On Prime Rings with Derivations

Abstract

This paper identifies all derivations associated with two well-known non-commutative prime rings and provides some remarks on one of these derivations, called the prime derivation. Also, it presents two results for some classes of non-commutative prime rings, regarding when the images of derivations on these rings are subrings of them. The paper also includes some illustrative examples and concludes with a set of questions for further exploration of this topic.

1. Introduction

The study of the properties of prime rings has played a significant role in the study of algebra. Many researchers have studied the properties of prime rings or other algebraic structures related to them using maps defined on these structures. One of the important maps is called a derivation. Its idea comes from the well-known function, the derivative function on real-valued functions. The researchers have studied different kinds of these derivations, such as usual derivations [1,2,3,4], multiplicative derivations [5], semiderivations [6], generalized derivations [7,8,9], $\sigma$-derivations [10], ($\sigma$,$\tau$)-derivations [11], etc.

Recall that a derivation D on a ring R is an additive mapping that satisfies $D\left(xy\right)=xD\left(y\right)+D\left(x\right)y$ for all $x,y\in R$. It is known that $D\left(0\right)=0$, and if R has an identity 1, then $D\left(1\right)=0$. An inner derivation on R induced by $a\in R$ is a derivation on R defined as follows: $D\left(x\right)=ax-xa$ for all $x\in R$. For more properties of the derivations, see [1,5]. If U is a non-empty subset of a ring R, then we say that U satisfies the 3-prime condition if, for all $x,y\in R$, the equation $xUy=\left\{0\right\}$ implies $x=0$ or $y=0$ (see [3]). In the theory of rings, we know that if S is an integral domain, then $M_n\left(S\right)$, which is the set of all square matrices of order n and their entries belong to S, a non-commutative prime ring for all positive integers $n⩾2$.

Throughout this paper, R is a ring, but we focus on the prime rings. It is worth noting that the primeness of rings has equivalent definitions. We use the following definition of primeness: R is prime if and only if $rRs=\left\{0\right\}$, where $r,s\in R$, implies that $r=0$ or $s=0$. If r is a non-zero central element of a prime ring R, then it is not a zero divisor [2]. In the second section of this paper, we include a definition of prime derivations and state some of their properties. In the third section, we do three things:

(1)

Find all possible derivations on two well-known prime rings, $M_2\left({\mathbb{Z}}_2\right)$ and $M_2\left({\mathbb{Z}}_3\right)$.

(2)

Investigate the prime derivations on $M_2\left({\mathbb{Z}}_2\right)$ and $M_2\left({\mathbb{Z}}_3\right)$ and show that the first ring has only one prime derivation, while the second has six prime derivations.

(3)

Give two results on the relationship between the derivation D and ($D\left(R\right)$ as a subring of R), where $R=M_2\left(S\right)$ such that S is an integral domain.

In the final section, we list some questions we received from the previous sections.

2. Preliminaries

Assume there is a prime ring R with a derivation D. One might ask whether the condition $aD\left(R\right)b=\left\{0\right\}$ for some $a,b\in R$ implies $a=0$ or $b=0$. The following remark answers this question.

 Remark 1. 

Let R be a prime ring with a derivation D. It can be shown that if $aD\left(R\right)b=\left\{0\right\}$ for some $a,b\in R$, then

 (i) 

$a=0$, $b=0$ or $aD\left(a\right)=D\left(b\right)b=0$.

 (ii) 

$aD\left(a\right)=0$ or $D\left(b\right)b=0$.

 (iii) 

$a=0$, $b=0$, or $D\left(ba\right)=0$. Moreover, if $aD\left(R\right)a=\left\{0\right\}$ for some $a\in R$, then $a=0$ or $D\left(a\right)=0$.

In the previous remark, we cannot remove the case $aD\left(a\right)=D\left(b\right)b=0$ in (i) or $D\left(ba\right)=0$ in (iii) as shown in Example 2. However, this remark leads us to introduce the following definition.

 Definition 1. 

Let R be a ring with a non-zero derivation D. Then we say that D is a prime derivation if $aD\left(R\right)b=\left\{0\right\}$ for some $a,b\in R$ implies that $a=0$ or $b=0$.

From the above definition, the zero derivation is not prime. Therefore, any prime ring that does not have a non-zero derivation does not have a prime derivation. For example, finite fields do not have non-zero derivations [6] and therefore do not have prime derivations. We can also note that D is a prime derivation if and only if $D\left(R\right)$ satisfies the 3-prime condition.

Now, we have the following useful remark that links prime derivations to the primeness of rings.

 Remark 2. 

Let R be a ring with a non-zero derivation D:

 (i) 

If D is prime, then R is a prime ring. Hence, all non-prime rings have no prime derivations.

 (ii) 

If R is a ring without zero divisors, then D is a prime derivation.

 Example 1. 

Let $R=\mathbb{R}\left[x\right]$ the polynomial ring with real coefficients and with the usual derivative D as a derivation on it. Then D is a prime derivation on R.

From Definition 1, it is clear that if R is a ring with a prime derivation D such that $D\left(R\right)$ is a subring of R, then $D\left(R\right)$ is prime. From the previous example, it is clear that $D\left(R\right)$ is a prime subring of R. However, we are unable to find any example in this case for non-commutative rings. In the last section, We pose a question about this case. On the other hand, we can find a prime ring R with a non-zero derivation D such that $D\left(R\right)$ is a subring of R but D is not prime, as shown in Example 4 (cases 2, 4 and 6).

For commutative rings, we have the following observation.

 Remark 3. 

(i) Let R be a commutative ring with a prime derivation D. Then R is a commutative prime ring and therefore it has no non-zero zero divisors. So, the converse of Remark 2(ii) is true if R is commutative.

(ii) Let R be a commutative prime ring. Then it is without zero divisors and therefore any non-zero derivation on it (if any) is a prime derivation according to Remark 2(ii). So, for prime derivations, we only need to check for non-commutative prime rings since they always have non-zero inner derivations.

In the next definition, we give the definition of a lower identity. We need it in Example 4 in the next section.

 Definition 2. 

Let R be a ring with identity $e\in R$. Then we say that $e^{\prime }\in R$ is a lower identity if $e^{\prime }x=x{e}^{\prime }=x$ for all $x\in R-\{e,e^{\prime }\}$ and $e^{\prime }{e}^{\prime }=e$.

3. Main Results and Main Examples

We begin with an example demonstrating that the converse of Remark 2(i) is false.

 Example 2. 

Let $R=M_2\left(\mathbb{Z}\right)$, the ring of all matrices of order 2 over integer numbers. Then R is a non-commutative prime ring. Take the inner derivation D induced by $\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]$. So, for all integers $x,y,z,w$, we have

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}0& -y\\ z& 0\end{array}\right].$$

Observe that

$$\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\left[\begin{array}{cc}0& -y\\ z& 0\end{array}\right]\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

Thereby, D is not a prime derivation, and the converse of Remark 2(i) is not true. Observe that

$$\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right]D\left(\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right]\right)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=D\left(\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]\right)\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]$$

and $aD\left(a\right)=D\left(b\right)b=0$ as shown in Remark 1(i), where $a=\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right]$ and $b=\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]$. Also,

$$D\left(\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right]\right)=D\left(\left[\begin{array}{cc}2& 0\\ 0& 0\end{array}\right]\right)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$$

and $D\left(ba\right)=0$, as shown in Remark 1(iii).

On the other hand, observe that

$$\begin{array}{ccc}\hfill D\left(ab\right)& =& D\left(\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]\right)=D\left(\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]\right)\hfill \\ & =& \left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\ne \left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\hfill \end{array}$$

The following interesting example demonstrates that an inner derivation on a prime ring can be prime.

 Example 3. 

Let $R=M_2\left(\mathbb{Z}\right)$. Then R is a non-commutative prime ring. We want to define a derivation on R that is a prime derivation.

