reduction to characteristic $p$

Suppose we are given a $\mathbb{C}$-algebra endomorphism $\phi$ of $A_n(\mathbb{C})$. Since the algebra $A_n(\mathbb{C})$ is finitely generated over $\mathbb{C}$, the endomorphism $\phi$ is actually defined over a ring $R$ which is finitely generated algebra over $\mathbb{Q}$.

By a specialization argument we may assume $R=\mathfrak{O}(\mathfrak{K})[1/f]$, where $\mathfrak{K}$ is a finite extension field of $\mathbb{Q}$, $\mathfrak{O}(\mathfrak{K})$ is the ring of all algebraic integers in $\mathfrak{K}$, $f$ is a non zero element of $\mathfrak{O}(\mathfrak{K})$.

For almost all (that is, all except finite number of) prime ideals $\mathfrak{p}$ of $\mathfrak{O}(\mathfrak{K})$, we obtain an algebra endomorphism $\phi_\mathfrak{p}$ of an algebra $A_n(k(\mathfrak{p}))$ over $k(\mathfrak{p})$ where $k(\mathfrak{p})=\mathfrak{O}(\mathfrak{K})/\mathfrak{p}$ is a field of a positive characteristic $p$.

Lemma 2   $\phi$ is invertible if and only if homomorphisms $\phi_\mathfrak{p}$ are invertible for all except finite number of primes $\mathfrak{p}\in \operatorname{Spec}(\mathcal O(\mathfrak{K}))$.

The ``only if'' part is clear. To prove ``if'', we suppose on the contrary that $\phi$ is not invertible. This means, as we mentioned, that $\phi$ is not surjective. Then there exists a nonzero linear functional ¹ $\chi$ such that $\chi \circ \phi=0$. It is easy to see that $\chi$ defines a non zero linear functional $\chi_\mathfrak{p}$ on $A_n(k(\mathfrak{p}))$ for all except finite number of primes $\mathfrak{p}$, and that $\chi_pe\circ \phi_\mathfrak{p}=0$. this is a contradiction, and the lemma is proved. $\qedsymbol$

It is thus worthwhile to study $A_n(k)$ and its automorphism when $k$ has a positive characteristic.

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