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Consider algebraic number fields $\mathbb{Q} \subseteq K \subseteq L$ with rings of integers $\mathbb{Z}\subseteq \mathcal{O}_K \subseteq \mathcal{O}_L$. If $0 \neq \mathfrak{p} \trianglelefteq \mathcal{O}_K$ is a prime ideal, then the ideal $\mathfrak{p}\mathcal{O}_L$ has a factorisation into prime ideals of $\mathcal{O}_L$, say $\mathfrak{p}\mathcal{O}_L = \prod_{i=1}^r \mathfrak{P}_i^{e_i}$. If additionally $L/K$ is a Galois extension, then $\mathrm{Gal}(L/K)$ operates transitively on $\{ \mathfrak{P}_1 , ... , \mathfrak{P}_r \}$ and all $e_i$ are equal.

I think this is well known, just as is Clifford's theorem from ordinary representation theory of finite groups which states the following.

Let $G$ be a finite group with a normal subgroup $N \trianglelefteq G$ and let $K\subseteq \mathbb{C}$ be a field which is invariant under complex conjugation and which is a splitting field for all group algebras $KH$ for all subgroups $H\leq G$. Denote by $( \cdot , \cdot )$ the scalar product on the space of class functions of $G$ and by $\chi_N$ the restriction of $\chi$ to $N$. If $\chi$ is an irreducible character of $G$ and $\vartheta$ is an irreducible character of $N$ such that $(\chi_N , \vartheta) =: e > 0$ and $\vartheta_1 , ... , \vartheta _t$ are the $G$-conjugates of $\vartheta$, then $\chi_N = e \sum_{i=1}^t \vartheta_i$, which is to say that the irreducible constituents of $\chi_N$ form an orbit under conjugation and they all occur with the same multiplicity $e$.

Now these two statements seem remarkably similar but my fellow students and I have not been able to decide whether there really is some conncetion between these two theorems or if what we see is a mere coincidence.

My question is: Is there some "deeper reason" that we encounter similar situations in representation theory and number theory?

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I see it as a testament to the universality of transitive group actions in modern mathematics. –  Matt Aug 28 '12 at 18:17
    
The $e$ occurring in Clifford theory is even called 'ramification index' by some authors (e.g. Isaacs). However, the analogy to the number theoretic situation is normally seen the other way round: Given an irreducible character $\vartheta$ of a normal subgroup $N$, the induced character decomposes $\vartheta^G = \sum_{\chi} e(\chi) \chi$. These $e(\chi)$'s need not be equal. –  ladisch Aug 30 '12 at 15:11

1 Answer 1

It seems not too mysterious, since the statements about prime ideals can be recast into the language of representation theory. Given a commutative ring $R$, the spectrum of prime ideals corresponds to the irreducible $R$-representations. That is to say, the simple $R$-modules are precisely the quotients $R/P$ where $P$ is a prime ideal.

So it seems likely that a general statement of Clifford's theorem would capture the number theory situation.

I can take you part of the way there: In Clifford theory, I'm pretty sure that you can replace the field $K$ with a commutative ring $R$, and talk about $R[G]$- and $R[N]$-modules, and basically the same Clifford theorem holds (see [Clifford theory - Encyclopedia of Mathematics])1.

In particular, if $R$ is $\mathcal{O}_K$, $G$ is the Galois group of $L/K$ and $N$ is the trivial subgroup, we get a Clifford theory statement about $\mathcal{O}_K[G]$-modules and simple $\mathcal{O}_K$-modules (which correspond to the prime ideals of $\mathcal{O}_K$).

Then when $\mathcal{O}_L$ has an integral normal basis, it is a rank 1 free $\mathcal{O}_K[G]$-module, so the Clifford theory statement would seem to apply here.

I suppose then some modest generalization to Clifford's theorem can be made to accommodate the cases when $\mathcal{O}_L$ does not have an integral normal basis.

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