Take the inner derivation D induced by $\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]$. So, for all integers $a,b,c,d$, we have

$$\begin{array}{ccc}\hfill D\left(\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\right)& =& \left[\begin{array}{cc}c-b& b-(a-d)\\ a-d-c& -(c-b)\end{array}\right].\hfill \end{array}$$

For some integers $e,f,g,h,x,y,w,z$, suppose that

$$\begin{array}{ccc}& & \left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}e& f\\ g& h\end{array}\right]\left[\begin{array}{cc}c-b& b-(a-d)\\ a-d-c& -(c-b)\end{array}\right]\left[\begin{array}{cc}x& y\\ w& z\end{array}\right]=\left[\begin{array}{cc}r& s\\ t& u\end{array}\right]\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill r& =& x\left(e\right(c-b)+f(a-d-c\left)\right)+w\left(e\right(b-(a-d))-f(c-b\left)\right)=0\hfill \\ \hfill s& =& y\left(e\right(c-b)+f(a-d-c\left)\right)+z\left(e\right(b-(a-d))-f(c-b\left)\right)=0\hfill \\ \hfill t& =& x\left(g\right(c-b)+h(a-d-c\left)\right)+w\left(g\right(b-(a-d))-h(c-b\left)\right)=0\hfill \\ \hfill u& =& y\left(g\right(c-b)+h(a-d-c\left)\right)+z\left(g\right(b-(a-d))-h(c-b\left)\right)=0.\hfill \end{array}$$

If $a=b=c=0$ and $d=1$, then we have the following equations.

$$xf=we$$

(1)

$$yf=ze$$

(2)

$$xh=wg$$

(3)

$$yh=zg$$

(4)

and if $a=c=d=0$ and $b=1$, then we have the following equations.

$$xe=we+wf$$

(5)

$$ye=ze+zf$$

(6)

$$xg=wg+wh$$

(7)

$$yg=zg+zh$$

(8)

Firstly, assume that

$$\left[\begin{array}{cc}x& y\\ w& z\end{array}\right]\ne \left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

and suppose that $w\ne 0$ ($x\ne 0$). Multiply Equation (5) by f (by e). Using Equation (1), we get that

$$\begin{array}{ccc}\hfill w{e}^2& =& wef+w{f}^2\hfill \\ \hfill (x{e}^2& =& xef+x{f}^2)\hfill \end{array}$$

So, $f^2+ef-e^2=0$ and hence

$$f={\displaystyle \frac{-e\pm \sqrt{e^2+4e^2}}{2}}={\displaystyle \frac{-1\pm \sqrt{5}}{2}}e\in \mathbb{Z}.$$

But this is true only if $e=0$. That means $f=0$ also from the same equation.

Now, multiply Equation (7) by h (by g). Using Equation (1), we get that

$$\begin{array}{ccc}\hfill w{g}^2& =& wgh+w{h}^2\hfill \\ \hfill (x{g}^2& =& xgh+x{h}^2)\hfill \end{array}$$

and in the same way as above, we can prove that $g=h=0$.

Finally, suppose that $y\ne 0$ ($z\ne 0$). Using Equations (2), (4), (6) and (8), and in the same way as above, we can prove that $e=f=g=h=0$. Hence

$$\left[\begin{array}{cc}e& f\\ g& h\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

Thus, D is a prime derivation. Since

$$\left[\begin{array}{cc}0& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

So, R has zero divisors and the converse of Remark 2(ii) is not true.

In the following interesting example, we study all possible derivations on the ring of $2\times 2$ matrices over ${\mathbb{Z}}_2$. We also provide some observations inside the example.

 Example 4. 

Let $R=M_2\left({\mathbb{Z}}_2\right)$. Then R is a non-commutative prime ring. Let D be a derivation on R. Taking $a,b,c,d\in {\mathbb{Z}}_2$, we can assume the following

$$D\left(\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right)=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right].$$

So,

$$\begin{array}{ccc}\hfill \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]& =& D\left(\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right)=D\left(\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right)=\left[\begin{array}{cc}0& b\\ c& 0\end{array}\right].\hfill \end{array}$$

Thus,

$$D\left(\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right)=\left[\begin{array}{cc}0& b\\ c& 0\end{array}\right]\in \left\{\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right],\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\right\}.$$

Also,

$$D\left(\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\right)=D\left(\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right)+D\left(\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right)=D\left(\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right)=\left[\begin{array}{cc}0& b\\ c& 0\end{array}\right].$$

In the same way, we can assume

$$D\left(\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\right)=D\left(\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\right)=\left[\begin{array}{cc}{a}^{\prime }& b^{\prime }\\ c^{\prime }& d^{\prime }\end{array}\right],$$

where $a^{\prime },b^{\prime },c^{\prime },d^{\prime }\in {\mathbb{Z}}_2$. So,

$$\begin{array}{ccc}\hfill \left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]& =& D\left(\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\right)=\left[\begin{array}{cc}{c}^{\prime }& a^{\prime }+d^{\prime }\\ 0& c^{\prime }\end{array}\right].\hfill \end{array}$$

Hence, $c^{\prime }=0$ and $a^{\prime }=d^{\prime }$. Therefore,

$$D\left(\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\right)=D\left(\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\right)=\left[\begin{array}{cc}{a}^{\prime }& b^{\prime }\\ 0& a^{\prime }\end{array}\right]$$

Now, observe that

$$\begin{array}{ccc}\hfill \left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]& =& D\left(\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right)=\left[\begin{array}{cc}c+a^{\prime }& 0\\ 0& 0\end{array}\right].\hfill \end{array}$$

Thus, $c=a^{\prime }$. That means

$$D\left(\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\right)=D\left(\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\right)=\left[\begin{array}{cc}c& b^{\prime }\\ 0& c\end{array}\right].$$

In the same way, we can assume

$$D\left(\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\right)=D\left(\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]\right)=\left[\begin{array}{cc}{a}^{″}& b^{″}\\ c^{″}& d^{″}\end{array}\right],$$

where $a^{″},b^{″},c^{″},d^{″}\in {\mathbb{Z}}_2$. So,

$$\begin{array}{ccc}\hfill \left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]& =& D\left(\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\right)=\left[\begin{array}{cc}{b}^{″}& 0\\ a^{″}+d^{″}& b^{″}\end{array}\right].\hfill \end{array}$$

So, $b^{″}=0$ and $a^{″}=d^{″}$ and

$$D\left(\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\right)=D\left(\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]\right)=\left[\begin{array}{cc}{a}^{″}& 0\\ c^{″}& a^{″}\end{array}\right].$$

In the same way as above, observe that

$$\begin{array}{ccc}\hfill \left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]& =& D\left(\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\right)=\left[\begin{array}{cc}{a}^{″}+b& 0\\ 0& 0\end{array}\right],\hfill \end{array}$$

and $a^{″}=b$. Also,

$$\begin{array}{c}\hfill \left[\begin{array}{cc}0& b\\ c& 0\end{array}\right]=D\left(\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right)=D\left(\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\right)=\left[\begin{array}{cc}{c}^{″}+b^{\prime }& b\\ c& 0\end{array}\right],\end{array}$$

and $c^{″}=b^{\prime }$. Hence,

$$D\left(\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\right)=D\left(\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]\right)=\left[\begin{array}{cc}b& 0\\ b^{\prime }& b\end{array}\right].$$

Completing this way and rewriting $c,b^{\prime },b$ as $a,b,c$, respectively, we have eight equations:

$$D\left(\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\right)=D\left(\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$$

$$D\left(\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\right)=D\left(\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right)=\left[\begin{array}{cc}0& c\\ a& 0\end{array}\right]$$

$$D\left(\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\right)=D\left(\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\right)=\left[\begin{array}{cc}a& b\\ 0& a\end{array}\right]$$

$$D\left(\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]\right)=D\left(\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]\right)=\left[\begin{array}{cc}c& 0\\ b& c\end{array}\right]$$

$$D\left(\left[\begin{array}{cc}0& 0\\ 1& 1\end{array}\right]\right)=D\left(\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right]\right)=\left[\begin{array}{cc}c& c\\ a+b& c\end{array}\right]$$

$$D\left(\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\right)=D\left(\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]\right)=\left[\begin{array}{cc}a+c& b\\ b& a+c\end{array}\right]$$

$$D\left(\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]\right)=D\left(\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]\right)=\left[\begin{array}{cc}a+c& b+c\\ a+b& a+c\end{array}\right]$$

$$D\left(\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]\right)=D\left(\left[\begin{array}{cc}0& 1\\ 0& 1\end{array}\right]\right)=\left[\begin{array}{cc}a& b+c\\ a& a\end{array}\right]$$

Now, we try to find all the possible eight cases:

Case 1:If $a=b=c=0$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\right\},$$

and D is the zero derivation. In fact, D is the inner derivation induced by $\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$ or $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$.

