Algebra endomorphisms and centers of Weyl algebras

Lemma 4   Let $k$ be a field of characteristic $p>0$. For any $k$-algebra endomorphism $\phi$ of $A_n(k)$,

1. $\pi_{a,b}\circ\phi$ is a surjective homomorphism for all $(a,b)\in k^{2n}$.
2. $\phi(Z_n(k))\subset Z_n(k)$.

We may assume that $k$ is an algebraically closed field.

(1) The kernel of $\pi_{a,b}\circ\phi$ is an ideal of $A_n(k)$ and therefore, by a dimension argument, is one of the maximal ideals of $A_n(k)$. It also follows that $\pi_{a,b}\circ\phi$ is surjective.

(2) The result (1) implies that for any $(a,b)\in k^{2n}$ we have

$$\phi(Z_n(k))\subset I_{a,b}+k$$

Thus the claim deduces to the following sublemma.

Sublemma 1  

$$\bigcap_{(a,b)\in k^{2n}} (I_{a,b}+k ) =Z_n(k)$$

[proof of sublemma]

The left hand side may be identified with a section of $\widetilde{A_n(k)}$ which reduces to 0 at each $k$-valued point when we regard it as a section of $\widetilde{(A_n(k)/Z_n(k))}$. Since $k$ is an infinite field, this implies that $f\in Z_n(k)$.

Corollary 1   $A_n(k)$ is generated by $\phi(A_n(k))$ and $Z(A_n(k))$.

[proof] We may assume $k$ is algebraically closed. Let $B$ be the algebra generated by $\phi(A_n(k))$ and $Z(A_n(k))$. Then the claim (1) of the previous lemma shows that for any maximal ideal $\mathfrak{m}$ of $Z(A_n(k))$, we have an isomorphism

$$B/\mathfrak{m}B \cong A_n(k)/\mathfrak{m}A_n(k).$$

as $Z(A_n(k))$-modules. Since $A_n(k)$ is finitely generated module over $Z(A_n(k))$, we see immediately that $M=A_n(k)$ as required.

Corollary 2   Let $\overline{A_n(k)}$ be a copy of $A_n(k)$. Let $\phi:\overline{A_n(k)}\to A_n(k)$ be a $k$-homomorphism. Let $Z_n(k)=Z(A_n(k))$, $\overline{Z_n(k)}=Z(\overline{A_n(k))}$ be the center of the algebras $A_n(k)$ $\overline{A_n(k)}$, respectively. Then the natural homomorphism

$$\overline{A_n(k)}\otimes_{\overline{Z_n(k)}} Z_n(k) \to A_n(k)$$

is an isomorphism.

By the corollary above we already know that it is surjective. Since both hand sides are free $Z_n(k)$-modules of rank $p^{2n}$, and since $Z_n(k)$ is an integral domain, the map is a bijection. [The surjection admits a splitting. Then we consider determinants.]

$\qedsymbol$

Corollary 3   An algebra homomorphism $\phi:A_n(k)\to A_n(k)$ is surjective if and only if its restriction to the center

$$\phi\vert _{Z(A_n(k))}^{Z(A_n(k))}:Z(A_n(k)) \to Z(A_n(k))$$

is surjective.

By the birational case of the Jacobian conjecture (which is already known to be true), we conclude that

Corollary 4   $\phi:A_n(k)\to A_n(k)$ is bijective if and only if the following three conditions hold.

1. $\psi= \phi\vert _{Z(A_n(k))}^{Z(A_n(k))}:Z(A_n(k)) \to Z(A_n(k))$ is injective.
2. $\psi$ is birational.
3. The Jacobian $\det D(\psi)$ is a nonzero constant.

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