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 nth 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.


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