C_2$ with $S\subseteq \mathrm{Sing}\left(C_2\right)$.

Example 1.

Fix an integer $n\ge 3$ and integers $a\ge 2$, $b\ge 2$, $v\ge u\ge 2$, e and w with $0\le e\le ab-a-b+1$ and $0\le w\le 1+vu-v-u(u+1)/2$. Let $Q\cong {\mathbb{P}}^1\times {\mathbb{P}}^1$ be a smooth quadric surface. There is an irreducible $C_1\in \left|{\mathcal{O}}_Q(a,b)\right|$ with exactly e nodes and no other singular point ([26], Cor. 2.14). Let $F_1$ be the Hirzebruch surface with a section of the ruling, h, with minimal self-intersection $-1$. We take as a basis of the Picard group of $F_1$ the irreducible curve h and a fiber f of the ruling of $F_1$. There is an integral and nodal $C_2\in \left|{\mathcal{O}}_{F_1}(uh+vf)\right|$ with exactly w nodes ([26], Cor. 2.14).

(i) 

First, assume n is odd. Let $W_1\subset {\mathbb{P}}^n$ be the image of Q by the embedding induced by the complete linear system $|{\mathcal{O}}_Q(1,(n-1)/2)|$. In this embedding, the image $D_1$ of the curve $C_1$ has degree $a(n-1)/2+b$ and it spans ${\mathbb{P}}^n$. See Proposition 2 for a discussion of multiple curves or reducible curves containing $\mathrm{Sing}\left(D_1\right)$ and with no component contained in $W_1$.

(ii) 

Now, assume n is even. Let $W_2$ be the image of $F_1$ by the complete linear system $|h+(n/2\left)f\right|$. In this embedding, the image $D_2$ of the curve $C_2$ has degree $un/2-u+v$ and it spans ${\mathbb{P}}^n$. See Proposition 3 for a discussion of multiple or reducible curves that are singular at all points of $\mathrm{Sing}\left(D_2\right)$ with no component contained in $W_2$.

Proposition 2.

Take $n\ge 3$ as odd and let $D_1\subset W_1\subset {\mathbb{P}}^n$ be the curve described in Example 1 for the integers a, b and e. Set $S:=\mathrm{Sing}\left(D_1\right)$. Let $C\subset {\mathbb{P}}^n$ be a local Cohen–Macaulay one-dimensional scheme such that $S\subseteq \mathrm{Sing}\left(C\right)$ and no irreducible component of $C_{\mathrm{red}}$ is contained in $W_1$. Then $\text{deg}\left(C\right)\ge e$.

Proof. 

Since $W_1$ is scheme-theoretically cut out by quadrics, the theorem of Bezout gives $2\text{deg}\left(C\right)\ge 2e$. □

Proposition 3.

Take $n\ge 4$ as even and let $D_2\subset {\mathbb{P}}^n$ be the curve described in Example 1 for the integers u, v and w. Set $S:=\mathrm{Sing}\left(D_2\right)$. Let $C\subset {\mathbb{P}}^n$ be a local Cohen–Macaulay one-dimensional scheme such that $S\subseteq \mathrm{Sing}\left(C\right)$ and no irreducible component of $C_{\mathrm{red}}$ is contained in $W_2$. Then, $\text{deg}\left(C\right)\ge w$.

Proof. 

Since $W_2$ is scheme-theoretically cut out by quadrics, the theorem of Bezout gives the inequality $2\text{deg}\left(C\right)\ge 2w$. □

Proposition 4.

Fix an integral variety $X\subseteq {\mathbb{P}}^r$ of dimension $n\ge 2$ and a positive integer s. There are integers $d_0$ and $d_1$ only depending on X and s with the following properties. Fix any $S\in S(X_{\mathrm{reg}},s)$. For any curve $C\subset X$ set $\text{deg}\left(C\right):=\text{deg}\left({\mathcal{O}}_C\left(1\right)\right)$.

(i) 

For every integer $d\ge d_0$, there is an integral curve $C\subset X$ such that $\text{deg}\left(C\right)=d\text{deg}\left(X\right)$ and $S=\mathrm{Sing}\left(C\right)$.

(ii) 

For every integer $d\ge d_1$, there is an integral curve $C\subset X$ such that $\text{deg}\left(C\right)=d\text{deg}\left(X\right)$, $S=\mathrm{Sing}\left(C\right)$ and at each $p\in S$, the Zariski tangent space of C at p has dimension n.

Moreover, for any closed algebraic set $T\subset X$ such that $dimT\le n-2$, there are curves C in (i) or (ii) with the additional property that $C\cap T=\varnothing$.

Proof. 

Taking (if $r>n$) the complete intersection of X with $r-n$ general hyperplanes, we reduce to the case $n=2$. For $n=2$, (i), (ii) and the “Moreover” part follow from a theorem of Bertini, Serre vanishing theorem for very ample line bundles and the Castelnuovo–Mumford Lemma. □

Remark 4.

Look at Proposition 4

(a) 

The case $n=r$, and hence, $\text{deg}\left(X\right)=1$, shows why we only required in Question 2 all integers $\text{deg}\left(C\right)$ larger than the one of a minimal degree solution.

(b) 

Fix an integer $m\ge 2$ and let ${\nu }_m:{\mathbb{P}}^2\to {\mathbb{P}}^r$, $r=\left({\displaystyle \genfrac{}{}{0pt}{}{m+2}{2}}\right)-1$, denote the order m Veronese embedding of ${\mathbb{P}}^2$. Set $X:={\nu }_m\left({\mathbb{P}}^2\right)$. In this case, all the curves on X have degree $\equiv 0(modm)$, and so, not all higher degrees may be obtained from curves contained in X. At the minimum, we need to restrict the integer $\text{deg}\left(C\right)$ to integers for which X has a curve.

Question 5: Is the obstruction discussed in part (b) of Remark 4, i.e., that the integer $\text{deg}\left(C\right)$ is achieved by at least one curve on X, a sufficient condition if $d\gg 0$?

Theorem 2.

Fix integers $a>b>3$. Let $Q\subset {\mathbb{P}}^3$ be a smooth quadric. Let $A\subset Q$ be the union of a distinct elements $L_1,\dots ,L_a$ of $|{\mathcal{O}}_Q(1,0)|$. Let $B\subset Q$ be the union of b distinct elements $R_1,\dots ,R_b$. Let S be the union of all $L_i\cap R_j$, $i=1,\dots ,a$, $j=1,\dots ,b$.

(a) 

$A\cup B$ is the only minimal degree reduced curve with S contained in its singular locus.

(b) 

$2B$ is the only minimal degree local Cohen–Macaulay curve with S contained in its singular locus.

(c) 

$2a+2b$ is the minimal degree of an irreducible curve C singular at each point of S, and there is $C\subset Q$ with $\text{deg}\left(C\right)=2a+2b$, $S=\mathrm{Sing}\left(C\right)$ and C nodal.

(d) 

Every local Cohen–Macaulay curve $T\subset {\mathbb{P}}^3$ with $S\subseteq \mathrm{Sing}\left(T\right)$ and no irreducible component contained in Q has a degree of at least $ab$.

(e) 

For all integers $x\ge 2a+2b$, there is an integral curve $X\subset Q$ such that $S=\mathrm{Sing}\left(X\right)$, X is nodal and $\text{deg}\left(X\right)=x$.

Proof. 

Note that $\#S=ab$. Obviously, both $A\cup B$ and $2B$ are singular at each point of S. Let $J\subset {\mathbb{P}}^3$ be an irreducible curve not contained in Q. Let e be the number of singular points of J contained in S and f the number of smooth points of J contained in S. The theorem of Bezout gives $2\text{deg}\left(J\right)\ge 2e+f$. We easily obtain that for non-integral curves of minimal degree it is sufficient to look at components contained in Q. The intersection theory of Q gives (a) and (b). For part (d), it is sufficient to use the theorem of Bezout and that $\text{deg}(T\cap Q)\ge 2ab$. Note that $2a+2b<ab$, because $b\ge 4$ and $a\ge 5$.

Claim 1: Fix $p\in Q\setminus A\cup B$. We have $h^1(Q,{\mathcal{I}}_{2S\cup 2p}(2a,2b))=0$

Proof of Claim 1: We use the residual exact sequence of the curve $R_1\cup \cdots \cup R_b$ and the cohomology of its connected components and obtain $h^1(Q,{\mathcal{I}}_{2S\cup 2p}(2a,2b))=h^1(Q,{\mathcal{I}}_{S\cup 2p}(2a,b))$. Then, we use the residual exact sequence of $L_1\cup \cdots \cup L_a$ to obtain $h^1(Q,{\mathcal{I}}_{S\cup 2p}(2a,b))=h^1(Q,{\mathcal{I}}_{2p}(a,b))$. The latter integer is 0, because $a>0$, $b>0$, and hence, ${\mathcal{O}}_Q(a,b)$ is very ample.

Since $\text{deg}(L_i\cap 2S)=2b$ and $L_i$ is not an irreducible component of C, $v\ge 2b$. Using $R_j$, we see that $u\ge 2a$. Hence, $d\ge 2a+2b$. Thus, to conclude the proof of part (c) of the theorem, it is sufficient to prove the following claim.

Claim 2: A general $D\in |{\mathcal{I}}_{2S\cap Q,Q}(2a,2b)|$ is integral, nodal and smooth outside S.

Proof of Claim 2: Let $\mathcal{B}\subset Q$ denote the scheme-theoretic base locus of the linear system $|{\mathcal{I}}_{2S\cap Q,Q}(2a,2b)|$. The scheme $\mathcal{B}$ is contained in the scheme-theoretic intersection of A and B. Hence, $\mathcal{B}$ is contained in the union of $ab$ degree 4 connected zero-dimensional schemes, one for each point of S. The theorem of Bertini gives that C is smooth outside S. Since the union of $A\cup B$ and any $E\in |{\mathcal{O}}_Q(a,b)|$ with $E\cap S=\varnothing$ is nodal at each point of S, the curve $C_1\cup C_2$ is nodal. Claim 1 gives that the rational map induced by the linear system $|{\mathcal{I}}_{2S\cap Q,Q}(2a,2b)|$ has a 2-dimensional image. Hence, the irreducibility of a general $D\in |{\mathcal{I}}_{2S\cap Q,Q}(2a,2b)|$ follows from a theorem of Bertini, usually called the second theorem of Bertini ([24], 6.3).

Fix any integers $x\ge 2a+2b$. To prove part (e), it is sufficient to observe that Claim 2 holds if instead of $(2a,2b)$, we take $(2a,x-2a)$. □

We leave to the interested reader the small modifications needed in the statement of Theorem 2 for the case $a=b>4$. Since $Q\cong {\mathbb{P}}^1\times {\mathbb{P}}^1$, Theorem 2 would fit well in Section 5.

Theorem 3.

Fix integers $n\ge 2$ and $\delta >0$. There is an integer $x_0\left(\delta \right)$ such that for all $x\ge x_0\left(\delta \right)$, there is $S\subset {\mathbb{P}}^n$ such that $\#S=x$, and 2 is the minimal degree of a local Cohen–Macaulay curve A singular at all points of S, while any reduced curve $C\subset {\mathbb{P}}^n$ with $S\subseteq \mathrm{Sing}\left(C\right)$ has degree of at least $\delta +1$. Moreover, if $x\ge n-1+n{x}_0\left(\delta \right)$, there is $S_1$ spanning ${\mathbb{P}}^n$ such that $\#S_1=x$. There is a local Cohen–Macaulay curve $A_1$ singular at all points of $S_1$ and $\text{deg}\left(A_1\right)\le 2n$, while any reduced curve $C_1\subset {\mathbb{P}}^n$ with $S_1\subseteq \mathrm{Sing}\left(C_1\right)$ has degree of at least $n\delta$.

Proof. 

Fix a line $L\subset {\mathbb{P}}^n$. For the first part, take as $x_0$ any integer $\ge 2\delta$ and as S any subset of L with $\#S=x$. Take as A a suitable double structure on L (Remark 16). Let $C\subset {\mathbb{P}}^n$ be a reduced curve whose singular locus contains S. Let $H\subset {\mathbb{P}}^n$ be a general hyperplane containing L. First, assume that L is not an irreducible component of C. Since H is a general hyperplane containing L and $\text{deg}(L\cap 2S)=2x$, the theorem of Bezout gives $\text{deg}\left(C\right)\ge 2x$. Now, assume that L is an irreducible component of C. Let $C^{\prime }$ denote the union of the other irreducible components of C. We have ${\mathrm{Res}}_H\left(2S\right)=S$. Since L is smooth, $C^{\prime }\supset S$. The theorem of Bezout gives $\text{deg}\left(C^{\prime }\right)\ge x$ and, hence, $\text{deg}\left(C\right)\ge x+1$.