Case 2:If $a=b=0,c=1$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}\alpha & \beta \\ 0& \alpha \end{array}\right]|\alpha ,\beta \in {\mathbb{Z}}_2\right\}$$

and $D\left(R\right)$ is a commutative subring of R with identity and with a lower identity, which is $\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$. Also, $D^2\left(R\right)=\left\{0\right\}$. But D is not prime since

$$\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]D\left(R\right)\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

This also shows that $D\left(R\right)$ is not a prime ring. Observe that D is the inner derivation induced by $\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$ or $\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$ and for all $x,y,z,w\in {\mathbb{Z}}_2$

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}z& x+w\\ 0& z\end{array}\right].$$

Case 3:If $a=c=0,b=1$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}0& \beta \\ \alpha & 0\end{array}\right]|\alpha ,\beta \in {\mathbb{Z}}_2\right\}$$

and D is not a prime derivation, since

$$\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]D\left(R\right)\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

Observe that $D^2\left(R\right)=D\left(R\right)$. In fact, $D\left(A\right)=A$ for all $A\in D\left(R\right)$ and D is the inner derivation induced by $\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$ or $\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]$ and for all $x,y,z,w\in {\mathbb{Z}}_2$

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}0& y\\ z& 0\end{array}\right].$$

Case 4:If $a=1,b=c=0$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}\alpha & 0\\ \beta & \alpha \end{array}\right]|\alpha ,\beta \in {\mathbb{Z}}_2\right\}$$

and D is not a prime derivation, since

$$\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]D\left(R\right)\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

But $D\left(R\right)$ is a commutative subring of R with identity. It also has a lower identity, which is $\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]$. Note that $D^2\left(R\right)=\left\{0\right\}$, while D is the inner derivation induced by $\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]$ or $\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]$ and for all $x,y,z,w\in {\mathbb{Z}}_2$

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}y& 0\\ x+w& y\end{array}\right].$$

Case 5:If $a=b=1,c=0$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}\alpha & \alpha \\ \beta & \alpha \end{array}\right]|\alpha ,\beta \in {\mathbb{Z}}_2\right\}$$

and D is not prime, since

$$\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]D\left(R\right)\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

Observe that $D\left(A\right)=A$ for all $A\in D\left(R\right)$. D is the inner derivation induced by $\left[\begin{array}{cc}0& 0\\ 1& 1\end{array}\right]$ or $\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right]$ and for all $x,y,z,w\in {\mathbb{Z}}_2$

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}y& y\\ x+z+w& y\end{array}\right].$$

Case 6:If $a=c=1,b=0$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}\alpha & \beta \\ \beta & \alpha \end{array}\right]|\alpha ,\beta \in {\mathbb{Z}}_2\right\},$$

and D is not prime, since

$$\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]D\left(R\right)\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

But $D\left(R\right)$ is a commutative subring of R with identity and with a lower identity, which is $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$. Observe that $D^2\left(R\right)=\left\{0\right\}$. In fact, D is the inner derivation induced by $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$ or $\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$ and for all $x,y,z,w\in {\mathbb{Z}}_2$

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}z+y& x+w\\ x+w& z+y\end{array}\right].$$

Case 7:If $a=0,b=c=1$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}\alpha & \beta \\ \alpha & \alpha \end{array}\right]|\alpha ,\beta \in {\mathbb{Z}}_2\right\}$$

and D is not prime, since

$$\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]D\left(R\right)\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

Observe that $D\left(A\right)=A$ for all $A\in D\left(R\right)$ and D is the inner derivation induced by $\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]$ or $\left[\begin{array}{cc}0& 1\\ 0& 1\end{array}\right]$ and for all $x,y,z,w\in {\mathbb{Z}}_2$

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}z& x+y+w\\ z& z\end{array}\right].$$

Case 8:If $a=b=c=1$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}\alpha & \alpha +\beta \\ \beta & \alpha \end{array}\right]|\alpha ,\beta \in {\mathbb{Z}}_2\right\}$$

and D is a prime derivation that is the only one. Indeed, suppose that $a_o,b_o,c_o,d_o,a^{\prime },b^{\prime },c^{\prime },d^{\prime }\in {\mathbb{Z}}_2$ such that

$$\left[\begin{array}{cc}{a}_o& b_o\\ c_o& d_o\end{array}\right]D\left(R\right)\left[\begin{array}{cc}{a}^{\prime }& b^{\prime }\\ c^{\prime }& d^{\prime }\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$$

So,

$$\begin{array}{ccc}\hfill \left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]& =& \left[\begin{array}{cc}{a}_o& b_o\\ c_o& d_o\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}{a}^{\prime }& b^{\prime }\\ c^{\prime }& d^{\prime }\end{array}\right]\hfill \\ & =& \left[\begin{array}{cc}{b}_o{a}^{\prime }+a_o{c}^{\prime }& b_o{b}^{\prime }+a_o{d}^{\prime }\\ d_o{a}^{\prime }+c_o{c}^{\prime }& d_o{b}^{\prime }+c_o{d}^{\prime }\end{array}\right]\hfill \end{array}$$

(9)

$$\begin{array}{ccc}\hfill \left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]& =& \left[\begin{array}{cc}{a}_o& b_o\\ c_o& d_o\end{array}\right]\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]\left[\begin{array}{cc}{a}^{\prime }& b^{\prime }\\ c^{\prime }& d^{\prime }\end{array}\right]\hfill \\ & =& \left[\begin{array}{cc}{a}_o{a}^{\prime }+b_o{a}^{\prime }+b_o{c}^{\prime }& a_o{b}^{\prime }+b_o{b}^{\prime }+b_o{d}^{\prime }\\ c_o{a}^{\prime }+d_o{a}^{\prime }+d_o{c}^{\prime }& c_o{b}^{\prime }+d_o{b}^{\prime }+d_o{d}^{\prime }\end{array}\right]\hfill \end{array}$$

(10)

$$\begin{array}{ccc}\hfill \left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]& =& \left[\begin{array}{cc}{a}_o& b_o\\ c_o& d_o\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}{a}^{\prime }& b^{\prime }\\ c^{\prime }& d^{\prime }\end{array}\right]\hfill \\ & =& \left[\begin{array}{cc}{a}_o{a}^{\prime }+a_o{c}^{\prime }+b_o{c}^{\prime }& a_o{b}^{\prime }+a_o{d}^{\prime }+b_o{d}^{\prime }\\ c_o{a}^{\prime }+c_o{c}^{\prime }+d_o{c}^{\prime }& c_o{b}^{\prime }+c_o{d}^{\prime }+d_o{d}^{\prime }\end{array}\right]\hfill \end{array}$$

(11)

Now, suppose $a_o=1$. From (11), $0=a^{\prime }+c^{\prime }+b_o{c}^{\prime }$. Then, either $a^{\prime }=c^{\prime }=1$ and $b_o=0$, or $a^{\prime }=0$ and $c^{\prime }=b_o=1$, or $a^{\prime }=c^{\prime }=0$. In the first two cases, the equation $b_o{a}^{\prime }+a_o{c}^{\prime }=0$ from (9) gives $1=0$, which is a contradiction. So, $a^{\prime }=c^{\prime }=0$. Also, we have from (11) that $b^{\prime }+d^{\prime }+b_o{d}^{\prime }=0$. Using the equation $b_o{b}^{\prime }+a_o{d}^{\prime }=0$ from (9) and in the same way as above, we conclude that $b^{\prime }=d^{\prime }=0$. In the same way, using the other equations in (11) and the other equations in (9), we can show that if $c_o=1$, then $a^{\prime }=b^{\prime }=c^{\prime }=d^{\prime }=0$. Using the same technique and Equation (10) and the same equations in (9), we can show that if $b_0=1$ or $d_o=1$, then $a^{\prime }=b^{\prime }=c^{\prime }=d^{\prime }=0$. Therefore, D is prime.

But $D\left(R\right)$ is not a subring of R because

$$\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\notin D\left(R\right).$$

Observe that $D\left(A\right)=A$ for all $A\in D\left(R\right)$ and D is the inner derivation induced by $\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]$ or $\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]$, and for all $x,y,z,w\in {\mathbb{Z}}_2$

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}z+y& x+y+w\\ x+z+w& z+y\end{array}\right].$$

Also, $A^2=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ for all $A\in D\left(R\right)-\left\{0\right\}$.

