# Base Point Freeness, Uniqueness of Decompositions and Double Points for Veronese and Segre Varieties

## Abstract

**:**

## 1. Introduction

**Theorem**

**1.**

## 2. Preliminaries

**Remark**

**1.**

## 3. Veronese Varieties

**Remark**

**2.**

**Remark**

**3.**

**Remark**

**4.**

**Lemma**

**1.**

**Proof.**

**Theorem**

**2.**

**Lemma**

**2.**

**Proof.**

**Proof**

**of**

**Theorem**

**2:**

**Claim**

**1:**

**Proof**

**of**

**Claim**

**1:**

**Proof**

**of**

**Theorem**

**1:**

## 4. Base Point Freeness

**Assumption**

**1.**

**Remark**

**5.**

**Assumption**

**2.**

## 5. Tensors

**Theorem**

**3.**

- 1 .
- $deg\left(B\right)=1$ and either $k=1$, ${n}_{1}\le 2$ or $k=2$ and ${n}_{1}={n}_{2}=1$.
- 2 .
- $deg\left(B\right)=2$ and either $k=1$, ${n}_{1}\le 2$ or $k=2$ and ${n}_{1}={n}_{2}=1$.
- 3 .
- $k=2$, ${n}_{1}\le deg\left(B\right)-1$ and $A,B,q$ are as in Remark 10.
- 4 .
- $deg\left(B\right)=3$, $Y={\left({\mathbb{P}}^{1}\right)}^{3}$ and $A,B,q$ are as in Remark 11.
- 5 .
- $deg\left(B\right)=3$, $Y={\left({\mathbb{P}}^{1}\right)}^{4}$ and $A,B,q$ are as in Example 9.
- 6 .
- $deg\left(B\right)=3$, $Y={\mathbb{P}}^{2}\times {\mathbb{P}}^{1}\times {\mathbb{P}}^{1}$ and A, B are as in Example 3.
- 7 .
- $deg\left(B\right)=3$ and $(Y,q,B)$ is as in Examples 4, 5, 6, 7 or 8.

**Remark**

**6.**

**Remark**

**7.**

**Remark**

**8.**

**Example**

**1.**

**Example**

**2.**

**Proposition**

**1.**

- 1 .
- $k=1$, ${n}_{1}=1$;
- 2 .
- $k=1$, ${n}_{1}=2$;
- 3 .
- $k=2$, ${n}_{1}={n}_{2}=1$.
- (a)
- In case (1) we may take $deg\left(Z\right)=1$ (with $\nu \left(Z\right)=\left\{q\right\}$), $deg\left(Z\right)=2$ and reduced or $deg\left(Z\right)=2$ and Z an arrow.
- (b)
- In case (2) $\nu (v\cup Z)$ is the complete intersection of two conics in the plane $\langle \nu (W\cup Z)\rangle $.
- (c)
- In case (3) there are ${\infty}^{2}$ reduced Z (parametrized by a plane minus a line) and ${\infty}^{1}$ arrows Z (parametrized by a line minus the point corresponding to v).

**Proof.**

**Remark**

**9.**

**Remark**

**10.**

**Proposition**

**2.**

- 1.
- $k=1$ and ${n}_{1}=1$ (here $q\in \nu \left(Y\right)$, say $q=\nu \left({o}^{\prime}\right)$);
- 2.
- $k=2$, ${n}_{1}={n}_{2}=1$;
- 3.
- $k=2$, ${n}_{1}+{n}_{2}=3$; $k={n}_{1}={n}_{2}=2$;
- 4.
- $k=3$, ${n}_{1}={n}_{2}={n}_{3}=1$.
- (i)
- In case (1) the schemes Z are as follows:
- (i1)
- $deg\left(Z\right)=1$ and Z is the point of Y such that $q=\nu \left(Z\right)$;
- (i2)
- Z is any degree 2 subscheme.