For the “Moreover” part, take any integer $x_0\ge 2n\delta$ and as $S_1$ the union of S and the union $S_2$ of $n-1$ general points of ${\mathbb{P}}^n$. We take as $A_1$ the union of A and $n-1$ general conics with, as their singular point, a different point of $S_2$. The theorem of Bezout gives the lower bound on the integer $\text{deg}\left(C_1\right)$. □

Proposition 5.

Fix integers $d\ge 2s>0$ and $(d,s)\ne (2,1)$. Fix $S\in S({\mathbb{P}}^2,s)$. Then, there is an integral and nodal curve $C\in |{\mathcal{O}}_{{\mathbb{P}}^2}\left(d\right)|$ such that $S=\mathrm{Sing}\left(C\right)$.

Proof. 

Since the case $s=1$ and $d\ge 3$ is obvious, we may assume $s\ge 2$ and use induction on s. Take a general $C\in |{\mathcal{I}}_{2S}\left(d\right)|$. Fix $p\in S$ and $o\in {\mathbb{P}}^2\setminus C$. Set L as the line spanned by p and o. Set $S^{\prime }:=S\setminus S\cap L\subseteq S\setminus \left\{p\right\}$. Since $\text{deg}(L\cap (2S\cup \left\{o\right\})\le d+1$, we have $h^1(L,{\mathcal{I}}_{(L\cap 2S)\cup \left\{o\right\},L}\left(d\right))=0$. Hence, the residual exact sequence of L and induction on the integer $d-s$ give $h^1\left({\mathcal{I}}_{2S\cup \left\{o\right\}}\left(d\right)\right)=0$. Thus $h^1\left({\mathcal{I}}_{2S}\left(d\right)\right)=0$, and ${\mathcal{I}}_{2S}\left(d\right)$ has no base points outside d. Hence, a theorem of Bertini gives $S=\mathrm{Sing}\left(C\right)$. Now, assume that o is not contained in any line spanned by 2 of the points of S. Using the residual exact sequence of L and $h^1\left({\mathcal{I}}_{2S^{\prime }\cup \left\{o\right\}}(d-1)\right)=0$ by the inductive assumption, we obtain $h^1\left({\mathcal{I}}_{2S\cup 2o}\left(d\right)\right)=0$. Hence, the rational map induced by the linear system $|{\mathcal{I}}_{2S}\left(d\right)|$ has a 2-dimensional image. The second theorem of Bertini ([24], 6.3) gives that C is integral. Since we may easily obtain a reducible $D\in |{\mathcal{I}}_{2S}\left(d\right)|$, which is nodal at each point of S, C is nodal. □

Remark 5.

Fix positive integers s and d. We look at degree d plane curves containing a prescribed set S with $\#S=s$. In this case, all curves are connected. If S is general, then $|{\mathcal{I}}_S\left(d\right)|\ne \varnothing$ if and only if $\left({\displaystyle \genfrac{}{}{0pt}{}{d+2}{2}}\right)>s$, and if this condition is satisfied, then we find a smooth plane curve containing S.

Remark 6.

Fix positive integers s and d. We look at degree d curves whose singular locus contains a prescribed set S with $\#S=s$. Obviously, we need $d>1$. Consider a general S such that $\#S=s$, We have 2 exceptional degrees, $d=2,4$. In all other cases, we have a degree d plane curve singular at S if and only if $3s<\left({\displaystyle \genfrac{}{}{0pt}{}{d+2}{2}}\right)$, and with $(d,s)\ne (6,9)$, we may even find $C\subset {\mathbb{P}}^2$ with degree d, $S=\mathrm{Sing}\left(C\right)$, C is irreducible and with only nodes as its singular points [27].

Now, assume $d=4$. If $s\le 4$, we find an integral and nodal plane quartic. If $d=s=2$, the unique solution is a double line, while if $d=2$ and $s=1$, the solutions are reducible.

Now we are required to find a curve C with singular locus containing an arbitrary set with cardinality s. If we require that C be irreducible, then the theorem of Bezout implies that we need $d\ge 2s$ and $d\ne 2$ if $s=1$. The converse is Proposition 5.

Now, we allow that C is reducible but with no multiple component. For $d=2$, we need $s=1$. By the case of C being irreducible, we may assume $x>(d-1)(d-2)/2$. The theorem of Bezout implies that no line contains $\lceil (d+1)/2\rceil$ points of S unless it is an irreducible component of C. In this way (taking unions of a suitable degree $d-1$ curve $C_1$ and a line transversal to $C_1$), we cover all curves with $x\le d(d-1)/2$, case $x=d(d-1)/2$ only being covered by the nodal unions of d lines.

Proposition 6.

Fix $n>1$, $p_1,p_2\in {\mathbb{P}}^n$, $p_1\ne p_2$, and let $L\subset {\mathbb{P}}^n$ be the line spanned by $p_1$ and $p_2$. Then, there is a local Cohen–Macaulay curve $A\subset {\mathbb{P}}^n$ such that $\text{deg}\left(A\right)=2$, $A_{\mathrm{red}}=L$, and ${\mathbb{P}}^n$ is the minimal projective space containing A. Moreover, we may take as A a locally complete intersection with local embedding dimension 2 at each point of L.

Proof. 

Fix an integer $d\ge n$ and $S\subset L$ such that $\#S=n-1$. For each $p\in S$, let $v_p\subset {\mathbb{P}}^n$ be a general connected degree 2 zero-dimensional scheme with p as its reduction. Set $Z:=L\cup {\bigcup }_{p\in S}{v}_p$. Obviously, $h^0\left({\mathcal{I}}_Z\left(d\right)\right)\ge \left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-d-n$. Let $T\subset {\mathbb{P}}^n$ be the intersection of $n-1$ general elements of $|{\mathcal{I}}_Z\left(d\right)|$. Using d distinct hyperplanes, we see that L is the set-theoretic base locus of $|{\mathcal{I}}_Z\left(d\right)|$ The theorem of Bertini gives that T is smooth outside L. Take a degree 2 zero-dimensional scheme with, as its reduction, a point of L and $v\ne v_p$ for all $p\in S$. Taking a union of hyperplanes, we first see that v is not contained in the scheme-theoretic base locus of ${\mathcal{I}}_Z\left(d\right)$ and then that T is smooth outside S. Since S is finite and T is general, using unions of hyperplanes, we see that T is smooth. Hence, T is a smooth surface containing L. Let $A\subset T$ be the double of L as a Cartier divisor of T. Since T is a smooth surface, A is a locally complete intersection with local embedding dimension 2. Since $A\supset Z$, ${\mathbb{P}}^n$ is the minimal projective space containing A. □

4. Fully Singular Curves in Projective Spaces

Remark 7.

Fix a finite set $S\subset {\mathbb{P}}^n$, $n\ge 2$, and let e (resp. $e_1$, resp. f) be the minimal integer such that $|{\mathcal{I}}_{2S}\left(e\right)|\ne \varnothing$ (resp. $|{\mathcal{I}}_{2S}\left(e_1\right)|$ contains a hypersurface with no multiple component, resp. $|{\mathcal{I}}_{2S}\left(f\right)|$ contains an irreducible hypersurface). Using the theorem of Bertini, it is easy to check that f is the minimal degree of a hypersurface containing an integral curve C that is fully singular at all points of S. Moreover, e (resp. $e_1$) is the minimal degree of a hypersurface containing a local Cohen–Macaulay curve $C_1$ (resp. reduced curve $C_2$) that is fully singular at all points of S.

Theorem 4.

Fix integers $n\ge 3$ and $s>0$. Take $S\in S({\mathbb{P}}^n,s)$ and any integer $d\ge ns+n$. Then there is an integral and rational degree d curve that is fully singular at all points of S and spanning ${\mathbb{P}}^n$.

Proof. 

We adapt the proof of Theorem 1 with the following modifications. Let $D\subset {\mathbb{P}}^d$ be a rational normal curve. Take $ns$ distinct points $p_{i,j}\in D$, $i=1,\dots ,s$, $j=1,\dots ,n$. For $i=1,\dots ,s$, set $F_i:=\langle \{p_{i,1},\dots ,p_{i,n}\}\rangle$ and $W:=\langle F_1\cup \cdots \cup F_s\rangle$. We take a general n-dimensional linear subspace $M\subset {\mathbb{P}}^d$ and we identify it with ${\mathbb{P}}^n$, and so, we write $S\subset M$ and $(2S,M)$ instead of $2S$. Order the points $o_1,\dots ,o_s$ of S. Set $V_i:=\langle F_i\cup \left\{o_i\right\}\rangle$. Let $L_i\subset V_i$ be a general $(n-1)$-dimensional linear space. Take $W_1:=\langle L_1\cup \cdots \cup L_s\rangle$. We have $dim{W}_1=sn-1$, and let $V\subset {\mathbb{P}}^n$ be a general $(d-n-1)$-dimensional linear space containing $W_1$. Since $W_1\cap {\sigma }_2\left(D\right)$ is finite, V is a general linear subspace containing $W_1$ and $d-dimV>3=dim{\sigma }_2\left(D\right)$; the set $V\cap {\sigma }_2\left(D\right)$ is finite, concluding the proof. □

The proofs of Proposition 1 and Theorem 4 give the following result.

Proposition 7.

Fix integers $n\ge 2$, $s>0$ and $g\ge 0$. Take any $S\in S({\mathbb{P}}^n,s)$ and any smooth genus g curve X. Then, for all integers $d\ge g+n+ns$, there is an integral degree d curve $C\subset {\mathbb{P}}^n$ that is fully singular at each point of S, spanning ${\mathbb{P}}^n$ and with X as its normalization.

Theorem 5.

Fix integers $n\ge 2$, $d\ge 3$ and x such that $0<(n+1)x\le \left({\displaystyle \genfrac{}{}{0pt}{}{n+d-1}{n}}\right)$ and $(n,d)\notin \left\{\right(2,5),(4,4),(3,5),(4,5\left)\right\}$. Let S be a general element of $S({\mathbb{P}}^n,x)$. Let $C\subset {\mathbb{P}}^n$ be the intersection of $n-1$ general elements of $|{\mathcal{I}}_{2S}\left(d\right)|$. Then, C is an integral curve of degree $d^{n-1}$ that is fully singular at each point of S and smooth outside S.

Proof. 

Since $d-1\ge 3$, $n\ge 2$, $(n,d)\notin \left\{\right(2,5),(4,4),(3,5),(4,5\left)\right\}$ and $x\le \left({\displaystyle \genfrac{}{}{0pt}{}{n+d-1}{n}}\right)$, the Alexander–Hirschowitz theorem [6,7,8] gives $h^1\left({\mathcal{I}}_{2S}(d-1)\right)=0$. Hence, the Castelnuovo–Mumford Lemma gives that $|{\mathcal{I}}_{2S}\left(d\right)|$ is globally generated. Hence, the theorem of Bertini gives $dimC=1$ and that C is smooth outside S. Since $2S\subset C$, C is fully singular at S. By the second theorem of Bertini ([24], Th. 6.3; [25], Th. 5.3), to prove that C is irreducible, it is sufficient to prove that the rational map $\varphi$ induced by $|{\mathcal{I}}_{2S}\left(d\right)|$ has an image of dimension $>1$. We will even prove that the image has dimension n. Fix a general $p\in {\mathbb{P}}^n\setminus S$. To prove that $dim\mathrm{Im}\left(\varphi \right)=n$, it is sufficient to prove that $\varphi$ has injective differential at p. Since $h^1\left({\mathcal{I}}_{2S}\left(d\right)\right)=0$, this differential is injective if and only if $h^1\left({\mathcal{I}}_{2S\cup 2p}\left(d\right)\right)=0$. Fix $o\in S$ and set $S^{\prime }:=S\setminus \left\{o\right\}$. Since $h^1\left({\mathcal{I}}_{2S}(d-1)\right)=0$, $h^1\left({\mathcal{I}}_{2S^{\prime }\cup \left\{o\right\}}(d-1)\right)=0$ and $h^0\left({\mathcal{I}}_{2S^{\prime }\cup \left\{o\right\}}(d-1)\right)\ge n>0$. Since p is general, $h^0\left({\mathcal{I}}_{2S^{\prime }\cup \{o,p\}}(d-1)\right)=h^0\left({\mathcal{I}}_{2S^{\prime }\cup \left\{o\right\}}(d-1)\right)-1$ and, hence, $h^0\left({\mathcal{I}}_{2S^{\prime }\cup \{o,p\}}(d-1)\right)=0$. Take a general hyperplane H containing $\{o,p\}$. Since $S\cup p$ is general, $H\cap (2S\cup 2p)=(2o,H)\cup (2p,H)$. hence, $h^1(H,{\mathcal{I}}_{H\cap (2S\cup 2p),H}\left(d\right))=0$. Since $2S^{\prime }\cup \{o,p\}$ is the residual of $2S\cup 2p$ with respect to H, the residual exact sequence of H gives $h^1\left({\mathcal{I}}_{2S\cup 2p}\left(d\right)\right)=0$. □

Corollary 1.

