# Counterexample of Schur lemma

I'm studying basic representation theory using GTM$$42$$, encountering a problem on Schur lemma:

Schur Lemma. Given $$G$$ a group, $$V$$ a vector space on $$\mathbb C$$ and an irreducible representation $$\rho : G \to Gl(V)$$. If a linear map $$f : V \rightarrow V$$ commutes with all $$\rho_g, g\in G$$, then $$f=\lambda$$ for some $$\lambda\in \mathbb C$$.

Proof of this lemma depends on the fact that $$\mathbb C$$ is algebraiclly closed. Now I'm asked to give a counterexample of this lemma when $$V$$ is on $$\mathbb Q$$ but cannot find one (not familiar with subgroups of $$Gl(n,\mathbb Q)$$). Moreover I'm wondering if a counterexample exists for any field not algebraiclly closed.

Any help will be appreciated.

Schur's lemma relies on the existence of enough eigenvalues. So, take $$f$$ to not have eigenvalues in the base field. Consider for example rotation of the rational plane $$f = \begin{bmatrix} 0&-1\\1&0\end{bmatrix}$$. This $$f$$ has order 4, so I'll take $$G = \mathbb{Z}/4\mathbb{Z}$$ with $$1\mapsto f$$.
Now $$\mathbb{Q}^2$$ is an irreducible (because any proper invariant subspace should be one dimensional) representation of $$G$$ and $$f\colon\mathbb{Q}^2\to\mathbb{Q}^2$$ is $$G$$-linear, but is not scalar multiplication. Incidentally, this also gives an example of an irreducible representation of an abelian group not of dimension 1. (The same applies over any field without a square root of $$-1$$).
• This is a nice answer. You can generalize to $G=\mathbb{Z}/n\mathbb{Z}$ acting on the $\mathbb{Q}$-vector space $\mathbb{Q}(\omega_n)$, where $\omega_n$ is a primitive $n$th root of unity. The matrix construction comes from realizing $\mathbb{Q}(\omega_n)\cong \mathbb{Q}[x]/\langle \Phi_n(x)\rangle$, where $\Phi_n$ is the $n$th cyclotomic polynomial. Oct 21, 2020 at 15:55