- (ii)
- In case (2) there are ${\infty}^{2}$ schemes Z formed by 2 points and ${\infty}^{1}$ schemes Z which are arrows.
- (iii)
- In cases (3) and (4) there are at least ${\infty}^{2}$ schemes Z formed by 2 points and at least ${\infty}^{1}$ schemes Z which are arrows; in this case Y is not the minimal multiprojective space such that $q\in \langle \nu \left(Y\right)\rangle $.

**Proof.**

- (a)
- Assume for the moment ${n}_{i}>1$ for some i, say ${n}_{1}>1$ and hence ${n}_{1}=2$. Since Y is the minimal multiprojective space containing Y, $deg\left({\pi}_{1}\left(A\right)\right)=3$ and ${\pi}_{1}\left(A\right)$ is linearly independent. Take ${H}_{1}\in \left|{\mathcal{I}}_{v}\left({\epsilon}_{1}\right)\right|$ and ${H}_{2}\in \left|{\mathcal{I}}_{u}\left({\epsilon}_{2}\right)\right|$. The scheme $E:={Res}_{{H}_{1}\cup {H}_{2}}\left(W\right)$ is contained in Z.
- (a1)
- Assume $E\ne \varnothing $. Since $Z\cap A=\varnothing $, quoting ([33], Lemma 5.1) we obtain ${h}^{1}\left({\mathcal{I}}_{E}(0,0,1,\cdots ,1)\right)=0$. Since $E\subseteq Z$ and ${\mathcal{O}}_{Y}(0,0,1,\cdots ,1)$ is globally generated, we obtain $E=Z$ and $deg\left({\pi}_{i}\left(Z\right)\right)=1$ for all $i>2$. Thus, $|{\mathcal{I}}_{Z}\left({\epsilon}_{3}\right)|\ne \varnothing $. Fix $M\in |{\mathcal{I}}_{Z}\left({\epsilon}_{3}\right)|$. Since $Z\cap A=\varnothing $ and $A\u2288M$, ([33], Lemma 5.1) gives ${h}^{1}\left({\mathcal{I}}_{{Res}_{M}\left(W\right)}\left({\widehat{\epsilon}}_{3}\right)\right)>0$. Since ${Res}_{M}\left(W\right)\subseteq A$ and ${\pi}_{1|A}$ is an embedding with linearly independent image, we obtain a contradiction.
- (a2)
- Assume $E=\varnothing $, i.e., assume $W\subset {H}_{1}\cup {H}_{2}$. By step (a1) we may also assume $W\subseteq {H}_{1}\cup {H}_{3}$ with ${H}_{3}\in \left|{\mathcal{I}}_{u}\left({\epsilon}_{3}\right)\right|$. Set $\left\{{M}_{1}\right\}$:= $|{\mathcal{I}}_{\{o,u\}}\left({\epsilon}_{1}\right)|$ and take ${M}_{2}\in \left|{\mathcal{I}}_{o}\left({\epsilon}_{2}\right)\right|$ and ${M}_{3}\in \left|{\mathcal{I}}_{o}\left({\epsilon}_{3}\right)\right|$. Since $A\subset ({M}_{1}\cup {M}_{2})\cap ({M}_{1}\cup {M}_{3})$, we also obtain $Z\subset {M}_{1}\cup {M}_{2}$ and $Z\subset {M}_{1}\cup {M}_{3}$. Assume that either $Z\cap {H}_{1}=\varnothing $ or $Z\cap {M}_{1}=\varnothing $, say $Z\cap {H}_{1}=\varnothing $. We get $Z\subset {H}_{2}$. The residual exact sequence of ${H}_{2}$ and ([33], Lemma 5.1) give a contradiction because $deg\left({\pi}_{1}\left(A\right)\right)=3$ and ${\pi}_{1}\left(A\right)$ is linearly independent. Assume that either $Z\subset {H}_{1}$ or $Z\subset {M}_{1}$, say $Z\subset {H}_{1}$. The residual exact sequence of ${H}_{1}$ gives a contradiction. Thus, $deg(Z\cap {H}_{1})=deg(Z\cap {M}_{1})=1$. Using ${H}_{1}$ we obtain ${h}^{1}\left({\mathcal{I}}_{{Res}_{{H}_{1}}\left(W\right)}\left({\widehat{\epsilon}}_{1}\right)\right)>0$ with ${Res}_{{H}_{1}}\left(W\right)$ the union of u and a point of ${A}_{red}$, call it e. Since ${h}^{1}\left({\mathcal{I}}_{{Res}_{{H}_{1}}\left(W\right)}\left({\widehat{\epsilon}}_{1}\right)\right)>0$, ${\pi}_{i}\left(u\right)={\pi}_{i}\left(e\right)$ for all $i>1$. Take $R\in |{\mathcal{I}}_{u}\left({\epsilon}_{3}\right)|$. Using R we obtain ${h}^{1}\left({\mathcal{I}}_{v\cup {Res}_{R}\left(Z\right)}\left({\widehat{\epsilon}}_{3}\right)\right)>0$. Using ${H}_{2}$ we obtain ${h}^{1}\left({\mathcal{I}}_{v\cup {Res}_{{H}_{2}}\left(Z\right)}\left({\widehat{\epsilon}}_{2}\right)\right)>0$. Thus, ${\pi}_{i}\left({Res}_{{H}_{2}}\left(Z\right)\right)\subseteq {\pi}_{i}\left(v\right)$ for all $i\ne 2$. Since $dim|{\mathcal{O}}_{Y}\left({\epsilon}_{2}\right)|+dim|{\mathcal{O}}_{Y}\left({\epsilon}_{3}\right)|\ge 2$, there are $T\in |{\mathcal{O}}_{Y}\left({\epsilon}_{2}\right)|$ and ${T}^{\prime}\in \left|{\mathcal{O}}_{Y}\left({\epsilon}_{2}\right)\right|$ such that $Z\subset T\cup {T}^{\prime}$. First assume $A\u2288T\cup {T}^{\prime}$. Since $A\cap Z=\varnothing $, ([33], Lemma 5.1) gives ${h}^{1}\left({\mathcal{I}}_{{Res}_{T\cup {T}^{\prime}}\left(A\right)}(1,0,0,\cdots )\right)>0$. Since $deg\left({\pi}_{1}\left(A\right)\right)=3$ and ${\pi}_{1}\left(A\right)$ is linearly independent, we obtained a contradiction. Now assume $A\subset T\cup {T}^{\prime}$. If $Z\subset T$ (resp. $Z\subset {T}^{\prime}$) we may take instead of ${T}^{\prime}$ (resp. T) a general element of its complete linear system and obtain a contradiction, because Y is the minimal multiprojective space containing A. Thus, $T\cap Z\ne \varnothing $ and ${T}^{\prime}\cap Z\ne \varnothing $. One of the two divisors T or ${T}^{\prime}$, say T, contains v. Set $\left\{a\right\}:={Res}_{T}\left(Z\right)$. Please note that $a\in {T}^{\prime}$. We obtain ${h}^{1}\left({\mathcal{I}}_{\{u,a\}}\left({\widehat{\epsilon}}_{2}\right)\right)>0$, i.e., ${\pi}_{i}\left(a\right)={\pi}_{i}\left(u\right)$ for all $i\ne 2$. If $a=e$ we obtain $a=u$, a contradiction. Please note that $a=e$ if Z is connected. Assume $a\ne e$ and hence $Z=\{a,e\}$. Since $A\u2288T$, $u\in {T}^{\prime}$. Hence ${\pi}_{3}\left(a\right)={\pi}_{3}\left(u\right)$. Since $a\ne e$, we obtain ${Res}_{{T}^{\prime}}\left(W\right)\subseteq v$. Hence the residual exact sequence of ${T}^{\prime}$ gives a contradiction.