Fix integers $n\ge 2$, $d\ge 5$ and x such that $x(n+1)=\left({\displaystyle \genfrac{}{}{0pt}{}{n+d-1}{n}}\right)$ and $(n,d)\notin \left\{\right(2,5),(4,4),(3,5),(4,5\left)\right\}$. Let $C\subset {\mathbb{P}}^n$ be the intersection of $n-1$ general elements of $|{\mathcal{I}}_{2S}\left(d\right)|$. Then, C is an integral curve of degree $d^{n-1}$ that is fully singular at each point of S and smooth outside S. Moreover, any local Cohen–Macaulay curve that is fully singular at each point of S is described, even set-theoretically, by at least $n-1$ equations, each of them of a degree at least d.

Proof. 

The first part is a particular case of Theorem 5. The “Moreover” part is true, because a curve in ${\mathbb{P}}^n$ needs at least $n-1$ equations and $h^0\left({\mathcal{I}}_{2S}(d-1)\right)=0$ by the Alexander–Hirschowitz theorem. □

Remark 8.

The interested reader may easily see for each low n which integers d have the property that $\left({\displaystyle \genfrac{}{}{0pt}{}{n+d-1}{n}}\right)\equiv 0(modn+1)$; hence, there is x satisfying Corollary 1. Furthermore, $n=3$ if and only if $d\equiv 1,2(mod3)$. If $n+1$ is a prime, then $\left({\displaystyle \genfrac{}{}{0pt}{}{n+d-1}{n}}\right)\equiv 0(modn+1)$ if and only if $(d-1)/(n+1)\notin \mathbb{Z}$.

Theorem 6.

Fix integers $n\ge 3$, $d\ge 2n+1$ and x such that $0<(n+1)x\le \left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-2$. Let S be a general element of $S({\mathbb{P}}^n,x)$. Then, the base locus $\mathcal{B}$ of ${\mathcal{I}}_{2S}\left(d\right)$ contains no hypersurface.

Proof. 

By Theorem 5, we may assume

$$(n+1)x\ge \left({\displaystyle \genfrac{}{}{0pt}{}{n+d-1}{n}}\right)+1$$

(1)

The theorem of Alexander–Hirschowitz ([6,7,8,9]) gives $h^1\left({\mathcal{I}}_{2S}\left(d\right)\right)=0$ and, hence, $h^0\left({\mathcal{I}}_{2S}\left(d\right)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-(n+1)x$. Assume that $\mathcal{B}$ contains a hypersurface, and call T the union of the hypersurfaces contained in $\mathcal{B}$. Set $e:=\text{deg}\left(T\right)$. Obviously, $0<e<d$. For a general $S\in S({\mathbb{P}}^n,x)$, either $S\subset T$ or $S\cap T=\varnothing$. The latter is impossible, because $h^0\left({\mathcal{I}}_{2S}(d-e)\right)=0$ by (1), the Alexander–Hirschowitz theorem and the assumption $d\ge 6$. Thus $S\subset T$.

Since S is a general element of the irreducible variety $S({\mathbb{P}}^n,x)$ and T is uniquely determined by S, either $2S\subset T$ or S is contained in the smooth locus of T. The former is excluded, because $h^0\left({\mathcal{I}}_{2S}\left(e\right)\right)=0$. Hence, ${\mathrm{Res}}_T\left(2S\right)=S$. Thus, the residual sequence of T gives the following exact sequence:

$$0\to {\mathcal{I}}_S(d-e)\to {\mathcal{I}}_{2S}\left(d\right)\to {\mathcal{I}}_{2S\cap T,T}\left(d\right)\to 0$$

(2)

Let ${\rho }_T:H^0\left({\mathcal{I}}_{2S}\left(d\right)\right)\to H^0(T,{\mathcal{I}}_{2S\cap T,T}\left(d\right))$ denote the restriction map, i.e., the map induced in cohomology from (2). Since $T\subseteq \mathcal{B}$, ${\rho }_T$ is the zero-map. Hence, $h^0\left({\mathcal{I}}_S(d-e)\right)=h^0\left({\mathcal{I}}_{2S}\left(d\right)\right)\ge 2$. Since S is general and $S\subset T$, $x<\left({\displaystyle \genfrac{}{}{0pt}{}{n+e}{n}}\right)$. We claim that

$$x=\left({\displaystyle \genfrac{}{}{0pt}{}{n+e}{n}}\right)-1$$

(3)

Assume $x\le \left({\displaystyle \genfrac{}{}{0pt}{}{n+e}{n}}\right)-2$ and take a general $T_1\in \left|{\mathcal{I}}_S\left(e\right)\right|$. Since S is contained in the smooth locus of T and $T_1$ is general, S is contained in the smooth locus of $T_1$. Hence, we obtain an exact sequence (2) with $T_1$ instead of T. Since $h^0\left({\mathcal{I}}_S(d-e)\right)=h^0\left({\mathcal{I}}_{2S}\left(d\right)\right)\ge 2$, the long cohomology exact sequence of (2) with $T_1$ instead of T gives ${\rho }_{T_1}=0$. Hence, $T_1\subseteq \mathcal{B}$. Since $T_1\ne T$ and T is the union of all hypersurfaces contained in $\mathcal{B}$, we obtain a contradiction proving (3). Since $h^0\left({\mathcal{I}}_S(d-e)\right)\ge 2$, we obtain $d\ge 2e+1$. Since $(n+1)x\ge \left({\displaystyle \genfrac{}{}{0pt}{}{n+d-1}{n}}\right)+1$, (3) gives

$$(n+1)\left({\displaystyle \genfrac{}{}{0pt}{}{n+e}{n}}\right)-(n+1)\ge \left({\displaystyle \genfrac{}{}{0pt}{}{n+2e}{n}}\right)+1$$

(4)

Set $\varphi (n,e):=\left({\displaystyle \genfrac{}{}{0pt}{}{n+2e}{n}}\right)$ and $\psi (n+e):=(n+1)\left({\displaystyle \genfrac{}{}{0pt}{}{n+e}{n}}\right)$. The inequality (4) shows that to prove the theorem, by contradiction, it is sufficient to prove that $\varphi (n,e)\ge \psi (n,e)$. First, assume $e\le n-1$. By (3) we have $x\le \left({\displaystyle \genfrac{}{}{0pt}{}{2n-1}{n}}\right)-1$ and hence $(n+1)x\le (n+1)\left({\displaystyle \genfrac{}{}{0pt}{}{2n-1}{n}}\right)-n-1$. Since $d\ge 2n+1$, (1) gives $(n+1)x\ge \left({\displaystyle \genfrac{}{}{0pt}{}{3n}{n}}\right)+1$, a contradiction. Now assume $e\ge n$. We have $u(n,e):=\varphi (n,e)/\psi (n,e)={\displaystyle \frac{(n+2e)!n!e!}{n!\left(2e\right)!(n+1)(n+e)!}}={\displaystyle \frac{(n+2e)!e!}{\left(2e\right)!(n+1)(n+e)!}}$. We have $u(e,e)={\displaystyle \frac{\left(3e\right)!e!}{\left(2e\right)!(e+1)!}}>1$. We fix e and see $u(n,e)$ as a function of n for all positive integers $n\le e$. We have $u(n,e)/u(n-1,e)={\displaystyle \frac{(n+2e)n}{(n+e)(n+1)}}>1$. We have $u(2,e)=(2e+2)(2e+1)/3(e+2)(e+1)>1$ (since $e\ge 3$), concluding the proof. □

Proposition 8.

Fix integers $d\ge 5$ and x such that $0<4x\le \left({\displaystyle \genfrac{}{}{0pt}{}{d+3}{3}}\right)-2$. Let S be a general element of $S({\mathbb{P}}^3,x)$. Then there is a locally Cohen-Macaulay curve which is the complete intersection of 2 degree d surfaces and it is fully singular at all points of S.

Proof. 

By Theorem 5, we may assume $x>\left({\displaystyle \genfrac{}{}{0pt}{}{d+2}{3}}\right)$. The theorem of Alexander–Hirschowitz gives $h^1\left({\mathcal{I}}_{2S}\left(d\right)\right)=0$ and, hence, $h^0\left({\mathcal{I}}_{2S}\left(d\right)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{d+3}{3}}\right)-4x$. By a theorem of Bertini ([24], Th. 6.3; [25], Th. 5.3), it is sufficient to prove that ${\mathcal{I}}_{2S}\left(d\right)$ has no surface contained in its base locus. Assume that it is false, and call T the 2-dimension part of the base locus of ${\mathcal{I}}_{2S}\left(d\right)$. Set $e:=\text{deg}\left(T\right)$. Obviously, $0<e<d$. As in the proof of Theorem 6, it is sufficient to disprove the case $n=3$ of (4), i.e., to prove that

$$4\left({\displaystyle \genfrac{}{}{0pt}{}{e+3}{3}}\right)-4\le \left({\displaystyle \genfrac{}{}{0pt}{}{2e+3}{3}}\right)+4$$

(5)

This inequality is true for all $e\ge 3$. For $e=2$ (resp. $e=1$), we use the case $n=3$ of (3) and obtain $x=9$ (resp. $x=3$), contradicting the inequalities $x>\left({\displaystyle \genfrac{}{}{0pt}{}{d+2}{3}}\right)\ge \left({\displaystyle \genfrac{}{}{0pt}{}{7}{3}}\right)=35$. □

Remark 9.

Let W be a smooth and connected variety of dimension $n>1$. Fix $p\in W$ and let $C\subset W$ be a local Cohen–Macaulay one-dimensional scheme. Since W is smooth at p, C is fully singular at p if and only if $p\in C_{\mathrm{red}}$ and the Zariski tangent space of C at p has dimension n. Now, assume that C is reduced in a neighborhood of p and that there is a unique irreducible component, T, containing p. Let $T^{\prime }$ be the partial normalization of T in which we only normalize the point p. It is well-known that $p_a\left(T\right)-p_a\left(T^{\prime }\right)\ge n-1$ and that sometimes equality holds, e.g., if T has at p exactly n branches, each of them smooth and whose tangents span $T_{p}W$.

Remark 10.

Fix a set $S\subset {\mathbb{P}}^n$, $n\ge 2$, such that $\#S=n+3$ and any $n+1$ of the points of S are linearly independent. For instance, we may take as S a general element of $S({\mathbb{P}}^n,n+3)$. Now assume $n=3$; hence, $\#S=6$. Since S spans ${\mathbb{P}}^3$, 3 is the minimal degree of an integral curve containing S. Any degree 3 integral and non-degenerate curve of ${\mathbb{P}}^3$ is a rational normal curve. Hence, C is the unique degree 3 integral curve containing S. Now, we consider low degree reducible curves $D:=C_1\cup \cdots \cup C_s$, $s\ge 2$, containing S with $C_i$ irreducible. We may assume that each $C_i$ is either a conic or a line. Since any 4 of the points of S span ${\mathbb{P}}^3$, a line contains at most 2 points of S and a conic at most 3 points of S. Hence, if we require that D be connected, then $\text{deg}\left(D\right)\ge 4$, while if we allow non-connected curves D, we find exactly 15 reducible degree 3 curves containing S: we partition S in 3 disjoint sets of cardinality 2, and we take the 3 lines spanned by the 3 subsets of cardinality 2. If we allow that a curve $E\supset S$ has multiple components, nothing is gained because $E_{\mathrm{red}}\supset S$.

Theorem 7.

Take any $S\in S({\mathbb{P}}^3,6)$ such that no 4 of its points are coplanar, e.g., a general $S\in S({\mathbb{P}}^3,6)$.

(i) 

6 is the minimal degree of a local Cohen–Macaulay curve containing $2S$ and each minimal degree curve is either the union of 2 of the 16 degree 3 curves containing S (Remark 10) or is the double of one such curve.

(ii) 

7 is the minimal degree of an integral curve $C\supset 2S$. Moreover, all such minimal degree curves are contained in a quadric surface, $p_a\left(C\right)=6$, with exactly 6 singular points, each of them either an ordinary node or an ordinary cusp, and they are arithmetically Cohen–Macaulay. Moreover, there are nodal $C\supset 2S$ of degree 6.

