A matrix representation of Weyl algebra

From now on in this section we fix a prime number $p$ and we assume $k$ is a field of characteristic $p$ . We will simply write $A_n$ instead of $A_n(k)$ .

Let us define operators(matrices) $\{\mu_i\}_{i=1}^{2 n}$ acting on $p^n$ -dimensional vector space

$$V=k[x_1,x_2,\dots,x_n]/(x_1^p,x_2^p,\dots x_n^p)$$

by

$$\left . \begin{aligned} &\mu_i=\text{multiplication by $x_i$}\... ...rtial/\partial x_i. \end{aligned}\right \} \qquad i=1,2,\dots,n.$$

Let

$$S_n=k[T_1,T_2,\dots,T_{2n-1},T_{2n}]$$

be a polynomial ring of $2 n$ -variables over $k$ .

Then we have a faithful representation

$$\Phi: A_n \to M_{p^n}(S_n)$$

of the Weyl algebra $A_n$ by putting

$$\Phi(\gamma_i)=T_i\cdot 1_{p^n}+\mu_i.$$

Furthermore, for any $c=(c_1,c_2,\dots,c_{2n}) \in k^{2n}$ , we have by specialization the following representation of $A_n$ .

$$\Phi_c: A_n \to M_{p^n}(k)$$

by putting

$$\Phi_c(\gamma_i)=c_i\cdot 1_{p^n}+\mu_i.$$

We recall also that

LEMMA 1.3 (Part I, Lemma 7.3)   The center $Z_n$ of $A_n$ is isomorphic to a polynomial algebra with $2 n$ indeterminates. Namely,

$$Z_n=k[\gamma_1^p,\gamma_2^p,\dots, \gamma_{2n}^p] .$$

LEMMA 1.4 (Part I, Corollary 7.8)   Let $k$ be an algebraically closed field of characteristic $p\neq 0$ . Then every finite dimensional irreducible representation $\alpha:A_n(k)\to \operatorname{End}_k(V)$ of $A_n(k)$ is equivalent to a representation $\Phi_c$ for some $c\in k^{2n}$ .

LEMMA 1.5   Let $k$ be a field of characteristic $p$ . If $k$ contains infinite number of elements, then:

$$\bigcap_{c\in k^{2 n} } \operatorname{Ker}(\Phi_c)=0.$$

Here is another thing we need to know.

LEMMA 1.6   $\Phi$ gives a $k$ -algebra isomorphism

$$\hat{\Phi}:S_n\otimes_{Z_n} A_n \cong M_{p^n} (S_n).$$

PROOF.. One may easily verify that we have a well-defined homomorphism given by

$$\hat{\Phi}: S_n\otimes_{Z_n} A_n \ni f \otimes a \mapsto f \Phi(a)\ni M_{p^n} (S_n).$$

For any $j\in \{1,2,3,\dots,2 n\}$ , we have

$$\hat{\Phi} (1\otimes \gamma_j -T_j\otimes 1)=\mu_j.$$

So the image $\operatorname{Image}(\hat{\Phi})$ contains $\mu_j$ . Since we know that $\{\mu_j\}_{j=1}^{2 n}$ generates $M_{p^n}(k)$ as $k$ -algebra (Part I, Corollary 7.10) , we conclude that the map $\hat{\Phi}$ is surjective.

Now, both $S_n\otimes_{Z_n}A_n$ and $M_{p^n}(S_n)$ is free $S_n$ -modules of rank $p^{2 n}$ . So the map $\hat{\Phi}$ is generically injective. (That means, if we take the quotient field $Q(S_n)$ of $S_n$ and consider

$$1_{Q(S_n)}\otimes\hat{\Phi}: Q(S_n)\otimes_{S_n}(S_n\otimes_{Z_n} A_n )\to M_{p^n} (Q(S_n)).$$

then by an elementary theorem in linear algebra, we see that it is an isomorphism.)

Since $S_n$ (a polynomial algebra over $k$ ) is an integral domain, $S_n\otimes_{Z_n}A_n$ is $Z_n$ -torsion free. (That means,

$$S_n \otimes_{Z_n} A_n\to Q(S_n)\otimes_{S_n}(S_n\otimes_{Z_n} A_n )$$

is injective.)

So we see that $\hat{\Phi}$ is injective.

$\qedsymbol$