- (b)
- Assume $Y={\left({\mathbb{P}}^{1}\right)}^{k}$. All cases with $k\le 3$ are listed. Thus, we assume $k\ge 4$. Let ${e}_{1}$ be the maximal integer such that ${e}_{1}=deg(H\cap W)$ for some $i\in \{1,\cdots ,k\}$ and some $H\in |{\mathcal{O}}_{Y}\left({\epsilon}_{i}\right)|$. With no loss of generality we may assume $i=1$. Set ${W}_{1}:={Res}_{H}\left(W\right)$. Let ${e}_{2}$ be the maximal integer such that ${e}_{2}=deg(M\cap {W}_{1})$ for some $i\in \{2,\cdots ,k\}$ and some $M\in |{\mathcal{O}}_{Y}\left({\epsilon}_{i}\right)|$. With no loss of generality we may assume $i=2$. Set ${W}_{2}:={Res}_{H}\left({W}_{1}\right)$. Obviously ${e}_{1}\ge {e}_{2}$. Since Y is the minimal multiprojective space containing A, $1\le i\le 4$.
- (b1)
- Assume ${e}_{1}=4$. Quoting ([33], Lemma 5.1) we obtain a contradiction.
- (b2)
- Assume ${e}_{1}=2$. Thus, $1\le {e}_{2}\le 2$. If ${e}_{2}=2$ applying ([33], Lemma 5.1) to $H\cup M$ we obtain a contradiction. Assume ${e}_{2}=1$. The definition of ${e}_{1}$ gives that each ${\pi}_{i|{W}_{1}}$, $i>2$, is an embedding. Fix $D\in |{\mathcal{O}}_{Y}\left({\epsilon}_{3}\right)|$ intersecting ${W}_{2}$. Please note that $deg\left({Res}_{D}\left({W}_{2}\right)\right)=1$. The residual exact sequence of $H\cup M\cup D$ and ([33], Lemma 5.1) gives a contradiction.
- (b3)
- Assume ${e}_{1}=1$. The definition of ${e}_{1}$ gives that each ${\pi}_{i|W}$ is an embedding. Take $D\in |{\mathcal{O}}_{Y}\left({\epsilon}_{3}\right)|$ such that $D\cap {W}_{2}\ne \varnothing $ and set ${W}_{3}:={Res}_{D}\left({W}_{2}\right)$. Please note that $deg(D\cap W)=1$ and that $deg\left({W}_{3}\right)=2$. Since ${\pi}_{4|{W}_{2}}$ is an embedding, ${h}^{1}\left({\mathcal{I}}_{{W}_{2}}(0,0,0,1,\cdots )\right)=0$, contradicting ([33], Lemma 5.1).
- (b4)
- Assume ${e}_{1}=3$. If ${e}_{2}=1$ quoting ([33], Lemma 5.1) we obtain a contradiction. Now assume ${e}_{2}=2$, i.e., assume $W\subset H\cup M$. If there are $i\in \{3,\cdots ,k\}$ and $D\in |{\mathcal{O}}_{Y}\left({\epsilon}_{i}\right)|$ with $deg({W}_{1}\cap D)=1$, then quoting ([33], Lemma 5.1) with respect to $H\cup D$ we obtain a contradiction. Thus, we may assume $deg\left({\pi}_{i}\left({W}_{1}\right)\right)=1$ for all $i\ge 2$. Hence ${\pi}_{1|{W}_{1}}$ is an embedding. Set $E:={Res}_{M}\left(W\right)$. Since in step (b1) we excluded the case ${e}_{1}=4$ and ${W}_{1}\subset M$, we have $2\le deg(W\cap M)\le 3$ and hence $2\le deg\left(E\right)\le 3$. By assumption $E\subset H$ and hence $deg\left({\pi}_{1}\left(E\right)\right)=1$.