(iii) 

6 is the minimal degree of a reduced and connected curve $C\supset 2S$.

(iv) 

6 is the minimal degree of a reduced curve $C\supset 2S$.

Proof. 

Take $C\supset 2S$ and set $d:=\text{deg}\left(C\right)$.

(a)

In this step, we assume that C is irreducible. The curve C is an irreducible degree d space curve with at least 6 singular points; hence, with $p_a\left(C\right)\ge 6$. By Castelnuovo’s theory ([28], Th. 3.7), $d\ge 7$, and if $d=7$, then $p_a\left(C\right)=6$. C is contained in a quadric surface and it is arithmetically Cohen–Macaulay. Assume $d=7$ and, hence, $p_a\left(C\right)=6$. Since $S\subseteq \mathrm{Sing}\left(C\right)$ and $\#S=p_a\left(C\right)$, $S=\mathrm{Sing}\left(C\right)$ and each singular point of C is either an ordinary node or an ordinary cusp. To conclude the proof of this case, it is sufficient to find a degree 7 irreducible curve that is singular at all points of S. Since no 4 of the points of S are coplanar, ${\mathcal{I}}_S\left(2\right)$ is globally generated and there is a smooth quadric surface $Q\supset S$ such that Q contains no line spanned by 2 of the points of S. It is sufficient to find an irreducible $C\in |{\mathcal{O}}_Q(4,3)|$ that is singular at each point of S. For a general S, it is sufficient to quote Proposition 11. For a non-general S, it is sufficient to observe that $h^0(Q,{\mathcal{I}}_{2S\cap Q,Q}(4,3))=2$ and check the reducible elements of $|{\mathcal{O}}_Q(4,3)|$. It is sufficient to use that each element of $|{\mathcal{O}}_Q(1,0)|$ or $|{\mathcal{O}}_Q(0,1)|$ contains at most one element of S, every irreducible element of $|{\mathcal{O}}_Q(1,1)|$ contains at most 3 points of S, every irreducible element of $|{\mathcal{O}}_Q(1,2)|$ contains at most 5 points of S, and it is smooth if it contains at least 4 points, and so on.

(b)

In this step, we assume that C has no multiple components, but it is reducible, say $C=C_1\cup \cdots \cup C_s$ with $s\ge 2$ and each $C_i$ irreducible. Set $d_i:=\text{deg}\left(C_i\right)$. We may assume $d_1\ge \cdots \ge d_s$. Hence, $d=d_1+\cdots +d_s$. Taking the union of 2 different degree 3 curves containing S (Remark 10), we obtain a connected solution of degree 6. Now assume $d\le 5$. If $d_i\le 3$, then $C_i$ is smooth, if $d_i=4$, then $C_i$ has at most one singular point. The case $d_1=4$ is excluded, because a line contains at most 2 points of S. The case $d_1=3$ and $s=2$ is excluded, because any conic contains at most 3 points of S. A straightforward check handles the cases $d_1\le 2$.

(c)

In this step, we allow that C has multiple components. Taking a double of one of the degree 3 curves containing S, we obtain a solution for the case $d=6$. Now assume $d\le 5$. We have $T:=C_{\mathrm{red}}\supset S$. Hence $t:=\text{deg}\left(T\right)\ge 3$. First, assume $d=4$. Hence, C is the union of a degree 3 curve $T_1$ and a double structure on a line L. Since $\#(L\cap S)\le 2$, $T_1$ has at least 4 singular points, a contradiction. Now, assume $d=3$. Since $d<6$, T is reducible. By Remark 7, T is the union of 3 disjoint lines. Hence, no multiple structure at 2 of the connected components of T is singular at all points of S, a contradiction.

□

Theorem 8.

Fix integers $n\ge 3$ and $\delta >0$. There is an integer $x_1\left(\delta \right)$ such that for all $x\ge x_1\left(\delta \right)$, there is $S\subset {\mathbb{P}}^n$ with $\#S=x$; n is the minimal degree of a local Cohen–Macaulay curve A that is fully singular at all points of S, while any reduced curve $C\subset {\mathbb{P}}^n$ that is fully singular at all points of S has a degree of at least $\delta +1$. Moreover, if $x\ge n(\lceil (n-1)/2\rceil +n{x}_1\left(\delta \right)$, there is $S_1$ spanning ${\mathbb{P}}^n$ such that $\#S_1=x$; there is a local Cohen–Macaulay curve $A_1$ that is fully singular at all points of $S_1$ and $\text{deg}\left(A_1\right)\le 2n$, while any reduced curve $C_1\subset {\mathbb{P}}^n$ that is fully singular at all points of $S_1$ has a degree of at least $n\delta$.

Proof. 

Fix a line $L\subset {\mathbb{P}}^n$. Note that ${\mathcal{I}}_L^2$ is a local Cohen–Macaulay curve of degree n that is fully singular at all points of L. For the first part, take as $x_1$ any integer $\ge n\delta$ and as S any subset of L with $\#S=x$.

For the last part, add to ${\mathcal{I}}_L^2$ the double $2p_1,\dots ,2p_{n-1}$ of $n-1$ general points $p_1,\dots ,p_{n-1}$ and the union of ${\mathcal{I}}_L^2$ and the double of the union of $\lceil (n-1)/2\rceil$ lines containing the points $p_1,\dots ,p_{n-1}$. □

Proposition 9.

Fix integers $s>0$ and $d\ge 3$. Let $S\subset {\mathbb{P}}^2$ be any set with $\#S=s$. If $d\ge 2s$, there is an integral $C\in |{\mathcal{I}}_{2S}\left(d\right)|$. If $2s>d$, there is S such that $\#S=s$, and there is no integral $C\in |{\mathcal{I}}_{2S}\left(d\right)|$.

Proof. 

For the second part, we take s collinear points and apply the theorem of Bezout.

Now, assume $d\ge 2s$. All cases with $s\le 2$ are true by the case of general sets (Remark 5). Hence, the cases $d\le 5$ are true. Fix a line $L\subset {\mathbb{P}}^2$. For any $o\in {\mathbb{P}}^2\setminus L$, let ${\ell }_o:{\mathbb{P}}^2\setminus \left\{o\right\}\to L$ denote the linear projection from o. Take o not contained in any secant line of S. Thus, $o\notin S$ and $\#{\ell }_o\left(S\right)=\#S=s$. Set $S^{\prime }:={\ell }_o\left(S\right)$. Take the homogeneous coordinate $x_0,x_1,x_2$ such that $L=\{x_2=0\}$ and $o=[0:0:1]$. For each $a=[a_0:a_1:a_2]\ne o$, we have ${\ell }_o\left(a\right)=[a_0:a_1:0]$. For each non-zero constant t, let $h_t:{\mathbb{P}}^2\to {\mathbb{P}}^2$ denote the automorphism defined by the formula $h_t\left([x_0:x_1:x_2]\right)=[x_0:x_1:t{x}_2]$. Obviously, $h^0\left({\mathcal{I}}_{2S}\left(d\right)\right)=h^0\left({\mathcal{I}}_{h_t\left(S\right)}\left(d\right)\right)$ for all $t\ne 0$. The set $S^{\prime }$ is a flat limit of the sets $h_t\left(S\right)$, $t\ne 0$.

Claim 1: We have $h^0\left({\mathcal{I}}_{2S}\left(d\right)\right)=h^0\left({\mathcal{I}}_{2S^{\prime }}\left(d\right)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{d+2}{2}}\right)-3s$.

Proof of Claim 1: Claim 1 for $S^{\prime }$ (resp. S) is equivalent to $h^1\left({\mathcal{I}}_{2S^{\prime }}\left(d\right)\right)=0$ (resp. $h^1\left({\mathcal{I}}_{2S}\left(d\right)\right)=0$). The semicontinuity theorem for cohomology says that to prove Claim 1, it is sufficient to prove that $h^1\left({\mathcal{I}}_{2S^{\prime }}\left(d\right)\right)=0$. This cohomological vanishing is proven using the residual exact sequence of L and that $h^1(L,{\mathcal{I}}_{2S^{\prime }\cap L,L}\left(d\right))=h^1(L,{\mathcal{I}}_{S^{\prime }\cap L,L}(d-1))=0$.

Fix a general $C\in |{\mathcal{I}}_{2S^{\prime }}\left(d\right)|$. Since $S^{\prime }$ is a flat limit of a family of projectively equivalent subsets, Claim 1 implies that it is sufficient to prove the lemma for the set $S^{\prime }$. Let $\mathcal{B}$ denote the scheme-theoretic base locus of ${\mathcal{I}}_{2S^{\prime }}\left(d\right)$. Consider the set $\Delta$ of unions of s reducible conics, general with the only restriction that the singular locus of each of them is a different point of $S^{\prime }$ and, if $d>2s$, a general curve of degree $d-2$. The curves in $\Delta$ show that $\mathcal{B}=2S^{\prime }$. Since C is general, the theorem of Bertini gives $S^{\prime }=\mathrm{Sing}\left(C^{\prime }\right)$. Since C is general, the curves in $\Delta$ prove that C has a node at each point of $S^{\prime }$. Since $d>2$, the theorem of Bezout implies the irreducibility of C. □

Remark 11.

Fix $p\in {\mathbb{P}}^n$, $n>1$. There is a reduced degree d curve $C\subset {\mathbb{P}}^n$ spanning ${\mathbb{P}}^n$ and containing p if and only $2d\ge n+1$, and we may take as C a union of lines with at least one of them containing p. If $2d\ge n+2$, we may also find C with $p\in \mathrm{Sing}\left(C\right)$. There is a reduced and connected degree d curve $C\subset {\mathbb{P}}^n$ spanning ${\mathbb{P}}^n$ if and only $d\ge n$, and we may take as C a union of n general lines containing p, so that we may obtain a minimum degree that is also fully singular at p. All the degree n reduced curves that are fully singular at p are projectively equivalent and they span ${\mathbb{P}}^n$.

5. Multiprojective Spaces

Let $Y={\mathbb{P}}^{n_1}\times \cdots \times {\mathbb{P}}^{n_k}$ be a multiprojective space. Set $n:=dim\left(Y\right)=n_1+\cdots +n_k$ and $r:=-1+(n_1+1)\times \cdots \times (n_k+1)$. Let $\nu :Y\to {\mathbb{P}}^r$ denote the Segre embedding, i.e., the embedding of Y induced by the complete linear system $|{\mathcal{O}}_Y(1,\dots ,1)|$.

For each $i=1,\dots ,k$, let ${\pi }_i:Y\to {\mathbb{P}}^{n_i}$ denote the projection onto the i-th factor and ${ϵ}_i$ the element $(a_1,\dots ,a_k)\in {\mathbb{N}}^k$ such that $a_i=1$ and $a_j:=0$ for all $j\ne i$.

Take any irreducible curve $C\subset Y$. The multidegree $(a_1,\dots ,a_k)$ of C is defined taking $a_i:=\text{deg}\left({\mathcal{O}}_C\left({ϵ}_i\right)\right)$ for all i. It is easy to see that $\text{deg}\left({\mathcal{O}}_C\left({ϵ}_i\right)\right)=\text{deg}\left({\pi }_{i|C}\right)\text{deg}\left({\pi }_i\left(C\right)\right)$ with the conventions $\text{deg}\left({\pi }_{i|C}\right)=0$ if ${\pi }_i\left(C\right)$ is a point and $\text{deg}\left({\pi }_i\left(C\right)\right)=1$ if $n_i=1$ and ${\pi }_i\left(C\right)={\mathbb{P}}^1$. The multidegree of a reducible subcurve of Y is defined as the sum of the multidegrees of its irreducible components. Let $C\subset Y$ be a local Cohen–Macaulay one-dimensional scheme. Let $D_1,\dots ,D_s$ be the irreducible components of $C_{\mathrm{red}}$. Let $(a_{i1},\dots ,a_{ik})$ be the multidegree of $D_i$. Let $e_i$ denote the multiplicity of C at a a general point of $D_i$. The multidegree of C is the integers ${\sum }_{i=1}^s{e}_i(a_{i1},\dots ,a_{ik})$.

Remark 12.

Assume $k>1$. Let $S\subset Y$ be a finite set. Y is the minimal multiprojective space containing S if and only if ${\pi }_i\left(S\right)$ spans ${\mathbb{P}}^{n_i}$ for all i. Hence, $\#{\pi }_i\left(S\right)\ge n_i+1$ for all i. Take two distinct points $p,q$ of Y, say $p=(p_1,\dots ,p_k)$ and $q=(q_1,\dots ,q_k)$. If $\{p,q\}$ is general, then $p_i\ne q_i$ for all i; hence, the multidegree $(a_1,\dots ,a_k)$ of every connected curve containing $\{p,q\}$ satisfies $a_i>0$ for all i. Moreover, for all $s\ge 2$, the minimal multiprojective space ${\mathbb{P}}^{m_1}\times \cdots \times {\mathbb{P}}^{m_k}$ containing a general $S\subset Y$ with $\#S=s$ satisfies $m_i=min\{n_i,s-1\}$ for all $i=1,\dots ,k$.