In the same way as Example 4, we study all possible derivations on the ring of $2\times 2$ matrices over ${\mathbb{Z}}_3$ in the following example.

 Example 5. 

Let $R=M_2\left({\mathbb{Z}}_3\right)$. Then R is a non-commutative prime ring. Let D be a derivation on R. In the same way as in Example 4, we get

$$D\left(R\right)=\left\{\begin{array}{c}\left[\begin{array}{cc}0& b\\ c& 0\end{array}\right],\left[\begin{array}{cc}c& a\\ 0& -c\end{array}\right],\left[\begin{array}{cc}b& 0\\ -a& -b\end{array}\right],\\ \left[\begin{array}{cc}b& b\\ c-a& -b\end{array}\right],\left[\begin{array}{cc}c& a-b\\ -c& -c\end{array}\right],\left[\begin{array}{cc}b+c& a\\ -a& -(b+c)\end{array}\right],\left[\begin{array}{cc}b& -b\\ -(a+c)& -b\end{array}\right],\\ \left[\begin{array}{cc}c& a+b\\ c& -c\end{array}\right],\left[\begin{array}{cc}b+c& a+b\\ c-a& -(b+c)\end{array}\right],\left[\begin{array}{cc}b+c& a-b\\ -(a+c)& -(b+c)\end{array}\right],\\ \left[\begin{array}{cc}c-b& a\\ a& b-c\end{array}\right],\left[\begin{array}{cc}b-c& b-a\\ c-a& c-b\end{array}\right],\left[\begin{array}{cc}b-c& -(a+b)\\ -(a+c)& c-b\end{array}\right],\\ \mathrm{where}a,b,c\in {\mathbb{Z}}_3\end{array}\right\}$$

So, we get the following fourteen cases:

Case 1:If $a=b=c=0$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\right\},$$

and D is the zero derivation, which is the inner derivation induced by $\left[\begin{array}{cc}\alpha & 0\\ 0& \alpha \end{array}\right]$, where $\alpha \in {\mathbb{Z}}_3$.

Case 2:If $a=b=0$ and $c=1$ or $c=2$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}\beta & 0\\ \alpha & -\beta \end{array}\right]|\alpha ,\beta \in {\mathbb{Z}}_3\right\}$$

and D is not prime, since

$$\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]D\left(R\right)\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

Observe that D is the inner derivation induced by $\left[\begin{array}{cc}\beta & 0\\ \alpha & \beta \end{array}\right]$, where $\alpha \in \{1,2\}$ and $\beta \in {\mathbb{Z}}_3$. So, for all $x,y,z,w\in {\mathbb{Z}}_3$

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}-\alpha y& 0\\ \alpha (x-w)& \alpha y\end{array}\right]$$

Case 3:If $a=c=0$ and $b=1$ or $b=2$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}\beta & \alpha \\ 0& -\beta \end{array}\right]|\alpha ,\beta \in {\mathbb{Z}}_3\right\}$$

and D is not a prime derivation, since

$$\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]D\left(R\right)\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

In fact, D is the inner derivation induced by $\left[\begin{array}{cc}\beta & \alpha \\ 0& \beta \end{array}\right]$, where $\alpha \in \{1,2\}$ and $\beta \in {\mathbb{Z}}_3$. So, for all $x,y,z,w\in {\mathbb{Z}}_3$

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}\alpha z& \alpha (w-x)\\ 0& -\alpha z\end{array}\right].$$

Case 4:If $a=1$ or $a=2$ and $b=c=0$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}0& \beta \\ \alpha & 0\end{array}\right]|\alpha ,\beta \in {\mathbb{Z}}_3\right\}$$

and D is not a prime derivation, since

$$\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]D\left(R\right)\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

Notice that D is the inner derivation induced by $\left[\begin{array}{cc}\alpha +\beta & 0\\ 0& \beta \end{array}\right]$, where $\alpha \in \{1,2\}$ and $\beta \in {\mathbb{Z}}_3$. So, for all $x,y,z,w\in {\mathbb{Z}}_3$

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}0& \alpha y\\ -\alpha z& 0\end{array}\right]$$

Case 5:If $b=c=1$ or $b=c=2$ and $a=0$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}\alpha & \beta \\ \beta & -\alpha \end{array}\right]|\alpha ,\beta \in {\mathbb{Z}}_3\right\}$$

and D is a prime derivation. To see this, suppose that $a_o,b_o,c_o,d_o,a^{\prime },b^{\prime },c^{\prime },d^{\prime }\in {\mathbb{Z}}_3$, such that

$$\left[\begin{array}{cc}{a}_o& b_o\\ c_o& d_o\end{array}\right]D\left(R\right)\left[\begin{array}{cc}{a}^{\prime }& b^{\prime }\\ c^{\prime }& d^{\prime }\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$$

So,

$$\begin{array}{ccc}\hfill \left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]& =& \left[\begin{array}{cc}{a}_o& b_o\\ c_o& d_o\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 2\end{array}\right]\left[\begin{array}{cc}{a}^{\prime }& b^{\prime }\\ c^{\prime }& d^{\prime }\end{array}\right]\hfill \\ & =& \left[\begin{array}{cc}{a}_o{a}^{\prime }+2b_o{c}^{\prime }& a_o{b}^{\prime }+2b_o{d}^{\prime }\\ c_o{a}^{\prime }+2d_o{c}^{\prime }& c_o{b}^{\prime }+2d_o{d}^{\prime }\end{array}\right]\hfill \end{array}$$

(12)

$$\begin{array}{ccc}\hfill \left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]& =& \left[\begin{array}{cc}{a}_o& b_o\\ c_o& d_o\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}{a}^{\prime }& b^{\prime }\\ c^{\prime }& d^{\prime }\end{array}\right]\hfill \\ & =& \left[\begin{array}{cc}{b}_o{a}^{\prime }+a_o{c}^{\prime }& b_o{b}^{\prime }+a_o{d}^{\prime }\\ d_o{a}^{\prime }+c_o{c}^{\prime }& d_o{b}^{\prime }+c_o{d}^{\prime }\end{array}\right]\hfill \end{array}$$

(13)

Using (12), we have the following four equations

$$\begin{array}{ccc}\hfill a_o{a}^{\prime }& =& b_o{c}^{\prime }\hfill \\ \hfill a_o{b}^{\prime }& =& b_o{d}^{\prime }\hfill \\ \hfill c_o{a}^{\prime }& =& d_o{c}^{\prime }\hfill \\ \hfill c_o{b}^{\prime }& =& d_o{d}^{\prime }\hfill \end{array}$$

From (13), we have $b_o{a}^{\prime }+a_o{c}^{\prime }=0$. Multiplying by $a^{\prime }$, we get $b_o{\left(a^{\prime }\right)}^2+a_o{a}^{\prime }{c}^{\prime }=0$. Using the last four equations, $b_o{\left(a^{\prime }\right)}^2+b_o{\left(c^{\prime }\right)}^2=0$ and hence ${\left(a^{\prime }\right)}^2=-{\left(c^{\prime }\right)}^2$ or $b_o=0$. In the same way, we can get that from $b_o{b}^{\prime }+a_o{d}^{\prime }=0$ that $b_o{\left(b^{\prime }\right)}^2+a_o{b}^{\prime }{d}^{\prime }=0$ and then from the four equations we have that $b_o{\left(b^{\prime }\right)}^2+b_o{\left(d^{\prime }\right)}^2=0$ and hence ${\left(b^{\prime }\right)}^2=-{\left(d^{\prime }\right)}^2$ or $b_o=0$. Thus, if $b_o\ne 0$, then ${\left(a^{\prime }\right)}^2=-{\left(c^{\prime }\right)}^2$ and ${\left(b^{\prime }\right)}^2=-{\left(d^{\prime }\right)}^2$ which implies that $a^{\prime }=c^{\prime }=0$ and $b^{\prime }=d^{\prime }=0$. Now, suppose $a_o\ne 0$ and multiplying the equations $b_o{a}^{\prime }+a_o{c}^{\prime }=0$ and $b_o{b}^{\prime }+a_o{d}^{\prime }=0$ by $c^{\prime }$ and $d^{\prime }$, respectively, gives us $b_o{c}^{\prime }{a}^{\prime }+a_o{\left(c^{\prime }\right)}^2=0$ and $b_o{d}^{\prime }{b}^{\prime }+a_o{\left(d^{\prime }\right)}^2=0$. Using the four equations above, we have $a_o{\left(a^{\prime }\right)}^2+a_o{\left(c^{\prime }\right)}^2=0$ and $a_o{\left(b^{\prime }\right)}^2+a_o{\left(d^{\prime }\right)}^2=0$. Since $a_o\ne 0$, we conclude that $a^{\prime }=c^{\prime }=0$ and $b^{\prime }=d^{\prime }=0$ in the same way as above. Using the same technique, we can show that if $c_o\ne 0$ or $d_o\ne 0$, then $a^{\prime }=c^{\prime }=0$ and $b^{\prime }=d^{\prime }=0$. Therefore, D is prime.

