Throughout $k$ is a field, $\text{char}(k)=0$ and $A$ is a $k$-algebra.

Example 2. Suppose that $\delta$ is a derivation of $A$ which is not inner. If $A$ is $\delta$-simple, then $B=A[x;\delta]$ is simple. In partcular, if $A$ is simple, then $A[x; \delta]$ is simple too.

Proof. Suppose, to the contrary, that $B$ is not simple. So $B$ has some non-zero ideal $I \neq B.$ Let $n$ be the minimum degree of non-zero elements of $I.$ Let $J$ be the set of leading coefficients of elements of $I.$ Clearly $J$ is a left ideal of $A$ because $I$ is an ideal of $B.$ To see that $J$ is also a right ideal of $A$, let $a \in J$ and $b \in A.$ Then there exists

$f=ax^n + \text{lower degree terms} \in I.$

But, by Remark 5

$fb = abx^n + \text{lower degree terms} \in I$

and so $ab \in J,$ i.e. $J$ is also a right ideal. Next, we’ll show that $J$ is a $\delta$-ideal of $A$:

if $a_n \in J,$ then there exists some $f(x)=\sum_{i=0}^n a_ix^i \in I.$ Clearly $xf - fx \in I,$ because $I$ is an ideal of $B.$ Now

$xf - fx=\sum_{i=0}^n xa_i x^i - \sum_{i=0}^n a_ix^{i+1}=\sum_{i=0}^n (a_ix +\delta(a_i))x^i - \sum_{i=0}^n a_i x^{i+1}$

$= \sum_{i=0}^n \delta(a_i)x^i.$

So $\delta(a_n) \in J$, i.e. $J$ is a non-zero $\delta$-ideal of $A.$ Therefore $J=A$, because $A$ is $\delta$-simple. So $1 \in J,$ i.e. there exists $g(x)=x^n + b_{n-1}x^{n-1} + \cdots + b_0 \in I.$ Finally let $a \in A.$ Now, $ga - ag,$ which is an element of $I,$ is a polynomial of degree at most $n-1$ and the coefficient of $x^{n-1}$ is $b_{n-1}a - ab_{n-1} + n \delta(a)$, which has to be zero because of the minimality of $n.$ Thus, since $\text{char}(k)=0,$ we may let $c=\frac{b_{n-1}}{n}$ to get $\delta(a)=ca-ac.$ That means $\delta$ is inner. Contradiction! $\Box$

Lemma. If $A$ is simple and $\delta = \frac{d}{dx},$ then $A[x]$ is $\delta$-simple.

Proof. Let $I \neq \{0\}$ be a $\delta$-ideal of $A[x].$ Let $f=\sum_{i=0}^n a_ix^i, \ a_n \neq 0,$ be an element of $I$ of the least degree. Suppose $n > 0.$ Then, since $I$ is a $\delta$-ideal, we must have $\frac{df}{dx}=\sum_{i=0}^{n-1}ia_ix^{i-1} \in I.$ Hence $na_{n-1}=0,$ by the minimality of $n,$ and thus $a_n=0$ because $\text{char}(k)=0.$  This contradiction shows that $n=0$ and so $f=a_0 \in A \cap I.$ Hence $Aa_0A \subseteq I$ because $I$ is an ideal of $A[x]$ and $A \subset A[x].$ But $A$ is simple and so $Aa_0A = A,$ i.e. $1 \in Aa_0A \subseteq I$ and thus $I=A[x]. \Box$

Definition 5. The algebra $A[x][y, \frac{d}{dx}]$ is called the first Weyl algebra over $A$ and is denoted by $\mathcal{A}_1(A).$ Inductively, for $n \geq 2,$ we define $\mathcal{A}_n(A)=\mathcal{A}_1(\mathcal{A}_{n-1}(A))$ and we call $\mathcal{A}_n(A)$ the $n$th Weyl algebra over $A.$

Example 3. If $A$ is simple, then $\mathcal{A}_n(A)$ is simple for all $n.$ In particular, $\mathcal{A}_n(k)$ is simple.

Proof. By Remark 3 in part (1), $\delta = \frac{d}{dx}$ is a non-inner derivation of $A[x].$ So, $\mathcal{A}_1(A)$ is simple by the above lemma and Example 2 in part (1). The proof is now completed by induction over $n. \Box$

Example 4. In this example we do not need to assume that $\text{char}(k)=0.$ Let $A$ be a simple ring and $k$ its center. Let $B$ be a simple $k$-algebra but the center of $B$ may or may not be $k.$ The first part of the corollary in this post shows that $A \otimes_k B$ is simple. This is another way of constructing new simple rings from old ones.