Remark 13.

Fix integers $k\ge 2$ and $n_i>0$, $1\le i\le k$. Set $Y:={\mathbb{P}}^{n_1}\times \cdots \times {\mathbb{P}}^{n_k}$. Every line bundle of Y is of the form ${\mathcal{O}}_Y(a_1,\dots ,a_k)$ for some uniquely determined $(a_1,\dots ,a_k)\in {\mathbb{Z}}^k$. We consider the following partial order ≤ on ${\mathbb{Z}}^k$ and, hence, on the set of all line bundles on Y. Write $(a_1,\dots ,a_k)\le (b_1,\dots ,b_k)$ if and only if $a_i\le b_i$ for all i. With this partial order, Remarks 3 and 7 work for the multiprojective space Y with the modification that we have “a minimal”, not “the minimal”, because minimal solutions may not be unique, even in very simple cases (Remark 14).

Remark 14.

Fix integers $k\ge 2$ and $n_i>0$, $1\le i\le k$. Set $Y:={\mathbb{P}}^{n_1}\times \cdots \times {\mathbb{P}}^{n_k}$. Take $p\in Y$. The line bundles ${\mathcal{O}}_Y({ϵ}_i+{ϵ}_j)$ and $1\le i<j\le k$ are the minimal line bundles $\mathcal{L}$ such that there is a reduced hypersurface singular at p. If we add the line bundles $2{ϵ}_i$, $i=1,\dots ,k$, we obtain the one allowing non-reduced components. The linear systems $|{\mathcal{O}}_Y(2{ϵ}_i+{ϵ}_j)|$, $i\ne j$ have an irreducible hypersurface singular at p, and they are minimal with this property, the only minimal ones if $n_i=1$. If $n_i\ge 2$, there is an irreducible $T\in |{\mathcal{O}}_Y\left(3{ϵ}_i\right)|$ singular at p.

Question 6: Take $Y={\mathbb{P}}^{n_1}\times \cdots \times {\mathbb{P}}^{n_k}$, $k>1$, $s>0$, a general $S\in S(Y,s)$, $i\in \{1,\dots ,k\}$ and $\underline{a}=(a_1,\dots ,a_k)\in {\mathbb{N}}^k$. Give upper or lower bound for the minimal integer t such that there is an irreducible curve (resp. a reduced curve, resp. a reduced and connected curve, resp. a local Cohen–Macaulay one-dimensional scheme) C such that $S\subseteq \mathrm{Sing}\left(C\right)$ and C has multidegree $\underline{a}+t{ϵ}_i$. For any such reduced curve, C is $S=\mathrm{Sing}\left(C\right)$. Now, we restrict to integral curves. Are all larger t allowed?

Theorem 9.

Fix integers $k\ge 2$, $n_1\ge \cdots \ge n_k>0$, $d_1\ge 4$, $d_2\ge 4$ and $d_i\ge 3$ for all $i>2$. Set $Y:={\mathbb{P}}^{n_1}\times \cdots \times {\mathbb{P}}^{n_k}$ and $n:=n_1+\cdots +n_k$. Fix an integer $x>0$ such that $(x+1)\le {\prod }_{i=1}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n_i+d_i-1}{n_i}}\right)$. Take a general $S\in S(Y,x)$. Then, there is an irreducible curve $C\subset Y$ that is fully singular at each point of S, smooth outside S and that is the complete intersection of $n-1$ general elements of $|{\mathcal{I}}_{2S}(d_1,\dots ,d_s)|$.

Proof. 

We have $h^1\left({\mathcal{I}}_{2S}(d_1-1,\dots ,d_s-1)\right)=0$ ([17] for $k=2$, [16] for $k>2$). Hence, it is sufficient to use the Castelnuovo–Mumford Lemma as we did in the proof of Theorem 5. □

Remark 15.

Assume $n_1=1$. The paper ([18], Th. 3.1) allows the interested reader to extend Theorem 9 to almost all multidegrees $(d_1,\dots ,d_k)$ with $d_i\ge 2$ for all i. Other cases for arbitrary $n_1$ may be carried out using [14].

Proposition 10.

Fix positive integers k, $n_i$, $1\le i\le k$, s, g, and set $Y:={\mathbb{P}}^{n_1}\times \cdots \times {\mathbb{P}}^{n_k}$, and let $\nu :Y\to {\mathbb{P}}^r$, $r:={\prod }_{i=1}^k(n_1+1)-1$ be the Segre embedding of Y. Assume $n\ge 2$. Let X be a smooth genus g curve. There is an integer $d_0(Y,g,s)$ such that for every $d\ge d_0(Y,g,s)$ and every $S\in S(Y,s)$, there is an integral and curve $C\subset Y$ that is singular at all points of S, with X as its normalization and with $\text{deg}\left(\nu \right(C\left)\right)=d$.

Proof. 

The case $n_i>1$ for all i only requires us to use Proposition 1 instead of Theorem 1. For the integer $n_i=1$, it is sufficient that there are many degree $g+2s$ morphisms $f_i:X\to {\mathbb{P}}^1$. □

Theorem 10.

Fix positive integers k, $n_i$, $1\le i\le k$, s, and set $Y:={\mathbb{P}}^{n_1}\times \cdots \times {\mathbb{P}}^{n_k}$, and let $\nu :Y\to {\mathbb{P}}^r$, $r:={\prod }_{i=1}^k(n_1+1)-1$ be the Segre embedding of Y. Assume $n:=n_1+\cdots +n_k\ge 2$. There is an integer $d_0(Y,s)$ such that for every $d\ge d_0(Y,s)$ and every $S\in S(Y,s)$, there is an integral and rational curve $C\subset Y$ that is singular at all points of S and with $\text{deg}\left(\nu \right(C\left)\right)=d$.

Proof. 

For any integral curve $C\subset Y$, the integer $\text{deg}\left(\nu \right(C\left)\right)$ is the integer $\text{deg}\left({\mathcal{O}}_C(1,\dots ,1)\right)$. The case $k=1$ is true by our assumption $n\ge 2$ and Theorem 1. The case $k=n=2$ is true by Example 1 (in that example, we take a and b large after fixing s). Hence, we may assume $n\ge 3$. We take $d_0(Y,s)={\sum }_{i=1}s{b}_i$ with $b_i:=d_0(n_i,s)$ if $n_i\ge 2$ and $b_i=2s$ if $n_i=1$.

(a)

Assume $n_i\ge 2$ for all i. We take $d_0(Y,s)={\sum }_{i=1}^k{d}_0(n_i,s)$. We need to find a map $f:{\mathbb{P}}^1\to Y$ birational onto its image such that $C:=f\left({\mathbb{P}}^1\right)$ is singular at each point of S and $\text{deg}\left(\nu \right(C\left)\right)=d$. We have $f=(f_1,\dots ,f_k)$ with $f_i:{\mathbb{P}}^1\to {\mathbb{P}}^{n_i}$. We will find f with the additional condition that each $f_i$ is birational onto its image and $f_i\left({\mathbb{P}}^1\right)$ is singular at all points of ${\pi }_i\left(S\right)$. Fix an integer $d_i\ge d_1(n_i,s)$. Note that $d_i\ge d_0(n_i,c)$ for all $c<s$, hence, $d_i\ge d_0(n_i,\#{\pi }_i\left(S\right))$. By Theorem 1, there is a morphism $f_i:{\mathbb{P}}^1\to {\mathbb{P}}^{n_i}$ such that $\text{deg}\left(f_i\right)=1$, $\text{deg}\left(f_i\left({\mathbb{P}}^1\right)\right)=d_i$ and $f_i\left({\mathbb{P}}^1\right)$ are singular at all points of ${\pi }_i\left(S\right)$. We need more; we need to use the way we proved Theorem 1. We started with an ordering $o_1,\dots ,o_s$ of the points of S. We use it to write the points of ${\pi }_i\left(S\right)$, except we repeat ${\pi }_i\left(o_j\right)$e times if $\#\left({\pi }_i^{-1}\left({\pi }_i\left(o_j\right)\right)\right)=e$. We took any $2s$ distinct points of ${\mathbb{P}}^1$, say $p_{i,j}$, $i=1,\dots ,s$, $j=1,2$, and we used the same points for the construction of all $f_j$. Since $f_i\left(p_{h,1}\right)=f_i\left(p_{h,2}\right)$ for all i, $f\left(p_{h,1}\right)=f\left(p_{h,2}\right)$ for all $h=1,\dots ,s$. Thus, each point of S comes from at least 2 points of ${\mathbb{P}}^1$, and hence, the rational curve $f\left({\mathbb{P}}^1\right)$ is singular at each point of S.

(b)

Assume $n_i=1$ for some i. We take $d_i=2s$ and take as $f_i$ a degree 2 map $f_i:{\mathbb{P}}^1\to {\mathbb{P}}^1$ such that $f_i\left(p_{h,1}\right)=f_i\left(p_{h,2}\right)={\pi }_i\left(o_h\right)$ for all h. This is possible even if $\#\left({\pi }_i\left(S\right)\right)=1$, because we took $d_i=2s$. To check that $f\left({\mathbb{P}}^1\right)$ is singular at each point of S, it is sufficient to prove that f is birational onto its image. This is true by step (a) if $n_i>1$ for some i. Now, assume $n_i=1$ for all i. It is sufficient to use Example 1 (in that example we take a and b large after fixing s); to obtain that, we may take $(f_1,f_2)$ birational onto its image and, hence, $(f_1,\dots ,f_k)$ birational onto its image even if $k>2$.

□

The proof of Theorems 8 and 10 gives the following result.

Theorem 11.

Take $k>1$ and $i\in \{1,\dots ,n\}$ and a positive integer δ. Set $n:=n_1+\cdots +n_k$. There is an integer $x_1\left(\delta \right)$ such that for all $x\ge x_1\left(\delta \right)$, there is $S\in S({\mathbb{P}}^n,x)$ with n as the minimal degree of a local Cohen–Macaulay curve A that is fully singular at all points of S, while any reduced curve $C\subset {\mathbb{P}}^n$ with $S\subseteq \mathrm{Sing}\left(C\right)$ has a degree of at least $\delta +1$. Moreover, if $x\ge n-1+n{x}_0\left(\delta \right)$, there is $S_1$ spanning ${\mathbb{P}}^n$ such that $\#S_1=x$; there is a local Cohen–Macaulay curve $A_1$ that is fully singular at all points of $S_1$ and $\text{deg}\left(A_1\right)\le n^2$, while any reduced curve $C_1\subset {\mathbb{P}}^n$ with $S_1\subseteq \mathrm{Sing}\left(C_1\right)$ has multidegree $(a_1,\dots ,a_k)$ with $a_1+\cdots +a_k\ge n\delta$.

Lemma 1.

Assume $k>1$. Let $S\subset Y$ be a finite set such that $\#\left({\pi }_h\left(S\right)\right)>1$ for all $h=1,\dots ,k$. Let $C\subset Y$ be a connected and reduced curve such that $S\subset C$ and let $(a_1,\dots ,a_k)$ the multidegree of C. Then, $a_h>0$ for all $h=1,\dots ,k$. Now, assume $S\subseteq \mathrm{Sing}\left(C\right)$. Then, either $a_h\ge 2$, or there are at least $\#\left({\pi }_h\left(S\right)\right)$ distinct irreducible components $E_1,\dots ,E_{\#{\pi }_h\left(S\right)}$ of C such that ${\pi }_h\left(E_j\right)$ is a point for all j.

Proof. 

Take $h\in \{1,\dots ,k\}$. Since $\#{\pi }_h\left(S\right)>1$, $S\subset C$ and C are connected, there is at least one component, A, of C such that ${\pi }_h\left(A\right)$ is a curve. Set $(b_1,\dots ,b_k)$ as the multidegree of A. We have $a_i\ge b_i$ for all i. Hence, $a_h\ge b_h>0$. If $a_h\ge 2$, then the lemma is true. Now, assume $a_h=b_h=1$. Thus, $A\cong {\mathbb{P}}^1$, and ${\pi }_{h|A}:A\to {\pi }_h\left(A\right)$ is an isomorphism. Since $S\subseteq \mathrm{Sing}\left(C\right)$, each point of S is contained in another irreducible component of C. Since $a_h=1$, all the other irreducible components of C are mapped to a point of ${\mathbb{P}}^{n_h}$. Hence, there are at least $\#\left({\pi }_h\left(S\right)\right)$ such components. □

Now, we see how Proposition 9 may be generalized to the case of $k>1$ factors. First (as in the case of Proposition 9), we discuss the only case, ${\mathbb{P}}^1\times {\mathbb{P}}^1$, with $dimY=2$.