But $D\left(R\right)$ is not a ring, since

$$\left[\begin{array}{cc}1& 2\\ 2& 2\end{array}\right]\left[\begin{array}{cc}1& 1\\ 1& 2\end{array}\right]=\left[\begin{array}{cc}0& 2\\ 1& 0\end{array}\right]\notin D\left(R\right).$$

Observe that D is the inner derivation induced by $\left[\begin{array}{cc}\beta & \alpha \\ -\alpha & \beta \end{array}\right]$, where $\alpha \in \{1,2\}$ and $\beta \in {\mathbb{Z}}_3$. So, for all $x,y,z,w\in {\mathbb{Z}}_3$

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}\alpha (y+z)& \alpha (w-x)\\ \alpha (w-x)& -\alpha (y+z)\end{array}\right].$$

Case 6:If $a=c=1$ or $a=c=2$ and $b=0$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}\beta & \beta \\ \alpha & -\beta \end{array}\right]|\alpha ,\beta \in {\mathbb{Z}}_3\right\}$$

and D is not prime, since

$$\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]D\left(R\right)\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

In fact, D is the inner derivation induced by $\left[\begin{array}{cc}\beta & 0\\ \alpha & \alpha +\beta \end{array}\right]$, where $\alpha \in \{1,2\}$ and $\beta \in {\mathbb{Z}}_3$. So, for all $x,y,z,w\in {\mathbb{Z}}_3$

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}-\alpha y& -\alpha y\\ \alpha (x+z-w)& \alpha y\end{array}\right].$$

Case 7:If $a=b=1$ or $a=b=2$ and $c=0$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}\beta & \alpha \\ -\beta & -\beta \end{array}\right]|\alpha ,\beta \in {\mathbb{Z}}_3\right\}$$

and D is not prime since

$$\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]D\left(R\right)\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

We find that D is the inner derivation induced by $\left[\begin{array}{cc}\alpha +\beta & \alpha \\ 0& \beta \end{array}\right]$, where $\alpha \in \{1,2\}$ and $\beta \in {\mathbb{Z}}_3$. So, for all $x,y,z,w\in {\mathbb{Z}}_3$

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}\alpha z& \alpha (y+w-x)\\ -\alpha z& -\alpha z\end{array}\right].$$

Case 8:If $a=1,b=2$ or $a=2,b=1$ and $c=0$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}\beta & \alpha \\ \beta & -\beta \end{array}\right]|\alpha ,\beta \in {\mathbb{Z}}_3\right\}$$

and D is not prime, since

$$\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]D\left(R\right)\left[\begin{array}{cc}1& 2\\ 1& 2\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

D is also the inner derivation induced by $\left[\begin{array}{cc}\beta & \alpha \\ 0& \alpha +\beta \end{array}\right]$, where $\alpha \in \{1,2\}$ and $\beta \in {\mathbb{Z}}_3$. So, for all $x,y,z,w\in {\mathbb{Z}}_3$

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}\alpha z& \alpha (w-x-y)\\ \alpha z& -\alpha z\end{array}\right].$$

Case 9:If $a=1,c=2$ or $a=2,c=1$ and $b=0$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}\beta & -\beta \\ \alpha & -\beta \end{array}\right]|\alpha ,\beta \in {\mathbb{Z}}_3\right\}$$

and D is not prime, since

$$\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]D\left(R\right)\left[\begin{array}{cc}1& 2\\ 1& 2\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

Indeed, D is the inner derivation induced by $\left[\begin{array}{cc}\alpha +\beta & 0\\ \alpha & \beta \end{array}\right]$, where $\alpha \in \{1,2\}$ and $\beta \in {\mathbb{Z}}_3$. So, for all $x,y,z,w\in {\mathbb{Z}}_3$

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}-\alpha y& \alpha y\\ \alpha (x-z-w)& \alpha y\end{array}\right].$$

Case 10:If $b=1,c=2$ or $b=2,c=1$ and $a=0$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}\alpha & \beta \\ -\beta & -\alpha \end{array}\right]|\alpha ,\beta \in {\mathbb{Z}}_3\right\}$$

and D is not prime, since

$$\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right]D\left(R\right)\left[\begin{array}{cc}1& 2\\ 1& 2\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

But D is the inner derivation induced by $\left[\begin{array}{cc}\beta & \alpha \\ \alpha & \beta \end{array}\right]$, where $\alpha \in \{1,2\}$ and $\beta \in {\mathbb{Z}}_3$. So, for all $x,y,z,w\in {\mathbb{Z}}_3$

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}-\alpha (y-z)& -\alpha (x-w)\\ \alpha (x-w)& \alpha (y-z)\end{array}\right].$$

Case 11:If $a=b=c=1$ or $a=b=c=2$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}\beta & \alpha +\beta \\ \alpha & -\beta \end{array}\right]|\alpha ,\beta \in {\mathbb{Z}}_3\right\}$$

and D is not prime, since

$$\left[\begin{array}{cc}1& 2\\ 1& 2\end{array}\right]D\left(R\right)\left[\begin{array}{cc}1& 2\\ 1& 2\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

Notice that D is the inner derivation induced by $\left[\begin{array}{cc}\alpha +\beta & \alpha \\ -\alpha & \beta \end{array}\right]$, where $\alpha \in \{1,2\}$ and $\beta \in {\mathbb{Z}}_3$. So, for all $x,y,z,w\in {\mathbb{Z}}_3$

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}\alpha (y+z)& \alpha (-x+y+w)\\ \alpha (-x-z+w)& -\alpha (y+z)\end{array}\right].$$

Case 12:If $a=b=1$ and $c=2$ or $a=b=2$ and $c=1$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}\beta & -\alpha -\beta \\ \alpha & -\beta \end{array}\right]|\alpha ,\beta \in {\mathbb{Z}}_3\right\}$$

and D is prime. To see this, suppose that $a_o,b_o,c_o,d_o,a^{\prime },b^{\prime },c^{\prime },d^{\prime }\in {\mathbb{Z}}_3$, such that

$$\left[\begin{array}{cc}{a}_o& b_o\\ c_o& d_o\end{array}\right]D\left(R\right)\left[\begin{array}{cc}{a}^{\prime }& b^{\prime }\\ c^{\prime }& d^{\prime }\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$$

So,

$$\begin{array}{ccc}\hfill \left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]& =& \left[\begin{array}{cc}{a}_o& b_o\\ c_o& d_o\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 2\end{array}\right]\left[\begin{array}{cc}{a}^{\prime }& b^{\prime }\\ c^{\prime }& d^{\prime }\end{array}\right]\hfill \\ & =& \left[\begin{array}{cc}{a}_o{a}^{\prime }+2a_o{c}^{\prime }+2b_o{c}^{\prime }& a_o{b}^{\prime }+2a_o{d}^{\prime }+2b_o{d}^{\prime }\\ c_o{a}^{\prime }+2c_o{c}^{\prime }+2d_o{c}^{\prime }& c_o{b}^{\prime }+2c_o{d}^{\prime }+2d_o{d}^{\prime }\end{array}\right]\hfill \end{array}$$

