For which commutative rings $R$, the first Weyl algebra $A_1(R)$ is simple?

Let $R$ be a commutative ring. The first Weyl algebra over $R$ is the associative $R$-algebra generated by $x$ and $y$ subject to the relation $yx-xy=1$.

For which rings $R$, the first Weyl algebra $A_1(R)$ is simple?

If I am not wrong, it is necessary that $R$ will have characteristic zero. Is it true that for any intgral domain $D$ of characteristic zero, $A_1(D)$ is simple?

Theorem 2.19 is relevant for my question. It relies on Lemma 2.16.

Thank you very much.

Everything in the algebra can be rewritten to a representative of the form $\sum_i\sum_j\alpha_{ij}x^iy^j$ where $\alpha_{ij}\in R$.

If $R$ has a nontrivial ideal $M$, then we can consider the subset of elements such that $\alpha_{ij}\in M$, and it is not hard to show this is a nontrivial ideal.

Essentially you are just getting a homomorphism $A_1(R)\to A_1(R/M)$ from the reduction of $R$ to $R/M$.

So it would seem that it is necessary for $R$ to be simple if $A_1(R)$ is to be simple. I'm not positive about the converse, but it seems plausible.

• Thank you very much! Nice idea. Any proof/counterexample for the converse is welcome. – user237522 Feb 20 '18 at 18:45
• @user237522 The natural thing to do would be to investigate a ideals of $R$ consisting of some subset of coefficients of an ideal of $A_1(R)$. Due to the twisting action of the relation, I wasn't able to get an immediate answer along these lines. – rschwieb Feb 20 '18 at 19:34
• Thanks for your comments. – user237522 Feb 20 '18 at 20:28
• @user237522 I wanted to add: it's not hard to show things like "the coefficients of things in an ideal" and "the set of constant terms of things in an ideal" are ideals of $R$, but I don't see a clear way to reasoning that they are not equal to $R$. – rschwieb Feb 20 '18 at 20:33
• Thank you. (Perhaps there is another way to show the converse, for example, by applying somehow the above mentioned Lemma 2.16?). – user237522 Feb 20 '18 at 20:40