Proposition 11.

Set $Y:={\mathbb{P}}^1\times {\mathbb{P}}^1$. Take positive integers s, a and b such that $a\ge b$. Let $S\subset Y$ be a general subset such that $\#S=s$.

(a) 

We have $|{\mathcal{I}}_{2S}(a,b)|\ne \varnothing$ if and only if either $3s<(a+1)(b+1)$ or $b=2$, a is even and $s=a+1$. If $3s<(a+1)(b+1)$ and $b>2$, then a general $C\in |{\mathcal{I}}_{2S}(a,b)|$ is irreducible. If $b=2$, a is odd and $s=a+1$, then the only element of $|{\mathcal{I}}_{2S}(a,2)|$ is the double of the unique element of $|{\mathcal{I}}_S(a/2,1)|$. If $b=2$, then all elements of $|{\mathcal{I}}_{2S}(a,2)|$ are reducible.

(b) 

We have $|{\mathcal{I}}_{2A}(a,b)|\ne \varnothing$ for all $A\in S(Y,s)$ if and only if either $3s<(a+1)(b+1)$ or $b=2$, a is even and $s=a+1$. For each $A\in S(Y,s)$, there is an integral $C\in |{\mathcal{I}}_{2A}(a,b)|$ if and only if $b\ge 2s$. If $b\ge 2s$, a general $C\in |{\mathcal{I}}_{2A}(a,b)|$ is irreducible, nodal and $\mathrm{Sing}\left(C\right)=A$.

Proof. 

The first statement of part (a) follows from the classification of secant varieties of Segre–Veronese embeddings of ${\mathbb{P}}^1\times {\mathbb{P}}^1$[2,18]. For $b>2$, the second statement follows from the irreducibility of the Severi variety of irreducible nodal curves [29]. Now, assume $b=2$ and either a is odd or $s\le a$. We use induction on the integer s. Every irreducible element of $|{\mathcal{O}}_Y(t,1)|$, $t\ge 0$ has arithmetic genus 0; hence, it is smooth and rational. Either C is the union of 2 irreducible (and hence, smooth) curves $C_1\cup C_2$, one of bidegree $(\lfloor a/2\rceil ,1)$ and the other one of bidegree $\lceil a/2\rceil ,1)$ with $S\subseteq C_1\cap C_2$, or it has an element of $|{\mathcal{O}}_Y(0,1)|$ as one of its irreducible components. First, assume $C=C_1\cup C_2$. Since S is general, we have $h^0\left({\mathcal{I}}_S(\lfloor a/2\rfloor ,1)\right)=2(\lfloor a/2\rfloor +1)-s$ and $h^0\left({\mathcal{I}}_S(\lceil a/2\rceil ,1)\right)=2(\lceil a/2\rceil +1)-s$. Hence, $s\le \lfloor a/2\rfloor +1$ and $dim|{\mathcal{I}}_{2S}(a,2))=2a+4-2s<3a+3-3s$, a contradiction. Now, assume $C=C_3\cup R$ with $R\in |{\mathcal{O}}_Y(1,0)|$. A dimensional count and induction on a show that we may assume $S\cap R\ne \varnothing$. Since S is general, $\#(S\cap R)=1$. Set $\left\{p\right\}:=S\cap R$ and $S^{\prime }:=S\setminus \left\{p\right\}$. Since R is smooth, ${\mathrm{Res}}_R\left(2S\right)=2S^{\prime }\cup \left\{p\right\}$. Since the pair $(S,p)$ is general in $S(Y,s-1)\times S(Y,1)$, the inductive assumption on s gives $h^0\left({\mathcal{I}}_{2S^{\prime }\cup \left\{p\right\}}(a-1,b)\right)=h^0\left({\mathcal{I}}_{2S^{\prime }}(a-1,2)\right)-1=3a-2s+1.$ Since $h^0(L,{\mathcal{I}}_{2p}(a,2))=1$, the residual exact sequence of R gives a contradiction.

The first statement of part (b) follows from part (a) and the semicontinuity theorem for cohomology. Assume $3s<(a+1)(b+1)$ or $b=2$, a even and $s=a+1$. First assume $2s<b$. Take $L\in |{\mathcal{O}}_Y(1,0)|$ and any $S^{\prime }\subset L$ such that $\#S^{\prime }=s$. Fix a general $C\in |{\mathcal{I}}_{2S^{\prime }}(a,b)|$. Since $2s<b$, the intersection theory of divisors on Y gives that L is an irreducible component of C.

Now, assume $2s\le b$ and fix any $A\in S(Y,s)$. Since $s>0$, $b\ge 2$. Since $a\ge b$, $3s<(a+1)(b+1)$ and, hence, $|{\mathcal{I}}_{2A}(a,b)|\ne 0$. For all $p\in A$, let $A_p$ be the set of all $E\in |{\mathcal{I}}_p(1,1)|$. Note that p is the scheme-theoretic base locus of $A_p$ and that a general $E\in A_p$ is smooth and irreducible. Let $B_p$ be the set of all unions of 2 distinct elements of $A_p$. The scheme $2p$ is the scheme-theoretic base locus of the set $A_p$. Let $D={\cup }_{p\in A}{D}_p$ be a general union with $D_p$ general in $B_p$, and, if $(a,b)\ne (2s,2s)$, a general element of $|{\mathcal{O}}_Y(a-2s,b-2s)|$. Using s times the residual exact sequences of $D_p$, $p\in A$ and the cohomology of divisors of Y, we obtain $h^1\left({\mathcal{I}}_{2A}(a,b)\right)=0$, i.e., $h^0\left({\mathcal{I}}_{2A}(a,b)\right)=(a+1)(b+1)-3s$. Let $\mathcal{B}$ be the scheme-theoretic base locus of ${\mathcal{I}}_{2A}(a,b)$. Varying the elements of $D_p$, we see that $\mathcal{B}=2A$. The theorem of Bertini and the generality of C give $\mathrm{Sing}\left(C\right)=A$. Since D is nodal, C is nodal. Now, we use induction on s. Fix $p\in A$. In the case $s=1$, it is sufficient that a general $E\in |{\mathcal{I}}_{2p}(2,2)|$ is an irreducible nodal curve with arithmetic genus 1 and geometric genus 0. Now assume $s>1$ and that part (b) is true for the integer $s-1$. Since $s>1$, $a\ge b\ge 3$. Set $A^{\prime }:=A\setminus \left\{p\right\}$. By the inductive assumption, a general $F\in |{\mathcal{I}}_{2A^{\prime }}(2s-2,2s-2)|$ is irreducible, nodal and $\mathrm{Sing}\left(F\right)=A^{\prime }$. Since $2p$ is the scheme-theoretic base locus of ${\mathcal{I}}_{2p}(2,2)$, $p\notin A^{\prime }$ and F is general, we may find E such that $C^{\prime }:=E\cup F$ is nodal. Assume C is reducible. Since $C^{\prime }\in \left|{\mathcal{I}}_{2A}(a,b)\right|$ and C is reducible, we see that $C=C_1\cup C_2$ with $C_1$, $C_2$ irreducible and $C_1\in \left|{\mathcal{O}}_Y(2,2)\right|$. Note that $2(b-2)+2(a-2)=2(a+b)-8$ is the intersection number of $C_1$ and $C_2$. Since C is nodal and $\mathrm{Sing}\left(C\right)=A$, we obtain that $C_1$ and $C_2$ are smooth outside A, nodal and they intersect transversally at all points of $C_1\cap C_2$. Since $C_1$ is singular at p, $C_2$ is singular at each point of $A^{\prime }$, C is nodal and $\mathrm{Sing}\left(C\right)=A$, $C_1\cap C_2=\varnothing$, a contradiction. □

If we allow multiple components, we see that there is no “minimum of the multidegrees”, as shown by the following result.

Proposition 12.

Set $Y:={\mathbb{P}}^1\times {\mathbb{P}}^1$. Take positive integers s, a and b such that $a\ge b\ge 2$. Take a general $S\in S(Y,s)$ and any $A\in S(Y,s)$.

(a) 

We have $|{\mathcal{I}}_{2A}(2s,0)|\ne \varnothing$, $|{\mathcal{I}}_{2A}(0,2s)|\ne \varnothing$ and $|{\mathcal{I}}_{2A}(s,s)|\ne \varnothing$.

(b) 

We have $|{\mathcal{I}}_{2S}(2s-1,0)|=\varnothing$ and $|{\mathcal{I}}_{2S}(0,2s-1)|=\varnothing$. We have $|{\mathcal{I}}_{2S}(s-1,s)|\ne \varnothing$ if and only if either $s(s-1)>3s$ or $s=3$.

Proof. 

Set $A_1:={\pi }_1\left(A\right)$, $A_2:={\pi }_2\left(A\right)$, $S_1:={\pi }_1\left(S\right)$ and $S_2:={\pi }_2\left(S\right)$. We have $\#A_1\le s$, $\#A_2\le s$ and $\#S_1=\#S_2=s$. Part (a) follows taking the union of all $2L_p$, $2R_q$ and $L_p\cup R_q$ with $p\in A_1$, $\left\{L_p\right\}=\left|{\mathcal{I}}_p(1,0)\right|$, $q\in A_2$ and $\left\{R_q\right\}=\left|{\mathcal{I}}_q(0,1)\right|$. Since $\#S_1=\#S_2=s$, we obtain the first statement of part (b). The second statement of part (b) is obvious if $s=1$. It is true for $s=2$, because any element of $|{\mathcal{O}}_Y(1,2)|$ or $|{\mathcal{O}}_Y(1,2)|$ has arithmetic genus 0 and $\#S_1=\#S_2=s$. Thus, we may assume $s>2$. The second statement of part (b) is true by part (a) of Proposition 11. □

There are several papers on rational curves contained in a multiprojective space. For irreducible curves of arithmetic genus 1, smooth or singular, we give the following results.

Proposition 13.

Let E be an elliptic curve. Y contains an elliptic curve $C\subset Y$ of multidegree $(a_1,\dots ,a_k)$ such that Y is the minimal multiprojective space containing C and $C\cong E$ if and only if $a_i\ge n_i+1$ and either $k\ge 3$ or $k=2$ and $(n_1,n_2)\ne (1,1)$ or $k=2$, $n_1=n_2=1$ and $a_1=a_2=2$.

Proof. 

If $i\in \{1,\dots ,k\}$. Since Y is the minimal multiprojective space containing C, ${\pi }_i\left(C\right)$ spans ${\mathbb{P}}^{n_i}$. Hence $\text{deg}\left({\pi }_i\left(C\right)\right)\ge n_i$, and equality holds only if ${\pi }_i\left(C\right)$ is a rational normal curve. In the latter case $\text{deg}\left({\pi }_{i|C}\right)\ge 2$; hence, $a_i\ge 2n_i$. Let $f_i:C\to {\mathbb{P}}^{n_i}$ be a morphism induced by a general $(n_1+i)$-dimensional linear subspace of a degree $a_i$ line bundle on E. This linear system has no base point and $f_i$ is an embedding if either $n_i\ge 3$ or $n_i=2$ and $a_i=3$. Take $f=(f_1,\dots ,f_k)$. We saw that f is an embedding if either $n_i\ge 3$ for some i or $n_i=2$ and $a_i=3$. If $k=2$ and $n_1=n_2=2$, the classification of curves on ${\mathbb{P}}^1\times {\mathbb{P}}^1$ gives that f is an embedding if and only if $a_1=a_2=2$.

Now assume $k\ge 3$ and $n_1=n_2=1$. Taking $(f_1,f_2)$ and using the structure of curves on ${\mathbb{P}}^1\times {\mathbb{P}}^1$, we see that the morphism is birational onto its image, that the differential of f is non-zero everywhere, that $f\left(E\right)$ has at most nodes and that it is an embedding if $a_1=a_2=2$. Since $k\ge 3$, a dimensional count gives that f is injective. If $k\ge 3$ and $(n_1,n_2)\ne (1,1)$, the case just proved (taking a smaller vector space of sections for the first 2 factors) that f is an embedding. Now, assume $k=2$. The map f is an embedding if either $n_i\ge 3$ for some i on $n_1=n_2=1$ and $a_1=a_2=2$. Now assume $n_i=2$ for some i, say $n_1=2$. We obtain that $f_1\left(E\right)$ is nodal of degree $a_i+1$. Using $f_2$, we obtain that f is an embedding. □

Singular integral curves of arithmetic genus 1 have exactly one singular point, and this singular point is either an ordinary node or an ordinary cusp. Each of these types (ordinary node or ordinary cusp) gives a unique curve, up to isomorphism. The proof of Proposition 13 gives the following result.

