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Here $G$ is a finite group(not neccessarily abelian),then there is a statement in some representation book that $\mathbb{Z}[G]$ is integral over $\mathbb{Z}$.That is, every element in $\mathbb{Z}[G]$ satisfies a monic polynomial equation with coefficients in $\mathbb{Z}$.

How to get this result?

I worked with the case $G=S_3$ and found it is indeed this case, and I know it also holds for the abelian case trivially, yet I have no idea how to get the general result.

Will someone be kind enough to give me some hints on this?Thank you very much!

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    @user10676:Thank you very much!2011-11-14

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Marc's argument doesn't work as written; the obvious map $\mathbb{Z}[S_n] \to \mathcal{M}_n(\mathbb{Z})$ isn't injective. Indeed the latter is a free $\mathbb{Z}$-module on $n^2$ generators while the former is a free $\mathbb{Z}$-module on $n!$ generators...

Fortunately, there's an easy way out. $\mathbb{Z}[G]$ acts faithfully on itself by left multiplication ("Cayley's theorem for rings"), and this directly defines an injection $\mathbb{Z}[G] \to \mathcal{M}_{|G|}(\mathbb{Z})$.

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    @Marc: yes, the issue really is quite minor, but I thought it was worth pointing out.2012-06-01
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Corrected in response to comment by @Qiaochu Yuan.

Embed $G$ into the symmetric group $S_n$ for $n=\#G$ using Cayley's theorem (don't take a shortcut if $G$ is already a permutation group), and map the ring $\mathbf{Z}[S_n]$ homomorphically to the matrix ring $M_n(\mathbf{Z})$ using permutation matrices. Although the second map is not injective, the composed map $\mathbf{Z}[G]\to M_n(\mathbf{Z})$ is, as can be seen by looking at the first column.

Now apply the Cayley-Hamilton theorem.

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    Whoops, nevermind.2012-06-01
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The subset $A$ of $\Bbb Z[G]$ of integers elements over $\Bbb Z$ is a subring.

You want to show that $A = \Bbb Z[G]$. Thus is it enough to prove that the (canonical) generators of $\Bbb Z[G]$ are integers over $\Bbb Z$. This is trivial since they are all root of unity : for $g\in G$, $g^{|G|} = 1$.

Remark — The determinant trick presented in the other answers is often used to show that the subset of integer elements is a subring.

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    @MarcvanLeeuwen — Agreed ! The sake of my remark was to explain that whereas your answer looks very different from mine, they are not *that* much different.2012-06-01