Definition 1. A ring $R$ with 1 is called simple if $(0)$ and $R$ are the only two-sided ideals of $R.$

Remark 1. The center of a simple ring $R$ is a field.

Proof. Let $a$ be any non-zero element of the center of $R.$ Then $Ra$ is a non-zero two-sided ideal of $R$ and hence, since $R$ is simple, $Ra=R.$ Thus there exists some $r \in R$ such that $ra=1,$ i.e. $a$ is invertible. Since $a$ is in the center, $r$ is in the center too and we’re done. $\Box$

Obviously commutative simple rings are  just fields. So only non-commutative simple rings are interseting.

Note 1. If $k$ is the center of a simple ring $R,$ then $R$ is both a vector space over $k$ and a ring. So it is a $k$-algebra. The term simple algebras is commonly used instead of simple rings.

We now give the first example of simple rings.

Example 1. Let $R$ be a ring with 1 and let $M_n(R)$ be the ring of $n \times n$ matrices with entries from $R.$ It is a well-known fact, and easy to prove, that $J$ is a two-sided ideal of $M_n(R)$ if and only if $J=M_n(I)$ for some two-sided ideal $I$ of $R.$ In particular, $M_n(R)$ is simple if and only if $R$ is simple. So, for example, since every division ring $D$ is obviously simple, $M_n(D)$ is simple too. In fact, by the Artin-Wedderburn  theorem, a ring $R$ is simple and (left or right) Artinian if and only if $R = M_n(D)$ for some division ring $D$ and some integer $n \geq 1.$

Example 1 gives all simple rings which are Artinian. But what about simple rings which are not Artinian? We can find a family of non-Artinian simple rings by generalizing the concept of derivation in Calculus, as it follows.

Notation. From now on we will assume that $k$ is a field and $A$ is a $k$-algebra.

Definition 2. A $k$-linear map $\delta : A \longrightarrow A$ is called a derivation of $A$ if $\delta(ab)=\delta(a)b + a \delta(b),$ for all $a,b \in A.$ If there exists some $c \in A$ such that $\delta(a)=ca-ac,$ for all $a \in A,$ then $\delta$ is called an inner derivation.

Note that if $c \in A$, then the map $\delta : A \longrightarrow A$ defined by $\delta(a)=ca-ac$, for all $a \in A$, is a derivation.

Remark 2. If $\delta$ is a derivation of $A,$ then $\delta(1)=\delta(1 \cdot 1)=\delta(1)+\delta(1)$ and thus $\delta(1)=0.$ Therefore if $\alpha \in k,$ then $\delta(\alpha)=\alpha \delta(1)=0.$

Remark 3. Consider the polynomial algebra $R:=A[x]$ and define the map $\delta : R \longrightarrow R$ by $\displaystyle \delta(f(x))=\frac{df}{dx}.$ Then $\delta$ is a derivation which is not inner. The reason is that if there was $g \in R$ such that $\delta(f)=gf - fg,$ for all $f \in R,$ then, since $x$ is central, we’d get $1 = \delta(x) = gx-xg=0,$ which is non-sense.

Definition 3. Let $\delta$ be a derivation of $A.$ An ideal $I$ of $A$ is called a $\delta$ideal if $\delta(I) \subseteq I.$ If the only $\delta$-ideals of $A$ are $\{0\}$ and $A,$ we’ll say that $A$ is $\delta$simple.

Obviously if $A$ is simple, then $A$ is $\delta$-simple for every derivation $\delta.$

Remark 4. If $\delta$ is an inner derivation of $A$, then every ideal of $A$ is also a $\delta$-ideal. An obvious example of  a $\delta$-simple algebra is $A=k[x]$ with $\displaystyle \delta = \frac{d}{dx}.$

Definition 4. If $\delta$ is a derivation of $A,$ then we define the  $k$-algebra $A[x; \delta]$ to be the set of all polynomials in the indeterminate $x$ and left coefficients in $A$, with the usual addition and multiplication and the rule $xa=ax + \delta(a).$ The algebra $A[x; \delta]$ is also called a differential polynomial algebra. Note that if $\delta=0,$ i.e. $\delta(a)=0$ for all $a \in A,$ then $A[x; \delta]=A[x],$ the ordinary polynomial algebra.

Note 2. So a typical element of $A[x; \delta]$ is in the form $a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0,$ where $a_i \in A.$ When we multiply two of these polynomials we will have to use the rule given in Definition 4. For example

$x(ax+b)=(xa)x + xb = (ax + \delta(a))x + bx +\delta(b)=ax^2 + (\delta(a)+b)x + \delta(b).$

Of course, we need to prove that $A[x;\delta]$ is a $k$-algebra but we won’t do it here.

Remark 5. Let $a \in A$ and $m \geq 1$ be an inetger. Then an easy induction shows that in $A[x; \delta]$ we have

$\displaystyle x^ma=\sum_{i=0}^m \binom{m}{i} \delta^{m-i}(a)x^i,$

where $\delta^0(a)=a.$ So $x^m a$ is a (left) polynomial of degree $m$ with the leading coefficient $a.$

In part (2) we will give three important examples of simple rings.

In general, by Example 1, for any integer $n \ge 1$ and any division ring $D,$ the ring of $n \times n$ matrices with entries from $D$ is both simple and Artinian.