Proposition 14.

Let E be an integral singular curve of arithmetic genus 1. There is an embedding $f:E\to Y$ of multidegree $(a_1,\dots ,a_k)$ if and only if either $k=2$, $n_1=n_2=1$ and $a_1=a_2=2$ or $a_i\ge n_i+1$.

Proposition 15.

Fix $p\in Y$. There is an integral rational curve $C\subset Y$ of multidegree $(a_1,\dots ,a_k)$ such that $\left\{p\right\}=\mathrm{Sing}\left(C\right)$, and Y is the minimal multiprojective space containing C if and only if $a_i\ge n_i+1$ for all i.

Proof. 

The existence part is true by Proposition 14 and the homogeneity of Y. The non-existence part is true by the proof of Proposition 13, because C is not smooth and rational. □

Proposition 16.

Fix $p\in Y$. There is a reduced (resp. reduced and connected, resp. integral curve) $C\subset Y$ of multidegree $(a_1,\dots ,a_k)$ such that Y is the minimal multiprojective space containing C, and C is singular at p if and only if $2d_i\ge n_i+1$ for all i and $2d_j\ge n_j+2$ for at least one j (resp. $a_i\ge n_i$, resp. $a_i\ge n_i+1$ for all i).

Proof. 

The case “C is integral” is true by Proposition 15.

Now, assume that C is reduced and connected. Hence, each ${\pi }_i\left(C\right)$ is connected. Remark 11 gives $\text{deg}\left({\pi }_i\left(C\right)\right)\ge n_i$ for all i. For the existence part with $a_i=n_i$, we take a chain E of d${\mathbb{P}}^1$’s $L_1,\dots ,L_d$, i.e., E is nodal, and $L_i\cap L_j\ne \varnothing$ if and only if $|i-j|\le 1$. Working on each component $L_i$, we obtain an embedding $f=(f_1,\dots ,f_k)$ of E such that $f_i\left(L_j\right)$ is a line for all $i,j$ and $f(L_1\cap L_2)=p$. If $a_i>n_i$ for some i, we add to $f\left(E\right)$$a_1+\cdots +a_k-n_1-\cdots -n_k$ curves, $a_i-n_i$ of them of bidegree ${ϵ}_i$ and meeting $f\left(E\right)$.

Now, assume that C is reduced. In principle, we cannot use Remark 11 to obtain the “only if” part, because for some i and some connected component A of C, the set ${\pi }_i\left(A\right)$ may be a point and the point, $o_i$, may contribute to the assumption that ${\pi }_i\left(C\right)$ spans ${\mathbb{P}}^{n_i}$. However, for any such point, there is at least one connected component A of C such that ${\pi }_i\left(A\right)$ is a point. Taking $f_i$ not constant on A, but with $o_i\in \pi \left(f_i\left(A\right)\right)$, we see that they are necessary conditions.

For the existence part, we take a disjoint union F of a reducible conic $D_1$ and $d-2$ disjoint ${\mathbb{P}}^1$s, embed it with $f=(f_1,\dots ,f_k)$ and apply Remark 11. □

Remark 16.

Note that in the set up of Proposition 16, we may take C with a far lower multidegree if we drop the condition that Y is the minimal multiprojective space containing C. We will see that in the set-up of local Cohen–Macaulay curves (even just a locally complete intersection double structure on a smooth curve), we may find C such that C has a very small multidegree, “many” singular points and has Y as the minimal multiprojective space.

Proposition 17.

Assume $k>1$. Fix $i\in \{1,\dots ,k\}$ and $p_j\in {\mathbb{P}}^{n_j}$, $j\ne i$. Set $s:=2$ if $n_i=1$ and $s:=3$ if $n_i=1$. Take a general $S^{\prime }\subset {\mathbb{P}}^{n_1}$ such that $\#S^{\prime }=s$ and let $S\subset Y$ be the only set of cardinality s with ${\pi }_i\left(S\right)=S^{\prime }$ and ${\pi }_j\left(S\right)=p_j$ for all $j\ne i$. Then, there is a locally complete intersection curve $A\subset Y$ with $S\subset \mathrm{Sing}\left(A\right)$, A of multidegree $2{ϵ}_i$, ${\pi }_j\left(A_{\mathrm{red}}\right)=p_j$ for all $j\ne i$ and with Y as the minimal multiprojective space containing S.

Proof. 

Fix $E\subset {\mathbb{P}}^1$ such that $\#E=s$. There is a degree 1 embedding $f_i:{\mathbb{P}}^1\to {\mathbb{P}}^{n_i}$ such that $f_i\left(E\right)=S^{\prime }$. For each $j\ne j$, let $f_j:{\mathbb{P}}^1\to {\mathbb{P}}^{n_j}$ be the constant map with $f\left({\mathbb{P}}^1\right)=p_j$. Set $f:=f(1,\dots ,f_k):{\mathbb{P}}^1\to Y$ and $L:=f\left({\mathbb{P}}^1\right)$. Note that $f\left(E\right)=S$ and, hence, $S\subset L$. The curve L has bidegree ${ϵ}_i$. Fix general subsets $S_h\subset L$, $h=1,\dots ,k$ such that $\#S_h=n_h+1$ for all h. Set $F:=S_1\cup \cdots \cup S_k$. For each $p\in F$, let $v_p\subset Y$ be a general degree 2 connected zero-dimensional scheme. Set $Z:=L\cup {\bigcup }_{p\in E}{v}_p$. Since each $v_p$ is general, the scheme ${\pi }_j\left(Z\right)$, $j\ne i$, is the first infinitesimal neighborhood of $p_j$ in ${\mathbb{P}}^{n_j}$, and hence, ${\mathbb{P}}^{n_j}$ is the minimal projective space containing the scheme ${\pi }_j\left(Z\right)$. As in the proof of Proposition 6, we see that ${\mathbb{P}}^{n_i}$ is the minimal projective space containing the scheme ${\pi }_i\left(Z\right)$. Hence Y is the minimal multiprojective space containing Z. Set $d_i:=1+n_1+\cdots +n_k$ for all i. As in the proof of Proposition 6. we obtain that the intersection T of $n_1+\cdots +n_k-2$ general elements of $|{\mathcal{I}}_Z(d_1,\dots ,d_k)|$ is a smooth surface containing L and that the double A of L as a Cartier divisor of T is a solution of Proposition 17. □

Proposition 18.

Assume $k>1$. Fix $i\in \{1,\dots ,k\}$ and take a general $p=(p_1,\dots ,p_k)\in Y$. Then, there is a locally complete intersection curve $A\subset Y$ with $\left\{p\right\}\in \mathrm{Sing}\left(A\right)$, A of multidegree $2{ϵ}_i$, ${\pi }_j\left(A_{\mathrm{red}}\right)=p_j$ for all $j\ne i$ and with Y as the minimal multiprojective space containing S.

Proof. 

We apply the proof of Proposition 17 with $S=\left\{p\right\}$. Only in this case, we may take a general S. If $\#S>1$, we cannot take S general and prescribe that ${\pi }_j\left(A_{\mathrm{red}}\right)$ is a point for all $j\ne i$. □

Obviously, in Propositions 17 and 18 $2{ϵ}_i$ is the minimal multidegree of a local Cohen–Macaulay curve with at least one singular point, and the reduced curves may have only 1 singular point.

Example 2.

Assume $k>1$. Fix $i\in \{1,\dots ,k\}$, set $s:=2$ if $n_i>1$, and take $S^{\prime }$, S, L and A as in Proposition 17. Fix a general $o\in L$. Take $h\in \{1,\dots ,k\}$ with the only restriction that $h\ne i$ if $n_i=1$. Take a general curve D of multidegree ${ϵ}_h$ passing through o. The curve $A\cup D$ works for $S\cup \left\{o\right\}$, and it only has multidegree $2{ϵ}_i+{ϵ}_k$, while Y is the minimal multiprojective space containing $A\cup D$.

Remark 17.

Taking unions of solutions of Propositions 17 or 18, we obtain multiple curves with low multidegree, singular at a prescribed large set S and with low multidegree but sometimes not the lowest possible. Indeed, we need the condition that Y is the minimal multidegree containing A just for one single A. See Example 2.

6. On Questions 1, 2 and 3

Remark 18.

Fix integers n, d, $x>0$ and t such that $n\ge 3$, $0\le t\le n-1$, $d\ge 3$ and $(n,d)\notin \left\{\right(3,4),(4,3),(4,4\left)\right\}$. Take a general $S\in S({\mathbb{P}}^n,x)$. The theorem of Alexander–Hirschowitz gives $h^0\left({\mathcal{I}}_{2S}\left(d\right)\right)=max\{0,\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-(n+1)x\}$ [6,7,8]. Hence,

$$\alpha (n,d,t)\le \lfloor (\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-n+t)/(n+1)\rfloor .$$

Remark 19.

Fix integers n, d, $x>0$ and t such that $n\ge 3$, $0\le t\le n-1$, $d\ge 3$ and $(n,d)\notin \left\{\right(3,4),(4,3),(4,4\left)\right\}$. Take $A\in S({\mathbb{P}}^n,x)$ such that $h^1\left({\mathcal{I}}_{2A}\left(d\right)\right)=0$ and a general $S\in S({\mathbb{P}}^n,x)$. Let ${\mathcal{B}}_A$ (resp. ${\mathcal{B}}_S$) denote the base locus of $|{\mathcal{I}}_{2A}\left(d\right)|$ (resp. $|{\mathcal{I}}_{2S}\left(d\right)|$). Since $h^0\left({\mathcal{I}}_{2A}\left(d\right)\right)=h^0\left({\mathcal{I}}_{2S}\left(d\right)\right)$ and S is general, $dim{\mathcal{B}}_S\le dim{\mathcal{B}}_A$. Hence, $x\ge \alpha (n,d,dim{\mathcal{B}}_A)$. The linear space $H^0\left({\mathcal{I}}_{2A}\left(d\right)\right)$ is computable with a computer. Hence, by using a few random sets A for a fixed pair $(n,d)$ (not too large), it should be easy to check by computer the value of $\alpha (n,d,t)$.

Lemma 2.

Fix integers $n\ge 3$, $x>0$, $z>0$, $1\le t\le n-2$ and $d\ge 4$ such that $(n+1)(x+1)+z\le \left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-1$. Fix a set $A\subset {\mathbb{P}}^n$ such that $\#A=x$ and $h^1\left({\mathcal{I}}_{2A\cup 2p}\left(d\right)\right)=0$ and the base locus $\mathcal{B}$ of ${\mathcal{I}}_{2A}\left(d\right)$ has dimension $\le t$. Fix a general $E\subset {\mathbb{P}}^n$ such that $\#E=z$. Then, the base locus of ${\mathcal{I}}_{2A\cup E}\left(d\right)$ has dimension $\le t$.

Proof. 

