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I am reading my notes and I stumbled upon a proof that I dont fully understand, and I was hoping maybe someone could clear the details.

The main goal was to show that if $k$ is algebraically closed, then the group ring $kG\cong M_{n_1}(k)\times....\times M_{n_r}(k)$ (where the characteristic of the field does not divide $|G|$).

First we applied Maschkes theorem and obtained $kG\cong M_{n_1}(D_1)\times...\times M_{n_r}(D_r)$, second we wanted to prove that in this case each $D_i$ equals $k$.

This are my poorly taken notes:

Let $x\in D$ then $k[x]\in D$ is a finite dimensional $k$-algebra which is a domain (because $D$ is division): Either $k[x]/k$ is integral or $\forall y\in k[x], y\neq 0$ the map induced by left multiplication by $y$ is an injection, hence a surjection, hence an isomorphism. So $k[x]/k$ is finite with $k$ algebraically closed then $k=k[x]$ so $x\in k$.

So on the above proof I see if $k[x]/k$ was integral then its finite and then we are done, but I dont understand the "or" part.

Thanks

Daniel Montealegre
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1 Answers1

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I don't really understand what you wrote either. Here is the statement and then the proof: we claim that if $D$ is a finite-dimensional division algebra over an algebraically closed field $k$, then in fact $D \cong k$. As proof, if $x \in D$, then consider the inverse closed subring of $D$ generated by $k$ and $x$. This is a finite, hence algebraic, extension of $k$, hence must be equal to $k$.

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Qiaochu Yuan
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  • Thanks, I am now trying to change the proof a little bit. In case I now have that $k$ contains the primitive $m$th root of unity, where $m$ is the exponent of $G$, then I want to show that $k$ is the splitting field, so the beginning is the same and when we get to $k[x]/k$ we have that $x$ satisfies $x^m-1=0$, since $m$ is the exponent of the group, and since $k$ contains the primitive $m^{th}$ root of unity, we have that $x\in k$, so indeed $k[x]=k$. Is the idea correct? – Daniel Montealegre May 29 '13 at 01:01
  • Why do we have that $x$ satisfies $x^m - 1 = 0$? $x$ is just some element of a division ring $D$. This part of the proof is just about division rings. – Qiaochu Yuan May 29 '13 at 01:24
  • I was thinking something along the lines of given $x$ in $D$ (where $D$ appears as the entries of one of my matrices given my Maschke) creating an element in the RHS which we could map (by the isomorphism) to $g$, and since $g$ would satisfy $t^m-1$, then I would have that $x$ does as well, (but this is me spitballing) – Daniel Montealegre May 29 '13 at 04:24
  • You have that elements of $G$ map to elements of $M_{n_i}(D_i)$ but you can't conclude that they map to scalar matrices, so it's unclear how to get an element of $D$ this way. – Qiaochu Yuan May 29 '13 at 04:48
  • Yeah. I'll have to think about some other approach. Thanks though – Daniel Montealegre May 29 '13 at 05:01