We may assume that is an algebraically closed field.
(1) The composition is a representation of . By Lemma 1.4 we see that any irreducible sub representation of is equivalent to for some . By a dimensional argument, we conclude that itself is equivalent to . (In other words, there exists such that
holds.) Thus is surjective as required.
(2) Let , . For any , we have (using the same notation as above)
Since we know by (1) that is surjective, we see that belongs to the center of . In particular for any , we have
Thus
So as required.
Let be a -algebra endomorphism of . Then by restriction we obtain a homomorphism
Furthermore, if the base field is perfect, then may be uniquely extended to its -th root.
In precise, Let us write down like
Then is given by the following formula.
Here comes a geometric interpretation of endomorphisms of Weyl algebras.
(*) |
(**) |
and
as in the previous Corollary. We put .
We have an well-defined -algebra homomorphism
By using the isomorphism in , we obtain a -homomorphism
which is compatible with in the sense that it satisfies the commutative diagram (*) of the statement.
It remains to prove that the map is represented as (**). By pull-back, we obtain an -algebra homomorphism
Where the first isomorphism in the above line is the inverse of the following -algebra homomorphism.
by an argument similar to that in (I,Lemma 7.9), we see that there exists such that
holds. (See appendix for the detail.)