Since $h^1\left({\mathcal{I}}_{2A\cup 2p}\left(d\right)\right)=0$, $h^1\left({\mathcal{I}}_{2A}\left(d\right)\right)=0$ (Remark 1). Since E is general and $\text{deg}\left(2A\right)+\#E\le \left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)$, $h^1\left({\mathcal{I}}_{2A\cup E}\left(d\right)\right)=0$ and, hence, $h^0\left({\mathcal{I}}_{2A\cup E}\left(d\right)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{d+n}{n}}\right)-(n+1)x-z\ge n-t$. Let $\varphi :{\mathbb{P}}^n⤏{\mathbb{P}}^r$, $r:=\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-(n+1)x-1$ denote the rational map induced by $|{\mathcal{I}}_{2A}\left(d\right)|$. Since $h^1\left({\mathcal{I}}_{2A\cup 2p}\left(d\right)\right)=0$, the differential of $\varphi$ at a general point of ${\mathbb{P}}^n$ is injective. Hence, $dim\varphi ({\mathbb{P}}^n\setminus \mathcal{B})=n$. By the second theorem of Bertini ([24], 6.3), the restriction to ${\mathbb{P}}^n\setminus \mathcal{B}$ of the scheme-theoretic intersection of a general codimension $n-1$ linear subspace of $|{\mathcal{I}}_{2A}\left(d\right)|$ is an irreducible and smooth curve. We use induction on the integer z, starting the induction with the case $z=0$, in which $E=\varnothing$. Assume $z>0$ and fix $o\in E$. Set $E^{\prime }:=E\setminus \left\{o\right\}$. Let B denote the base locus of ${\mathcal{I}}_{2A\cup E^{\prime }}\left(d\right)|$. By the inductive assumption, $dimB\le t$. Let $\psi :{\mathbb{P}}^n\setminus B\to {\mathbb{P}}^{r-z+1}$ denote the morphism induced by $|{\mathcal{I}}_{2A\cup E^{\prime }}\left(d\right)|$. Since $h^1\left({\mathcal{I}}_{2A\cup 2p}\left(d\right)\right)=0$, $\text{deg}(2A\cup 2p\cup E^{\prime })=(n+1)(x+1)+z-1\le \left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)$, and $E^{\prime }$ is general, $h^1\left({\mathcal{I}}_{2A\cup 2p\cup E^{\prime }}\left(d\right)\right)=0$. Hence, the differential of $\psi$ at a general point of ${\mathbb{P}}^n$ is injective. Thus, $dim\psi ({\mathbb{P}}^n\setminus B)=n$. By the second theorem of Bertini ([24], 6.3), the restriction to ${\mathbb{P}}^n\setminus B$ of the scheme-theoretic intersection of general codimension $n-1$ linear subspace of $|{\mathcal{I}}_{2A}\left(d\right)|$ is an irreducible and smooth curve, C. Set $F:=E^{\prime }\cup \left\{a\right\}$. Let a be a general point of C. Since C is a general element of a covering family of curves of ${\mathbb{P}}^n$, F is a general element of $S({\mathbb{P}}^n,z)$. Thus, it is sufficient to prove that the base locus $B^{\prime }$ of ${\mathcal{I}}_{2A\cup F}\left(d\right)$ has dimension $\le t$. We have $B^{\prime }\subseteq B\cup C$. Since $t>0$, $dim{B}^{\prime }\le t$. □

Lemma 3.

Fix integers $n\ge 3$, $x>0$, $z\ge 0$, $1\le t\le n-2$ and $d\ge 4$ such that $(n+1)(x+1)+z\le \left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)$. Fix a set $A\subset {\mathbb{P}}^n$ and a hyperplane H such that $\#A=x$, $A\cap H=\varnothing$ and the base locus $\mathcal{B}$ of ${\mathcal{I}}_{2A}\left(d\right)$ has dimension $\le t$. Fix a general $p\in H$ and assume $h^1\left({\mathcal{I}}_{2A\cup 2p}\left(d\right)\right)=0$ and $h^0\left({\mathcal{I}}_{2A}(d-1)\right)+z\le h^0\left({\mathcal{I}}_{2A\cup 2p}\left(d\right)\right)$. Fix a general $E\subset H$ such that $\#E=z$. Then, the base locus of ${\mathcal{I}}_{2A\cup E}\left(d\right)$ has dimension $\le t$.

Proof. 

The case $z=0$ is a consequence of Lemma 2. Hence, we may assume $z>0$ and use induction on the integer z. Fix $o\in E$ and set $E^{\prime }:=E\setminus \left\{o\right\}$. Since $h^1\left({\mathcal{I}}_{2A\cup 2p}\left(d\right)\right)=0$, we have $h^1\left({\mathcal{I}}_{2A}\left(d\right)\right)=0$ (Remark 1) and, hence, $h^0\left({\mathcal{I}}_{2A}\left(d\right)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-(n+1)x$.

Claim 1: We have $h^0\left({\mathcal{I}}_{2A\cup p}(d-1)\right)=h^0\left({\mathcal{I}}_{2A}(d-1)\right)-1$.

Proof of Claim 1: By assumption, $h^0\left({\mathcal{I}}_{2A}(d-1)\right)>0$. For any $q\in {\mathbb{P}}^n\setminus A$, we have $h^0\left({\mathcal{I}}_{2A}(d-1)\right)-1\le h^0\left({\mathcal{I}}_{2A\cup q}(d-1)\right)\le h^0\left({\mathcal{I}}_{2A}(d-1)\right)$ (Remark 1). Assume $h^0\left({\mathcal{I}}_{2A\cup p}(d-1)\right)=h^0\left({\mathcal{I}}_{2A}(d-1)\right)$. Since $A\cap H=\varnothing$ and p is general in H, the residual exact sequence of H gives $h^0\left({\mathcal{I}}_{2A}(d-1)\right)=h^0\left({\mathcal{I}}_{2A}(d-2)\right)$, contradicting the inequality $h^0\left({\mathcal{I}}_{2A}(d-1)\right)>0$.

Claim 2: We have $h^1\left({\mathcal{I}}_{2A\cup 2p\cup E}\left(d\right)\right)=0$, i.e., E gives z independent conditions to $H^0\left({\mathcal{I}}_{2A\cup 2p}\left(d\right)\right)$.

Proof of Claim 2: The case $z=0$, i.e., $E=\varnothing$, is true by assumption. Hence, we may use induction even in the proof of Claim 2. By the inductive assumption $h^1\left({\mathcal{I}}_{2A\cup 2p\cup E^{\prime }}\left(d\right)\right)=0$; hence, $h^0\left({\mathcal{I}}_{2A\cup 2p\cup E^{\prime }}\left(d\right)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-(n+1)(x+1)-z+1>0$. Assume the inequality $h^1\left({\mathcal{I}}_{2A\cup 2p\cup E}\left(d\right)\right)>0$. Since o is general in H, H is contained in the base locus of $|{\mathcal{I}}_{2A\cup 2p\cup E^{\prime }}\left(d\right)|$. Since ${\mathrm{Res}}_H(2A\cup 2p\cup E^{\prime })=2A\cup p$, we obtain $h^0\left({\mathcal{I}}_{2A\cup 2p\cup E^{\prime }}\left(d\right)\right)=h^0\left({\mathcal{I}}_{2A\cup p}(d-1)\right)$. The inequality $h^0\left({\mathcal{I}}_{2A\cup 2p\cup E^{\prime }}\left(d\right)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-z+1>h^0({\mathcal{I}}_{2A\cup p}(d-1)$ (Claim 1) gives a contradiction.

Let B denote the base locus of ${\mathcal{I}}_{2A\cup E^{\prime }}\left(d\right)$. By the inductive assumption, we have $dimB\le t$. Let $\varphi :{\mathbb{P}}^n\setminus B\to {\mathbb{P}}^r$, $r=\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-(n+1)x-z$ be the morphism induced by $|{\mathcal{I}}_{2A\cup E^{\prime }}\left(d\right)|$. Since $h^1\left({\mathcal{I}}_{2A\cup 2p\cup E}\left(d\right)\right)=0$ (Claim 2 and Remark 1), the differential of $\varphi$ is injective at a general point of H. Thus, $dim\varphi ({\mathbb{P}}^n\setminus B)=n$ and $dim\varphi (H\setminus B\cap H)=n-1$. Hence, for a general $a\in H$, $dim{\varphi }^{-1}\left(\varphi \left(a\right)\right)\cap H)=0$. Hence, $dim{\varphi }^{-1}\left(a\right))\le 1\le t$. Since a is general in H, $F:=E^{\prime }\cup \left\{a\right\}$ is general in $S(H,z)$ and $t>0$, to prove the lemma, it is sufficient to observe that the base locus of ${\mathcal{I}}_{2A\cup F}\left(d\right)$ is contained in $B\cup {\varphi }^{-1}\left(p\right)$. □

The proof of the following result is an example of the use of the Horace Method [1,2]. We assume $n\ge 4$, because for $n=3$, Proposition 8 is stronger.

Theorem 12.

Fix integers $n\ge 4$, $d\ge 5$, $x>0$, $z>0$, $0<t\le n-2$ such that $(n+1)x+z\le \left({\displaystyle \genfrac{}{}{0pt}{}{n+d-1}{n}}\right)$, $\left({\displaystyle \genfrac{}{}{0pt}{}{n+d-2}{n}}\right)+z\le \left({\displaystyle \genfrac{}{}{0pt}{}{n+d-1}{n}}\right)$ and $z\le \alpha (n-1,d,t-1)$. If $n=4$, assume $d\ge 6$. Then, $x+z\le \alpha (n,d,t)$.

Proof. 

Fix a general pair $(A,H)\in S({\mathbb{P}}^n,x)\times {\mathbb{P}}^{n\vee }$. The theorem of Alexander–Hirschowitz [6,7,8] gives $h^1\left({\mathcal{I}}_{2A}(d-1)\right)=0$. Take a general $E\subset H$ such that $\#E=z$. By assumption, $h^1(H,{\mathcal{I}}_{(2E,H),H}\left(d\right))=0$. Consider the residual exact sequence

$$0\to {\mathcal{I}}_{2A\cup E}(d-1)\to {\mathcal{I}}_{2S}\left(d\right)\to {\mathcal{I}}_{H\cap 2E,H}\left(d\right)\to 0$$

(6)

Claim 1: We have $h^1\left({\mathcal{I}}_{2A\cup E}\left(d\right)\right)=0$.

Proof of Claim 1: The case $z=0$, i.e., $E=\varnothing$, is true by the generality of H and the Alexander–Hirschowitz theorem applied in degree $d-1$. Hence, we may use induction on the integer z. Fix $o\in E$ and set $E^{\prime }:=E\setminus \left\{o\right\}$. By the inductive assumption, $h^1\left({\mathcal{I}}_{2A\cup E^{\prime }}\left(d\right)\right)=0$ and, hence, $h^0\left({\mathcal{I}}_{2A\cup E^{\prime }}\left(d\right)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-(n+1)(x+1)-z+1>0$. Assume $h^1\left({\mathcal{I}}_{2A\cup E}\left(d\right)\right)>0$. Since o is general in H, H is contained in the base locus of $|{\mathcal{I}}_{2A\cup E^{\prime }}\left(d\right)|$. Since ${\mathrm{Res}}_H(2A\cup E^{\prime })=2A$, we obtain $h^0\left({\mathcal{I}}_{2A\cup E^{\prime }}\left(d\right)\right)=h^0\left({\mathcal{I}}_{2A}(d-1)\right)$. Since $h^0\left({\mathcal{I}}_{2A\cup E^{\prime }}\left(d\right)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-z+1>h^0({\mathcal{I}}_{2A\cup p}(d-1)$, we have a contradiction.

Since $z\le \alpha (n-1,d,t-1)$, we have $nz\le \left({\displaystyle \genfrac{}{}{0pt}{}{n+d-1}{n}}\right)-n+t$ and, hence, $z\le \left({\displaystyle \genfrac{}{}{0pt}{}{n+d-1}{n-1}}\right)$. Hence, the theorem of Alexander–Hirschowitz gives $h^1(H,{\mathcal{I}}_{2E\cap H,H}\left(d\right))=0$. Claim 1 and the long cohomology exact sequence of (6) gives $h^1\left({\mathcal{I}}_{2(A\cup E)}\left(d\right)\right)=0$. Hence, to prove that $y+z\le \alpha (n,d,t-1)$, it is sufficient to prove that the base locus, B, of ${\mathcal{I}}_{2(A\cup E)}\left(d\right)$ has dimension $\le t$. Assume that B has an irreducible component T of dimension $>t$. We have $dimT\cap H\ge t$. The long cohomology exact sequence of (6) and Claim 1 imply the surjectivity of the restriction map $\rho :H^0\left({\mathcal{I}}_{\left(2\right(A\cup E)}\left(d\right)\right)\to H^0(H,{\mathcal{I}}_{(2E\cap H,H}\left(d\right))$. Hence, $T\cap H$ is in the base locus of ${\mathcal{I}}_{(2E\cap H,H}\left(d\right)$, contradicting the assumption $z\le \alpha (n,d,t-1)$. □

7. Discussions

For a general finite subset S of a projective or a multiprojective space, we construct curves that are singular or with maximal Zariski tangent space in the ambient projective or multiprojective space and with minimal degree or “complexity”, i.e., the minimum possible number of multivariate equations, each of them of minimum degree. We do the same for all finite sets with a prescribed number of points. Our tools come from algebraic geometry, and some of the results are both existence and non-existence ones, e.g., all the ones with start with “The minimal degree”. In the asymptotic results, we gave explicit bounds. In the introduction, we listed the specific existence results that may give explicit solutions using “random” inputs.

We gave several open problems and discussed the following question (open for large x):

Question: Take $n\ge 3$, $d\ge 3$, $(n,d)\notin \left\{\right(3,4),(4,3),(4,4\left)\right\}$. Fix an integer x such that $(n+1)x<\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)$. Take a general $S\subset {\mathbb{P}}^n$ such that $\#S=x$. Compute the dimension of the base locus of the degree d hypersurfaces that are singular at all points of S.