(14)

$$\begin{array}{ccc}\hfill \left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]& =& \left[\begin{array}{cc}{a}_o& b_o\\ c_o& d_o\end{array}\right]\left[\begin{array}{cc}1& 0\\ 2& 2\end{array}\right]\left[\begin{array}{cc}{a}^{\prime }& b^{\prime }\\ c^{\prime }& d^{\prime }\end{array}\right]\hfill \\ & =& \left[\begin{array}{cc}{a}_o{a}^{\prime }+2b_o{a}^{\prime }+2b_o{c}^{\prime }& a_o{b}^{\prime }+2b_o{b}^{\prime }+2b_o{d}^{\prime }\\ c_o{a}^{\prime }+2d_o{a}^{\prime }+2d_o{c}^{\prime }& c_o{b}^{\prime }+2d_o{b}^{\prime }+2d_o{d}^{\prime }\end{array}\right]\hfill \end{array}$$

(15)

$$\begin{array}{ccc}\hfill \left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]& =& \left[\begin{array}{cc}{a}_o& b_o\\ c_o& d_o\end{array}\right]\left[\begin{array}{cc}0& 1\\ 2& 0\end{array}\right]\left[\begin{array}{cc}{a}^{\prime }& b^{\prime }\\ c^{\prime }& d^{\prime }\end{array}\right]\hfill \\ & =& \left[\begin{array}{cc}2b_o{a}^{\prime }+a_o{c}^{\prime }& 2b_o{b}^{\prime }+a_o{d}^{\prime }\\ 2d_o{a}^{\prime }+c_o{c}^{\prime }& 2d_o{b}^{\prime }+c_o{d}^{\prime }\end{array}\right]\hfill \end{array}$$

(16)

Using (16), we have the following equations, which we call the four equations

$$\begin{array}{ccc}\hfill a_o{c}^{\prime }& =& b_o{a}^{\prime }\hfill \\ \hfill a_o{d}^{\prime }& =& b_o{b}^{\prime }\hfill \\ \hfill c_o{c}^{\prime }& =& d_o{a}^{\prime }\hfill \\ \hfill c_o{d}^{\prime }& =& d_o{b}^{\prime }\hfill \end{array}$$

Also, from (14)

$$\begin{array}{ccc}\hfill a_o{a}^{\prime }& =& a_o{c}^{\prime }+b_o{c}^{\prime }\hfill \\ \hfill a_o{b}^{\prime }& =& a_o{d}^{\prime }+b_o{d}^{\prime }\hfill \\ \hfill c_o{a}^{\prime }& =& c_o{c}^{\prime }+d_o{c}^{\prime }\hfill \\ \hfill c_o{b}^{\prime }& =& c_o{d}^{\prime }+d_o{d}^{\prime }\hfill \end{array}$$

Suppose $a_o\ne 0$. Multiplying $a_o{a}^{\prime }=a_o{c}^{\prime }+b_o{c}^{\prime }$ by $a^{\prime }$, we have $a_o{\left(a^{\prime }\right)}^2=a_o{a}^{\prime }{c}^{\prime }+b_o{a}^{\prime }{c}^{\prime }$. Using the four equations, $a_o{\left(a^{\prime }\right)}^2=a_o{a}^{\prime }{c}^{\prime }+a_o{\left(c^{\prime }\right)}^2$. Thus, ${\left(a^{\prime }\right)}^2=a^{\prime }{c}^{\prime }+{\left(c^{\prime }\right)}^2$. If $a^{\prime }$ is non-zero, then ${\left(a^{\prime }\right)}^2=1$ and hence $a^{\prime }{c}^{\prime }+{\left(c^{\prime }\right)}^2=1$. For all values of $c^{\prime }$, the last equation is impossible. So, $a^{\prime }=0$ and hence, $c^{\prime }=0$. Multiplying $a_o{b}^{\prime }=a_o{d}^{\prime }+b_o{d}^{\prime }$ by $b^{\prime }$, we get $a_o{\left(b^{\prime }\right)}^2=a_o{b}^{\prime }{d}^{\prime }+b_o{b}^{\prime }{d}^{\prime }$ and using the four equations, $a_o{\left(b^{\prime }\right)}^2=a_o{b}^{\prime }{d}^{\prime }+a_o{\left(d^{\prime }\right)}^2$. Thus, ${\left(b^{\prime }\right)}^2=b^{\prime }{d}^{\prime }+{\left(d^{\prime }\right)}^2$ and in the same way as above, we get $b^{\prime }=d^{\prime }=0$. Using the same technique, we can show that if $c_o$ is non-zero, then $a^{\prime }=b^{\prime }=c^{\prime }=d^{\prime }=0$. For $b_o$ and $d_o$, we use (15) to obtain the following equations

$$\begin{array}{ccc}\hfill a_o{a}^{\prime }& =& b_o{a}^{\prime }+b_o{c}^{\prime }\hfill \\ \hfill a_o{b}^{\prime }& =& b_o{b}^{\prime }+b_o{d}^{\prime }\hfill \\ \hfill c_o{a}^{\prime }& =& d_o{a}^{\prime }+d_o{c}^{\prime }\hfill \\ \hfill c_o{b}^{\prime }& =& d_o{b}^{\prime }+d_o{d}^{\prime }\hfill \end{array}$$

Suppose $b_o\ne 0$. Multiplying $a_o{a}^{\prime }=b_o{a}^{\prime }+b_o{c}^{\prime }$ by $c^{\prime }$, yields $a_o{a}^{\prime }{c}^{\prime }=b_o{a}^{\prime }{c}^{\prime }+b_o{\left(c^{\prime }\right)}^2$ and using the four equations, $b_o{\left(a^{\prime }\right)}^2=b_o{a}^{\prime }{c}^{\prime }+b_o{\left(c^{\prime }\right)}^2$. Thus, ${\left(a^{\prime }\right)}^2=a^{\prime }{c}^{\prime }+{\left(c^{\prime }\right)}^2$ and by completing the same method mentioned above, we conclude $a^{\prime }=b^{\prime }=c^{\prime }=d^{\prime }=0$. We can do the same for $d_o$. Therefore, D is prime.

But $D\left(R\right)$ is not a subring of R since

$$\left[\begin{array}{cc}1& 2\\ 0& 2\end{array}\right]\left[\begin{array}{cc}2& 2\\ 2& 1\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1& 2\end{array}\right]\notin D\left(R\right).$$

Observe that D is the inner derivation induced by $\left[\begin{array}{cc}\alpha +\beta & \alpha \\ \alpha & \beta \end{array}\right]$, where $\alpha \in \{1,2\}$ and $\beta \in {\mathbb{Z}}_3$. So, for all $x,y,z,w\in {\mathbb{Z}}_3$

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}-\alpha (y-z)& -\alpha (x-y-w)\\ \alpha (x-z-w)& \alpha (y-z)\end{array}\right].$$

Case 13:If $a=c=1$ and $b=2$ or $a=c=2$ and $b=1$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}\beta & \beta -\alpha \\ \alpha & -\beta \end{array}\right]|\alpha ,\beta \in {\mathbb{Z}}_3\right\}$$

and D is prime using the same technique mentioned in case 12. But $D\left(R\right)$ is not a subring of R since

$$\left[\begin{array}{cc}1& 1\\ 0& 2\end{array}\right]\left[\begin{array}{cc}2& 0\\ 2& 1\end{array}\right]=\left[\begin{array}{cc}1& 1\\ 1& 2\end{array}\right]\notin D\left(R\right).$$

D is the inner derivation induced by $\left[\begin{array}{cc}\beta & \alpha \\ \alpha & \alpha +\beta \end{array}\right]$, where $\alpha \in \{1,2\}$ and $\beta \in {\mathbb{Z}}_3$. So, for all $x,y,z,w\in {\mathbb{Z}}_3$

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}-\alpha (y-z)& -\alpha (x+y-w)\\ \alpha (x+z-w)& \alpha (y-z)\end{array}\right].$$

Case 14:If $a=2$ and $b=c=1$ or $a=1$ and $b=c=2$, then

$$D\left(R\right)=\left\{\left[\begin{array}{cc}\beta & \alpha -\beta \\ \alpha & -\beta \end{array}\right]|\alpha ,\beta \in {\mathbb{Z}}_3\right\}$$

and D is not prime, since

$$\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right]D\left(R\right)\left[\begin{array}{cc}1& 1\\ 2& 2\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$$

We can see that D is the inner derivation induced by $\left[\begin{array}{cc}\alpha +\beta & -\alpha \\ \alpha & \beta \end{array}\right]$, where $\alpha \in \{1,2\}$ and $\beta \in {\mathbb{Z}}_3$. So, for all $x,y,z,w\in {\mathbb{Z}}_3$

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}-\alpha (y+z)& \alpha (x+y-w)\\ \alpha (x-z-w)& \alpha (y+z)\end{array}\right].$$

Finally, observe that we have two inner derivations for each case from 2 to 14. So, the total is 27 inner derivations and six of them are prime derivations.

