Galois representation and simple algebras

Question

I'm reading the book of Hida about Galois representation. He wants to introduct semisimple algebras. let $\rho:G\longrightarrow GL(n,E)$ a linear representation (G is a group and E a field). Let $R=R(\rho)$ be the E-subalgebra of $End_E(V(\rho))$ generated by $\rho(\sigma),\ \sigma\in G$. Then he obtains that $R=\prod_\mathfrak{m}R/\mathfrak{m}^N$ where the product is over the two-sided maximal ideals of R, and N is such that $I^N=0$, where $I$ is the radical of Jacobson of R. Then he says: if $V$ is irreducibile (I think simple as R-module?) then $I=\mathfrak{m}$ for a single maximal two-sided ideal, why? I really can't understand. Since R acts faithfully on V (again, why??), from $\mathfrak{m}V=0$ whe conclude $\mathfrak{m}=0$. Can someone help me to uderstand this? Thanks!

Answer 1

Recall that a representation $\rho:G\to\mathrm{GL}(n,E)$ determines an $E$-algebra homomorphism $$E[G]\to\mathrm{End}_E(V(\rho)).$$

Note that $R$ is the image of this map, so in particular, it acts faithfully on $V$.

Since $R = \prod_{\mathfrak m} R/\mathfrak m^N$, if there is more than one two-sided maximal ideal, then each $R/\mathfrak m^N$ will be a proper $R[G]$-submodule. Hence, since $V$ is irreducible, there can only be one such ideal, which must be the Jacobson radical by definition.