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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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    Use the fact that $x$ is integral over $Z$ iff $Z[x]$ is finitely generated as a $Z$-module.2011-11-14
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    If you pick $f=a_0+a_1 g_1 + \cdots + a_n g_n\in \mathbb{Z}[G]$, you might be able to show that $f^N$ is an integer (ie all group elements are $e_G$) for some $N$ (I suspect that you might need $N>\vert G \vert$).2011-11-14
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    @JackManey:I tried that, yet failed.Let $G=S_3$ and $g\in G$ has order $3$, and let $f=e+g+g^2$, then $f^2=3f$ ,which is not an integer(though in this case the problem is solved).2011-11-14
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    @user10676:how to show that it is finitely generated?Does that mean to show that $x^N=a_0+a_1x+a_2x^2+...+a_{n}x^n$ for large $N$?Can you give some more hints?2011-11-14
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    I find that I can copy the proof of Hamilton-Cayley theorem in linear algebra word by word here for the proof of this question.Let $f\in \mathbb{Z}[G]$ define a $\mathbb{Z}$-linear hom from $\mathbb{Z}[G]$ to itself, thus corresponding to a matrix $A$ with entries all integers under the basis $\{g|g\in G\}$ sending $a$ in $G$ to $f*a$ where the mulitiplication $*$ is the one in the algebra $\mathbb{Z}[G]$.Then the proof of the H-C theorem applies here using the companion matrix of $A$.2011-11-14
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    You have to know that a sub-$Z$-module of a finitely generated $Z$-module is also finitely generated (the proof is not trivial but you can find it in any book of commutative algebra). However your proof is correct, and it is exactly how to prove the fact I claimed.2011-11-14
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    @user10676:Thank you very much!2011-11-14

3 Answers 3

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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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    I've corrected my formulation; thank for the correction. Note that the map I used hasn't actually changed, and that it is the same as in this answer.2012-06-01
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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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    @Marc: this argument doesn't work (but it is easily fixed). See my answer.2012-05-31
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    If $G=S_n$ you still have the same problem right - shouldn't it be $M_{n\color{Red}!}(\Bbb Z)$?2012-06-01
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    @anon: I said $n=\#G$, so you shouldn't take $G=S_n$ (except if $n\in\{1,2\}$ ;-). Taking $G=S_m$ is OK and will give $n=m!$. I think I did warn somewhere against taking a shortcut for permutation groups, but I forgot where...2012-06-01
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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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    This is an important observation. However, I think "determinant trick" is usually used to designate a result that is both harder to state, and harder to prove than by a simple application of Cayley-Hamilton (rather its proof uses a variation of a certain proof of Cayley-Hamilton). Also its name suggests something unnatural is going on, which is not entirely false. So I think for this particular question it is helpful to know that one can do without any trickery. But of course I don't deny that general truths do have their utility...2012-06-01
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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