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I know that the irreducible representations of $\mathbb{Z}_n$ over $\mathbb{C}$ are the $n$ group homomorphisms $\mathbb{Z}_n \to \mathbb{C}$ sending $1$ to the roots of unity. There are $n$ of these and they're all $1$ dimensional since $\mathbb{Z}/n\mathbb{Z}$ is abelian. Over $\mathbb{Q}$ I am having some troubles since $\mathbb{Q}$ isn't algebraically closed. I thought it might be easier to find the simple (irreducible) $\mathbb{Q}[\mathbb{Z}_n]$ modules. We have $\mathbb{Q}[\mathbb{Z}_n] \cong \mathbb{Q}[X]/(X^n-1) \cong\mathbb{Q}[x]/\prod_{d \mid n} \Phi_d \cong \prod_{d \mid n} \mathbb{Q}[x]/\Phi_d$ the last isomorphism from CRT. But how exactly do I find all the simple $ \prod_{d \mid n}\mathbb{Q}[x]/\Phi_d$ modules? $\Phi_d$ is irreducible, so this is a product of fields (specifically of $\mathbb{Q}(\zeta_d)s)$ and so would the simple modules are just those $ \prod_{d \mid n}\mathbb{Q}[x]/\Phi_d$ modules isomorphic to some $\mathbb{Q}(\zeta_d)$? I understand how (the obvious module isomorphic to) $\mathbb{Q}(\zeta_d)$ is a simple $\mathbb{Q}[\mathbb{Z}_n]$ module, but how do I prove these are the only such, if this is true?

user1632940
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    See here for the general theory of representations over $\mathbb{Q}$: https://math.stackexchange.com/questions/676303/representation-theory-over-mathbbq/676338#676338 – Alex B. May 23 '25 at 17:52

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Finding a representation amounts to finding a rational square matrix $A$ such that $A^n=I.$ Such a rational matrix must have a characteristic polynomial equal to a product $\Phi_d(x)$ where $d\mid n,$ including possibly repetitions, and a minimal polynomial the product of distinct $\Phi_d(x),$ since the minimal polynomial must be a factor of $x^n-1.$

That latter fact means $A$ must be diagonalizable, since there are no repeated roots to the minimal polynomial.

If the characteristic polynomial of $A$ is not irreducible over $\mathbb Q[x],$ then, since $A$ is diagonalizable over $\mathbb C,$ we can split the representation.

So we get irreducible representations starting with $1\mapsto A$ where $A$ has characteristic polynomial $\Phi_d(x)$ for some $d\mid n.$ We get exactly one such representation, up to isomorphism, for each $d\mid n.$

GreginGre
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Thomas Andrews
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  • Why is it necessarily true that such a rational matrix must have characteristic polynomial equal to a product $\Phi_d(x)$ for $d \mid n$? – user1632940 May 24 '25 at 13:27
  • Because the minimal polynomial foe $A$ divides $x^n-1,$ which is a product if such $\Phi_d(x),$ which are known to be prime, so the characteristic polynomial can't have any different prime factors. If $p$ is the characteristic polynomial, and $m$ is the minimal polynomial then there is some $m$ such that $$m(x)\mid p(x)\mid m^k(x)$$ – Thomas Andrews May 24 '25 at 16:31
  • Essentially, the only roots of the characteristic polynomial has the same set of roots as the minimal polynomial, just with possibly more repetition. – Thomas Andrews May 24 '25 at 16:35