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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. – user10676 Nov 14 '11 at 8:53
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 $N>\vert G \vert$). – user5137 Nov 14 '11 at 9:07
@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). – user14242 Nov 14 '11 at 10:38
@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? – user14242 Nov 14 '11 at 11:56
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 $A$. – user14242 Nov 14 '11 at 12:02
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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. – Marc van Leeuwen Jun 1 '12 at 8:11
@Marc: yes, the issue really is quite minor, but I thought it was worth pointing out. – Qiaochu Yuan Jun 1 '12 at 13:56
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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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Thank you very much. – user14242 Nov 18 '11 at 0:37
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@Marc: this argument doesn't work (but it is easily fixed). See my answer. – Qiaochu Yuan May 31 '12 at 20:30
   
If $G=S_n$ you still have the same problem right - shouldn't it be $M_{n\color{Red}!}(\Bbb Z)$? – anon Jun 1 '12 at 8:16
@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... – Marc van Leeuwen Jun 1 '12 at 9:06
Whoops, nevermind. – anon Jun 1 '12 at 9:07
up vote 1 down vote

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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+1, Nice answer! – wxu Jun 1 '12 at 10:11
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... – Marc van Leeuwen Jun 1 '12 at 14:33
@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. – Lierre Jun 1 '12 at 18:08

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