- (b4.1)
- Assume $deg\left(E\right)=2$. Using M instead of H in the first part of step (b4) we obtain $deg\left({\pi}_{i}\left(E\right)\right)=1$ for all $i\ne 2$. Take ${H}_{i}\in \left|{\mathcal{O}}_{Y}\left({\epsilon}_{i}\right)\right|$, $3\le i\le k$, containing E. We obtain $W\subset M+{H}_{i}$. We also have $W\subset H\cup M$. Please note that the set $M\cap H\cap {H}_{3}\cap \cdots \cap {H}_{k}$ is a point and that $\nu (M\cap {H}_{3}\cap \cdots \cap {H}_{k})$ is a line of the Segre variety $\nu \left(M\right)$. We obtain ${W}_{red}\subset M$. Since $deg(W\cap M)=3$, we obtain $\#{W}_{red}=3$, i.e., Z is not reduce. Using H instead of M we obtain ${W}_{red}\subset H\cap M$. Using M and ${H}_{i}$, $3\le i\le k$, we obtain ${W}_{red}$ is contained in the point $H\cap M\cap {H}_{3}\cap \cdots \cap {H}_{k}$, absurd.
- (b4.2)
- Assume $deg\left(E\right)=3$. In the set-up of step (b4.1) we may also assume that $E\cap {H}_{i}=\varnothing $ for all $i=3,\cdots ,k$. Set ${H}_{2}:=M$. Let $w\subset E$ be a degree 2 scheme. Since $\nu $ is an embedding, there is ${i}_{w}\in \{2,\cdots ,k\}$ such that ${\pi}_{{i}_{w}|w}$ is an embedding. Let ${M}_{w}$ be an element of $|{\mathcal{O}}_{Y}\left({\epsilon}_{{i}_{w}}\right)|$ containing a point of ${w}_{red}$. Fix $i\in \{2,\cdots ,k\}\setminus \left\{{i}_{w}\right\}$ and set $F:={Res}_{{H}_{i}+{M}_{w}}\left(W\right)$. Since $deg(F\cap w)=1$, $1\le deg\left(F\right)\le 2$. By ([33], Lemma 5.1) we first obtain $deg\left(F\right)=2$ and then $deg\left({\pi}_{h}\left(F\right)\right)=1$ for all $h\notin \{i,{i}_{w}\}$. Either E is the union of 3 points, say $E=\{a,b,c\}$, or the union of a point a and an arrow z.
- (b4.2.1)
- Assume $E=\{a,b,c\}$. Thus, $E\ne A$. Take $w:=\{b,c\}$. Taking ${M}_{w}$ containing c we obtain ${\pi}_{h}\left(a\right)={\pi}_{h}\left(b\right)$ for all $h\notin \{i,{i}_{w}\}$. Taking ${M}_{w}$ containing b we obtain ${\pi}_{h}\left(a\right)={\pi}_{h}\left(c\right)$ for all $h\notin \{i,{i}_{w}\}$. Thus, $deg\left({\pi}_{h}\left(E\right)\right)=1$ for all $h\notin \{i,{i}_{w}\}$. Varying i we obtain $deg\left({\pi}_{h}\left(E\right)\right)=1$ for all $h\ne {i}_{w}$. In this case $\langle \nu \left(E\right)\rangle $ is a line contained in the ${i}_{w}$-ruling of the Segre $\nu \left(Y\right)$. Thus, $\nu \left(E\right)$ is linearly dependent. Since $k>2$, Proposition 1 gives $q\notin \nu \left(Y\right)$. Since $q\in \langle \nu \left(A\right)\rangle \cap \langle \nu \left(Z\right)\rangle $ and $q\notin \langle \nu \left({A}^{\prime}\right)\rangle $ for any ${A}^{\prime}\u228aA$, we obtain ${h}^{1}\left({\mathcal{I}}_{W}(1,\cdots ,1)\right)\ge 2$. Since $\nu \left(A\right)$ is linearly independent, ${h}^{1}\left({\mathcal{I}}_{W}(1,\cdots ,1)\right)=2$. Thus, $\nu \left(Z\right)$ is contained in the plane, $\langle \nu \left(A\right)\rangle $, which also contains the line $\langle \nu \left(E\right)\rangle \subset \nu \left(Y\right)$. Since $q\notin \langle \nu \left({Z}^{\prime}\right)\rangle $ for any ${Z}^{\prime}\u228aZ$, $q\notin \langle \nu \left({A}^{\prime}\right)\rangle $ for any ${A}^{\prime}\u228aA$ and $Z\cap A=\varnothing $, ${h}^{1}\left({\mathcal{I}}_{{W}^{\prime}}(1,\cdots ,1)\right)\le 1$ for all ${W}^{\prime}\ne W$. Thus, $\langle \nu \left(E\right)\rangle \cap \nu \left(W\right)=E$ (scheme-theoretically). Since $\nu \left(Y\right)$ contains no plane $\langle \nu \left(A\right)\rangle \u2288\nu \left(Y\right)$. Since $\nu \left(W\right)\subset \langle \nu \left(A\right)\rangle $ and $\nu \left(Y\right)$ is scheme-theoretically cut out by quadrics, $\nu \left(Y\right)\cap \langle \nu \left(A\right)\rangle $ is the union of 2 lines. Since Y is the minimal multiprojective space containing A, we obtain $k=2$ (([29], Proposition 5.2) or ([30], Proposition 1.1) or Theorem 1), a contradiction.
- (b4.2.2)
- Assume $E=z\cup \left\{a\right\}$ and set $\left\{b\right\}:={z}_{red}$. Take $w=z$. We obtain ${\pi}_{h}\left(a\right)={\pi}_{h}\left(b\right)$ for all $h\notin \{i,{i}_{w}\}$. Varying i we obtain ${\pi}_{h}\left(a\right)={\pi}_{h}\left(b\right)$ for all $h\ne {i}_{w}$. Take $w=\{a,b\}$, but call ${i}_{{w}^{\prime}}$ the integer associated with this degree 2 scheme. We obtain ${\pi}_{h}\left(a\right)={\pi}_{h}\left(b\right)$ for all $h\ne {i}_{{w}^{\prime}}$. Thus, ${i}_{w}={i}_{{w}^{\prime}}$. Thus, $\langle \{a,b\}\rangle $ is a line contained in the w-th ruling of $\nu \left(Y\right)$. If $\nu \left(E\right)\subset \langle \{a,b\}\rangle $ we conclude as in step (b4.2.1). Assume $\nu \left(E\right)\u2288\langle \{a,b\}\rangle $. Either $z=v$ or Z is connected and $Z=z$. Take $h\ne {i}_{w}$ and $D\in |{\mathcal{I}}_{a}\left({\epsilon}_{h}\right)|$. By construction $deg(W\cap ({H}_{{i}_{w}}\cup D))=4$. Thus, ${h}^{1}\left({\mathcal{I}}_{{Res}_{{H}_{{i}_{w}}\cup D}}\right)=0$, contradicting ([33], Lemma 5.1).

