I'm solving exercises of noncommutative ring theory and I have find across the following problem.

If $D$ and $D'$ are division rings and $M_m(D)\simeq M_n(D')$, show that $D\simeq D'$ and $m=n$.

I have already tried to attack in various ways and I understand that this exercise says that, by the Wedderburn-Artin theorem, a simple Artinian ring $R$ is a ring of matrices over a division ring $D$ unique up to isomorphism.

Any suggestion is appreciated. Thank you.

  • $\begingroup$ Once you have the isomorphism $D \simeq D'$, the equality $m=n$ should follow by an easy dimension argument. To see how $D$ is determined by the matrix ring $M_n(D)$, as far as I remember, you should look at the endomorphism ring of a simple module over your ring. It is "almost" D. $\endgroup$ – Torsten Schoeneberg Oct 6 '17 at 1:39
  • $\begingroup$ I should take $R=M_m(D)$ that is simple, then $R$ is primitive and $M=D^m$ is a faithful and simple $R$-module. Then I should use the Schur's lemma to get the division ring $End_RM=\Delta\simeq D$. Is that what you said? $\endgroup$ – M. Wolf Oct 6 '17 at 1:54
  • $\begingroup$ In fact, $\Delta=D$. $\endgroup$ – M. Wolf Oct 6 '17 at 2:05
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    $\begingroup$ Sort of. I think one gets the opposite ring $D^op$, but that still determines $D$. Now I wonder if it is a) important and b) easy to see that $D^m$ is, up to isomorphism, the only simple $R$-module. $\endgroup$ – Torsten Schoeneberg Oct 6 '17 at 3:15
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    $\begingroup$ What I mean is: If two rings $R$ and $R'$ are isomorphic, and if $M$ is (up to isomorphism) the only simple left-$R$-module, and $M'$ is (up to isomorphism) the only simple left-$R'$-module, then $End_R(M) \simeq End_{R'}(M')$. Apply this to $R = M_m(D)$ and $R' = M_n(D')$. (Btw, in my earlier comment, I wanted to write "the opposite ring $D^{op}$".) $\endgroup$ – Torsten Schoeneberg Oct 6 '17 at 21:57

(I just expanded on the answer of Torsten Schoeneberg in the comments above)

We have the following ring isomorphisms for any division ring $D$:

(i) $\Gamma: D^{o}\rightarrow \text{End}_D(D)=\{D\text{-module endomorphisms of }D\}\\d\mapsto \theta_d\quad \text{s.t.}\quad \theta_d(x)=xd$

(ii) $\Gamma: D^{o}\rightarrow \text{End}_{M_n(D)}(D^n)=\{M_n(D)\text{-module endomorphisms of }D^n\}\\d\mapsto \theta_d\quad \text{s.t.}\quad \theta_d(x_1,\ldots,x_n)=(x_1d,\ldots,x_nd)$

(iii) $\Gamma: M_n(D)^o\rightarrow M_n(D^o)\\A\mapsto A^T$

(iv) $\Gamma: M_n(D^o)\rightarrow \text{End}_D(D^n)\\A\mapsto \theta_A\quad \text{s.t.}\quad \theta_A(x_1,\ldots,x_n)=(x_1,\ldots,x_n).A$

Using the second ring isomorphism, we have the following argument:

Given two division rings $D_i$, $i=1,2$,

$D_i^n$ is the unique simple $M_n(D_i)$-module up to isomorphism, so, since $M_n(D_i)$ are isomorphic as rings by hypothesis, $D_i^n$ are isomorphic as $M_n(D_i)$-modules, so $D_1^o\cong \text{End}_{M_n(D_1)}(D_1^n)\cong \text{End}_{M_n(D_2)}(D_2^n)\cong D_2^o$, so $D_1\cong D_2$.

  • $\begingroup$ You left out the word "simple" in the last paragraph. $\endgroup$ – KCd Jun 21 at 15:25
  • $\begingroup$ Corrected it, thank you. $\endgroup$ – AgentSmith Jun 21 at 17:56

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