Schur’s lemma over $\mathbb{F}_p$

Question

I’m studying modular representation theory, and I got really stuck with the seemingly innocent statement.

Consider $\mathrm{GL}_{2}(\mathbb{F}_{p})$ and its center $Z$, which is just a set of all scalar matrices. Now let $\pi: \mathrm{GL}_{2}(\mathbb{F}_{p}) \to \mathrm{GL}(V) $ be an irreducible representation over $\mathbb{F}_p$. Show that any $z\in Z$ is also a scalar operator in representation ($\pi(z) = \alpha\cdot \mathsf{Id}_{V}$ for some $\alpha\in\mathbb{F}_p$).

And for algebraically closed field this is a one liner, as by commutative nature of $z$ we know that $\pi(z)$ is an endmorphism of $(\pi,V)$, for which Schur’s Lemma states exactly that it is scalar.

However this doesn’t work for $\mathbb{F}_p$, as not all endimorohisms of $V$ have eigenvalues. We can generalize Schur’s Lemma and say that if we have a finite group $G$ and it’s irreducible representation $(\pi,V)$, then we know that $\mathrm{End}_{\mathbb{F}_p[G]}(V)$ is a division ring. Since we’re over a finite field, $\mathrm{End}_{\mathbb{F}_p[G]}(V)$ is also a field, but I am not sure that it is exactly $\mathbb{F}_p$, why couldn’t it be a field $\mathbb{F}_{p^n}$?

And I’m also interested what is the best Schur’s Lemma adaptation for a finite field case.

Answer 1

Since we’re over a finite field, $\mathbb{End}_{\mathbb{F}_p[G]}(V)$ is also a field, but I am not sure that it is exactly $\mathbb{F}_p$, why couldn’t it be a field $\mathbb{F}_{p^n}$?

In general, it can. The smallest example is $G = C_3$ acting on $\mathbb{F}_2^2$ where a generator $g$ of $C_3$ acts by the matrix $\left[ \begin{array}{cc} -1 & -1 \\ 1 & 0 \end{array} \right]$. This is an irreducible representation of $C_3$, and its endomorphism ring is generated by $g$ itself (since $C_3$ is abelian); it is $\mathbb{F}_2[g]/(g^2 + g + 1) \cong \mathbb{F}_4$. This counterexample can also be defined over $\mathbb{F}_p$ for any $p \equiv 2 \bmod 3$.

The case of $GL_2(\mathbb{F}_p)$ is special because its center is $\mathbb{F}_p^{\times}$ which is cyclic of order $p - 1$. If we write $g$ for a generator, it follows that the matrix of $g$ in any linear representation satisfies $g^{p-1} = 1$, which factors as

$$\prod_{i=1}^{p-1} (g - i) = 0.$$

In other words, all of the eigenvalues of $g$ are defined over $\mathbb{F}_p$. Can you finish from here?

And I’m also interested what is the best Schur’s Lemma adaptation for a finite field case.

In general you just get that the endomorphism ring is $\mathbb{F}_{p^k}$ for some $k$, as you say, and every possible value of $k$ occurs (it suffices to take $G$ to be a cyclic group).