**Remark**

**11.**

**Example**

**3.**

**Lemma**

**3.**

**Proof.**

**Example**

**4.**

- 1.
- ${Y}^{\prime}$ is the minimal multiprojective space containing ${u}^{\prime}$.
- 2.
- ${Y}^{\prime}$ is not the minimal multiprojective space containing ${u}^{\prime}$, i.e., $deg({\pi}_{i}\left({u}^{\prime}\right)=1$ for exactly one $i\in \{1,2\}$; in this case we require ${n}_{i}=1$ and ${\pi}_{i}\left({u}^{\prime}\right)\ne {o}_{i}$.

**Example**

**5.**

**Example**

**6.**

**Example**

**7.**

**Example**

**8.**

**Lemma**

**4.**

**Proof.**

**Lemma**

**5.**

**Proof.**

**Remark**

**12.**

- (a)
- Assume ${\eta}_{i}\left(A\right)={\eta}_{i}\left(\{o,{o}^{\prime}\}\right)$. In this case ${\eta}_{i}\left(v\right)={\eta}_{i}\left(o\right)$. We obtain that $\langle \nu \left(v\right)\rangle $ is a line contained in Y. Thus, all points of $\langle \nu \left(A\right)\rangle $ have tensor rank at most 2. This is Example 7.
- (b)
- Assume ${\eta}_{i}\left(A\right)={\eta}_{i}\left(v\right)$. In this case ${\pi}_{h}\left(u\right)={\pi}_{h}\left(o\right)$ for all $h\ne i$. Thus, $\langle \nu \left(\right\{o,u\left\}\right)\rangle $ is a line contained in the i-th ruling of the Segre $\nu \left(Y\right)$, say $\langle \nu \left(\right\{o,u\left\}\right)\rangle =\nu \left(R\right)$. For each ${u}^{\prime}\in R\setminus \left\{o\right\}$ we have $\langle \nu \left(A\right)\rangle =\langle \nu (v\cup {u}^{\prime})\rangle $. In particular uniqueness fails for each q irredundantly spanned by $\nu \left(A\right)$. This case obviously occurs both with ${n}_{i}=1$ and with ${n}_{i}=2$. This is Example 8.
- (c)
- Fix a set $S\subset Y$ such that $\#S=3$ and Y is the minimal multiprojective space containing Y. Assume the existence of $i\in \{1,\cdots ,k\}$ such that ${\eta}_{i|S}$ is not injective. By ([26], Remark 1.10) each point of $\langle \nu \left(S\right)\rangle $ has tensor rank at most 2.

**Lemma**

**6.**

**Proof.**

- (a)
- Assume for the moment that Y is not the minimal multiprojective space containing B. Thus, there is $i\in \{1,\cdots ,k\}$ and $H\in |{\mathcal{O}}_{Y}\left({\epsilon}_{i}\right)|$ such that $B\subset H$ and hence ${Res}_{H}\left(W\right)\subseteq A$. Since $A\cap B=\varnothing $, ([33], Lemma 5.1) gives ${h}^{1}\left({\mathcal{I}}_{{Res}_{H}\left(W\right)}\left({\widehat{\epsilon}}_{1}\right)\right)=0$. Since $A\subseteq {Res}_{H}\left(W\right)$, either ${\eta}_{i|A}$ is not an embedding or $A={Res}_{H}\left(W\right)$ and $deg\left({\pi}_{j}\left(A\right)\right)=1$ for $k-2$ indices j, contradicting the minimality of Y. Thus, Y is the minimal multiprojective space containing B. If there is $i\in \{1,\cdots ,k\}$ such that ${\eta}_{i|B}$ is not an embedding, we apply Remark 12 to B.
- (b)
- By assumption we are not as in Example 7 or 8 for A or B. Thus, we may assume that all ${\eta}_{i|B}$ and all ${\eta}_{i|A}$ are embeddings. There are 4 degree 2 schemes $E\subset W$ formed by one point of ${A}_{red}$ and one point of ${B}_{red}$. Since $k>4$, there is $i\in \{1,\cdots ,k\}$ such that ${\eta}_{i|W}$ is an embedding. With no loss of generality we may assume $i=1$.
- (b1)
