Let $R$ and $S$ be commutative rings such that there is a homomorphism $\varphi: R \to S$, and let $A_n(R)$ be the $n$-th Weyl algebra over $R$. If we extend the scalars on $A_n(R)$, is it true that
\begin{equation} A_n(R) \otimes_R S \cong A_n(S) \end{equation}
as $S$-algebras?
I believe I have a partial proof. We have an $S$-algebra homomorphism $\theta: A_n(R) \otimes_R S \to A_n(S)$ given by $x_i \otimes s \mapsto sx_i$ and $y_i \otimes s \mapsto sy_i$, where $x_i$ and $y_i$ are the generators of the Weyl algebra (I'll omit checking well-definition and that this is a homomorphism). In the case where $R$ has characteristic $0$, this is surjective since any element in $A_n(S)$ has the form
\begin{equation} \sum_{i,j,k,l} s_{ijkl} x_i^k y_j^l = \theta \left (\sum_{i,j,k,l} x_i^k y_j^l \otimes s_{ijkl} \right ), \end{equation}
and where $S$ has characteristic $0$, I think injectivity is clear. I understand that in positive chracteristic, it's possible that, for instance, $x_i^k = x_i^{k'}$ but $k \neq k'$, and I'm not entirely sure how to deal with this. Is my argument correct so far, and does this extend to positive characteristic?
If this is not true in general, I am particularly interested in the case where $R$ is a finitely generated domain of characteristic $0$ and $S=\overline{\mathbb{F}}_p$.
Apologies if I have made any mistakes, my knowledge in this area is mostly self-taught and this is my first Stack Exchange post!