In Example 5, we could not show that $D\left(R\right)$ is a subring of R for all cases. In fact, we have the following result.

 Theorem 1. 

Let S be an integral domain such that $char\left(S\right)\ne 2$ and $R=M_2\left(S\right)$ which has a non-zero inner derivation D, then $D\left(R\right)$ is not a subring of R.

 Proof. 

Let D be the inner derivation on R induced by $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$. For all $x,y,z,w\in S$, we have

$$\begin{array}{c}\hfill D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}bz-cy& b(w-x)+(a-d)y\\ -c(w-x)+(a-d)(-z)& -(bz-cy)\end{array}\right]\end{array}$$

So,

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}r& 0\\ 0& r\end{array}\right]$$

where $r={\left(bz-cy\right)}^2-\left(bw+ay-bx-dy\right)\left(cw+az-cx-dz\right)$. If $b\ne 0$, then choose $z=1$ and $x=y=w=0$. So, $r=b^2\ne 0$. If $c\ne 0$, then choose $y=1$ and $x=z=w=0$ and, hence $r=c^2\ne 0$. So, in both cases, we can find values of x, y, z and w such that $r\ne 0$. Since $char\left(S\right)\ne 2$, $r\ne -r$ in this case and

$$\left[\begin{array}{cc}r& 0\\ 0& r\end{array}\right]\ne \left[\begin{array}{cc}r& 0\\ 0& -r\end{array}\right]$$

Hence,

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)\notin D\left(R\right)$$

in general and $D\left(R\right)$ is not a subring of R. Now, if $b=c=0$, then

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}0& (a-d)y\\ -(a-d)z& 0\end{array}\right]$$

In general, observe that

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}-yz{\left(a-d\right)}^2& 0\\ 0& -yz{\left(a-d\right)}^2\end{array}\right]\notin D\left(R\right)$$

except $a=d$. But for $a=d$ and $b=c=0$, we have $D=0$, a contradiction. So, in all cases $D\left(R\right)$ is not a subring of R. □

The next result discusses when $D\left(R\right)$ is a subring of $R=M_2\left(S\right)$, where S is an integral domain such that $char\left(S\right)=2$ and D is a non-zero inner derivation on R. Moreover, the result shows that if S is finite and $D\left(R\right)$ is a subring of R, then D is not prime.

 Theorem 2. 

Let S be an integral domain of the characteristic 2 and $R=M_2\left(S\right)$, which has a non-zero inner derivation D induced by $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$. Then $D\left(R\right)$ is a subring of R if and only if $a=d$. In this case, if S is finite, then D is not prime.

Proof. 

Suppose $D\left(R\right)$ is a subring of R. So, it is closed under multiplication. For all $x,y,z,w\in S$, we have

$$\begin{array}{c}\hfill D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}bz+cy& b(w+x)+(a+d)y\\ c(w+x)+(a+d)z& bz+cy\end{array}\right]\end{array}$$

Thus, $D\left(\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]\right)=\left[\begin{array}{cc}{s}_o& r_o\\ r_o& s_o\end{array}\right]$, where $r_o=a+d$ and $s_o=b+c$. Now,

$$\begin{array}{ccc}& & D\left(\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]\right)D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)\hfill \\ & =& \left[\begin{array}{cc}\left(z{r}_o+c\left(w+x\right)\right)r_o+\left(bz+cy\right)s_o& \left(y{r}_o+b\left(w+x\right)\right)s_o+\left(bz+cy\right)r_o\\ \left(z{r}_o+c\left(w+x\right)\right)s_o+\left(bz+cy\right)r_o& \left(y{r}_o+b\left(w+x\right)\right)r_o+\left(bz+cy\right)s_o\end{array}\right]\hfill \end{array}$$

But $D\left(\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]\right)D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)\in D\left(R\right)$. Thus, for all $x,y,z,w\in S$,

$$\begin{array}{ccc}\hfill \left(z{r}_o+c\left(w+x\right)\right)r_o+\left(bz+cy\right)s_o& =& \left(y{r}_o+b\left(w+x\right)\right)r_o+\left(bz+cy\right)s_o\hfill \\ & ⟹& r_o((z+y)r_o+\left(w+x\right)s_o)=0\hfill \end{array}$$

This means $r_o=0$ or $(z+y)r_o=\left(w+x\right)s_o$ for all $x,y,z,w\in S$. If $z=1$ and $x=y=w=0$, then $r_o=0$. So, in all cases $r_o=0$ and then $a=d$.

Conversely, suppose that $a=d$. Therefore,

$$D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)=\left[\begin{array}{cc}bz+cy& b(w+x)\\ c(w+x)& bz+cy\end{array}\right]$$

for all $x,y,z,w\in S$. Observe that

$$\begin{array}{c}\hfill D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)D\left(\left[\begin{array}{cc}{x}^{\prime }& y^{\prime }\\ z^{\prime }& w^{\prime }\end{array}\right]\right)=\left[\begin{array}{cc}b{z}^{″}+c{y}^{″}& b(w^{″}+x^{″})\\ c(w^{″}+x^{″})& b{z}^{″}+c{y}^{″}\end{array}\right]=D\left(\left[\begin{array}{cc}{x}^{″}& y^{″}\\ z^{″}& w^{″}\end{array}\right]\right)\end{array}$$

where

$$\begin{array}{ccc}\hfill z^{″}& =& z\left(b{z}^{\prime }+c{y}^{\prime }\right)+c\left(w+x\right)\left(w^{\prime }+x^{\prime }\right)\hfill \\ \hfill y^{″}& =& y\left(b{z}^{\prime }+c{y}^{\prime }\right)\hfill \\ \hfill w^{″}& =& \left(w+x\right)\left(b{z}^{\prime }+c{y}^{\prime }\right)\hfill \\ \hfill x^{″}& =& \left(bz+cy\right)\left(w^{\prime }+x^{\prime }\right)\hfill \end{array}$$

Thus, $D\left(R\right)D\left(R\right)\subseteq D\left(R\right)$. Since $D\left(A\right)-D\left(B\right)=D(A-B)$ for all $A,B\in R$, so $D\left(R\right)$ is a subring of R.

Now, assume that $D\left(R\right)$ is a subring of R. We need to check the primeness of D. Firstly, suppose that $b=c$. Observe that

$$\begin{array}{ccc}& & \left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]\hfill \\ & =& \left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]\left[\begin{array}{cc}bz+by& b(w+x)\\ b(w+x)& bz+by\end{array}\right]\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\hfill \end{array}$$

and D is not prime. If $b=0$, then

$$\begin{array}{ccc}& & \left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right]D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)\left[\begin{array}{cc}0& 0\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\hfill \end{array}$$

and if $c=0$, then

$$\begin{array}{ccc}& & \left[\begin{array}{cc}0& 1\\ 0& 1\end{array}\right]D\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\hfill \end{array}$$