- Assume that ${\pi}_{1|W}$ is an embedding. Take $M\in |{\mathcal{O}}_{Y}\left({\epsilon}_{1}\right)|$ containing a reduced connected component a of B. Please note that ${Res}_{M}\left(W\right)=W\setminus \left\{a\right\}$. Since $A\subset {W}^{\prime}$ and Y is the minimal multiprojective space containing Y, ${Y}_{1}$ is the minimal multiprojective space containing ${\eta}_{1}\left(A\right)\subset {\eta}_{1}\left({Res}_{M}\left(W\right)\right)$ and ${\nu}_{1}\left({\eta}_{1}\left(A\right)\right)$ is linearly independent by Remark 7. Since ${\eta}_{1|W}$ is an embedding, ${\eta}_{1|{Res}_{H}\left(W\right)}$ is an embedding. Thus, ${h}^{1}\left({\mathcal{I}}_{{Res}_{H}\left(W\right))}\left({\widehat{\epsilon}}_{1}\right)\right)={h}^{1}({Y}_{1},{\mathcal{I}}_{{\eta}_{1}({Res}_{H}\left(W\right)}(1,\cdots ,1))$. Since $A\cap B=\varnothing $ and ${\eta}_{1|W}$ is an embedding, ([33], Lemma 5.1) gives ${h}^{1}({Y}_{1},{\mathcal{I}}_{{\eta}_{1}(W\setminus \left\{a\right\})}(1,\cdots ,1))>0$. Since $k\ge 3$ and Y is the minimal multiprojective space containing Y, ${\nu}_{1}\left(A\right)$ is linearly independent by ([26], Lemma 4.4). Let ${W}_{1}\subseteq {\eta}_{1}(W\setminus \left\{a\right\})$ the minimal subscheme of ${\eta}_{1}(W\setminus \left\{a\right\})$ containing A such that ${h}^{1}({Y}_{1},{\mathcal{I}}_{{W}_{1}}(1,\cdots ,1))>0$. If $deg\left({W}_{1}\right)=4$ (resp. $deg\left({W}_{1}\right)=5$), Proposition 1 (resp. Proposition 2) gives that ${\eta}_{1}\left(A\right)$ depends on at most 2 (resp. 3) factors of ${Y}_{1}$. Thus, A depends on at most 4 factors of Y, a contradiction.
- (b2)
- Assume that ${\pi}_{1|W}$ is not an embedding. If there is $M\in |{\mathcal{O}}_{Y}\left({\epsilon}_{1}\right)|$ such that $M\cap B\ne \varnothing $ and $M\cap A=\varnothing $, then we may repeat the proof of step (b1). Since Y is the minimal multiprojective space containing B by step (a), we conclude if there is $M\in |{\mathcal{O}}_{Y}\left({\epsilon}_{1}\right)|$ such that $M\cap A\ne \varnothing $ and $M\cap B=\varnothing $. Thus, we may assume ${\pi}_{1}{\left(A\right)}_{red}={\pi}_{1}{|\left(B\right)}_{red}$. Since $A\cap B=\varnothing $ and $A\u2288H$, ([33], Lemma 5.1) gives ${h}^{1}\left({\mathcal{I}}_{{Res}_{H}\left(W\right))}\left({\widehat{\epsilon}}_{1}\right)\right)>0$. Since ${\eta}_{1|W}$ is an embedding, ${\eta}_{|{Res}_{H}\left(W\right)}$ is an embedding. Thus, ${h}^{1}\left({\mathcal{I}}_{{Res}_{H}\left(W\right))}\left({\widehat{\epsilon}}_{1}\right)\right)={h}^{1}({Y}_{1},{\mathcal{I}}_{{\eta}_{1}({Res}_{H}\left(W\right)}(1,\cdots ,1))$. Let ${W}_{1}\subseteq {\eta}_{1}\left({Res}_{H}\left(W\right)\right)$ be a minimal subscheme of such that ${h}^{1}({Y}_{1},{\mathcal{I}}_{{W}_{1}}(1,\cdots ,1))>0$. If ${W}_{1}={\eta}_{1}\left({Res}_{H}\left(W\right)\right)$, then Proposition 2 gives that ${Res}_{H}\left({W}_{1}\right)$ depends on at most 3 factors of Y, one of them being the first one.

**Example**

**9.**

**Observation**

**1:**

**Proof**

**of**

**Theorem**

**3.**

## 6. Speculations on the Higher Derivatives and Uniqueness

**Example**

**10.**

**Claim**

**1:**

**Proof**

**of**

**Claim**

**1:**

**Example**

**11.**

**Question**

**4.**

**Example**

**12.**

## Funding

## Institutional Review Board Statement

## Informed Consent Statement

## Conflicts of Interest

## References

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**MDPI and ACS Style**

Ballico, E.
Base Point Freeness, Uniqueness of Decompositions and Double Points for Veronese and Segre Varieties. *Symmetry* **2021**, *13*, 2344.
https://doi.org/10.3390/sym13122344

**AMA Style**

Ballico E.
Base Point Freeness, Uniqueness of Decompositions and Double Points for Veronese and Segre Varieties. *Symmetry*. 2021; 13(12):2344.
https://doi.org/10.3390/sym13122344

**Chicago/Turabian Style**

Ballico, Edoardo.
2021. "Base Point Freeness, Uniqueness of Decompositions and Double Points for Veronese and Segre Varieties" *Symmetry* 13, no. 12: 2344.
https://doi.org/10.3390/sym13122344