Now, If S is finite, then it is a finite field which is the Galois field $S=F_{2^n}=\{{\alpha }_o+{\alpha }_{1}a+\dots +{\alpha }_{n-1}{a}^{n-1}|{\alpha }_i\in {\mathbb{Z}}_2\&a^n=1+a\}$. From the above, the rest of the cases we have is happened when $b\ne c$ where b and c are both non-zero. That means we have $(2^n-1)(2^n-2)$ cases. We can cover all these cases by choosing ($c=a_{o}b$), ($c=a_o^{2}b$), ($c=a_o^{3}b$), …, ($c=a_o^{2^n-2}b$), where $a_o$ is a generator of the multiplicative group $(F_{2^n}^{\ast },·)$ and $F_{2^n}^{\ast }=F_{2^n}-\left\{0\right\}$. Let $c=a_o^{j}b$ where $1\le j\le 2^n-2$ Firstly, suppose that j is an odd number and take two natural numbers i and k such that $i-k=j$ and $i+k=2^n-1$. So,

$$\left[\begin{array}{cc}{a}_o^i& 1\\ 0& 0\end{array}\right]\left[\begin{array}{cc}b(z+a_o^{j}y)& b(w+x)\\ a_o^{j}b(w+x)& b(z+a_o^{j}y)\end{array}\right]\left[\begin{array}{cc}{a}_o^k& a_o^k\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}r& r\\ 0& 0\end{array}\right]$$

where $r=b\left(z+a_o^{j}y\right)+a_o^k\left(a_o^{i}b\left(z+a_o^{j}y\right)+a_o^{j}b\left(w+x\right)\right)+a_o^{i}b\left(w+x\right)$. Now,

$$\begin{array}{ccc}\hfill r& =& b\left(z+a_o^{j}y\right)+a_o^k{a}_o^{i}b\left(z+a_o^{j}y\right)+a_o^k{a}_o^{j}b\left(w+x\right)+a_o^{i}b\left(w+x\right)=0.\hfill \end{array}$$

and D is not prime. Now, suppose that j is an even number and take a natural number i such that $j=2i$. So,

$$\left[\begin{array}{cc}{a}_o^i& 1\\ a_o^i& 1\end{array}\right]\left[\begin{array}{cc}b(z+a_o^{j}y)& b(w+x)\\ a_o^{j}b(w+x)& b(z+a_o^{j}y)\end{array}\right]\left[\begin{array}{cc}1& 1\\ a_o^i& a_o^i\end{array}\right]=\left[\begin{array}{cc}r& r\\ r& r\end{array}\right]$$

where $r=a_o^i\left(b\left(z+y{a}_o^j\right)+b{a}_o^i\left(w+x\right)\right)+b{a}_o^j\left(w+x\right)+b{a}_o^i\left(z+y{a}_o^j\right)$. Now,

$$\begin{array}{ccc}\hfill r& =& a_o^{i}b\left(z+y{a}_o^j\right)+a_o^{2i}b\left(w+x\right)+b{a}_o^{2i}\left(w+x\right)+b{a}_o^i\left(z+y{a}_o^j\right)=0\hfill \end{array}$$

and D is not prime. This completes the proof. □

Observe that all possible derivations in Examples 4 and 5 are inner derivations. Is that true for all non-commutative prime rings of the form $M_n\left(S\right)$, where S is an integral domain? The answer is no. In the following example, we show that $M_2\left(S\right)$ can have a non-zero derivation, which is not inner.

 Example 6. 

Let $S=\mathbb{R}\left[x\right]$ the polynomial ring with real coefficients. So, it is an integral domain. Thus, $R=M_2\left(S\right)$ is a prime ring. Define a map D on R by

$$D\left(\left[\begin{array}{cc}f& g\\ h& i\end{array}\right]\right)=\left[\begin{array}{cc}{f}^{\prime }& g^{\prime }\\ h^{\prime }& i^{\prime }\end{array}\right]$$

where $f,g,h,i\in S$ and $f^{\prime }$ is the derivative of f. Clearly D is a non-zero additive mapping. Now, for all $f,g,h,i,f_1,g_1,h_1,i_1\in S$, we have

$$\begin{array}{ccc}\hfill D\left(\left[\begin{array}{cc}f& g\\ h& i\end{array}\right]\left[\begin{array}{cc}{f}_1& g_1\\ h_1& i_1\end{array}\right]\right)& =& \left[\begin{array}{cc}{f}^{\prime }{f}_1+f{f}_1^{\prime }+g^{\prime }{h}_1+g{h}_1^{\prime }& f^{\prime }{g}_1+f{g}_1^{\prime }+g^{\prime }{i}_1+g{i}_1^{\prime }\\ h^{\prime }{f}_1+h{f}_1^{\prime }+i^{\prime }{h}_1+i{h}_1^{\prime }& h^{\prime }{g}_1+h{g}_1^{\prime }+i^{\prime }{i}_1+i{i}_1^{\prime }\end{array}\right]\hfill \\ & =& \left[\begin{array}{cc}f& g\\ h& i\end{array}\right]D\left(\left[\begin{array}{cc}{f}_1& g_1\\ h_1& i_1\end{array}\right]\right)+D\left(\left[\begin{array}{cc}f& g\\ h& i\end{array}\right]\right)\left[\begin{array}{cc}{f}_1& g_1\\ h_1& i_1\end{array}\right]\hfill \end{array}$$

This shows that D is a derivation on R. Now, suppose D is an inner derivation. So, there is $\left[\begin{array}{cc}{f}_o& g_o\\ h_o& i_o\end{array}\right]\in R$ such that

$$\begin{array}{ccc}\hfill \left[\begin{array}{cc}{f}^{\prime }& g^{\prime }\\ h^{\prime }& i^{\prime }\end{array}\right]& =& D\left(\left[\begin{array}{cc}f& g\\ h& i\end{array}\right]\right)=\left[\begin{array}{cc}{f}_o& g_o\\ h_o& i_o\end{array}\right]\left[\begin{array}{cc}f& g\\ h& i\end{array}\right]-\left[\begin{array}{cc}f& g\\ h& i\end{array}\right]\left[\begin{array}{cc}{f}_o& g_o\\ h_o& i_o\end{array}\right]\hfill \\ & =& \left[\begin{array}{cc}{f}_{o}f+g_{o}h-(f{f}_o+g{h}_o)& f_{o}g+g_{o}i-(f{g}_o+g{i}_o)\\ h_{o}f+i_{o}h-(h{f}_o+i{h}_o)& h_{o}g+i_{o}i-(h{g}_o+i{i}_o)\end{array}\right]\hfill \end{array}$$

Now, take $f=1,g=h=i=0$. So,

$$\left[\begin{array}{cc}{f}^{\prime }& g^{\prime }\\ h^{\prime }& i^{\prime }\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& -g_o\\ h_o& 0\end{array}\right]$$

Thus, $g_o=h_0=0$. Now, suppose that $g=1,f=h=i=0$. Therefore,

$$\left[\begin{array}{cc}{f}^{\prime }& g^{\prime }\\ h^{\prime }& i^{\prime }\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}-h_o& f_o-i_o\\ 0& h_o\end{array}\right]$$

So, $f_o=i_o$. Hence, $\left[\begin{array}{cc}{f}_o& g_o\\ h_o& i_o\end{array}\right]=\left[\begin{array}{cc}{f}_o& 0\\ 0& f_o\end{array}\right]=f_o\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ which is a central element in R and hence D is the zero derivation, a contradiction. So, D is not inner.

It is clear that if D is a prime derivation, then $-D$ is also a prime derivation. We know that the set of all derivations is closed under addition. Is that true for the set of all prime derivations? The answer is no even if we add the zero derivation to this set, as the following example shows.

 Example 7. 

From Example 5, take two prime derivations $D_1$ and $D_2$. $D_1$ comes from Case 12 by taking $\alpha =1$ and $D_2$ comes from Case 13 by taking $\alpha =2$. For all $x,y,z,w\in {\mathbb{Z}}_3$, observe that

$$\begin{array}{ccc}\hfill (D_1+D_2)\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)& =& \left[\begin{array}{cc}0& 2y\\ -2z& 0\end{array}\right]=D_3\left(\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]\right)\hfill \end{array}$$

and $D_3$ is not prime from Case 4 by taking $\alpha =2$.

4. Discussion

At the end of this paper, we have some observations:

• In all examples, when we tried to prove that a non-zero derivation D is not a prime derivation, it was enough to choose non-zero elements x and y from $D\left(R\right)$ (instead of R) to show that $xD\left(R\right)y=\left\{0\right\}$. Is this true for all examples and hence D is a prime derivation if and only if ($xD\left(R\right)y=\left\{0\right\}$ implies $x=0$ or $y=0$, where $x,y\in D\left(R\right)$)?
• For the commutative prime ring $R=\mathbb{R}\left[x\right]$, the derivative function D is an example of a prime derivation on R such that $D\left(R\right)$ is a subring of R. However, we could not find any example of a prime derivation D such that $D\left(R\right)$ is a subring of a non-commutative ring R. Is this true for all non-commutative prime rings?
• If R is a prime ring that has non-zero derivations, should R have a prime derivation? This is true for rings without zero divisors that have non-zero derivations.
• If every non-zero derivation on R is a prime derivation, should R be without zero divisors?