# If the matrix rings are isomorphic, then the scalars rings are isomorphic

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.

• 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. – Torsten Schoeneberg Oct 6 '17 at 1:39
• 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? – M. Wolf Oct 6 '17 at 1:54
• In fact, $\Delta=D$. – M. Wolf Oct 6 '17 at 2:05
• 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. – Torsten Schoeneberg Oct 6 '17 at 3:15
• 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 $D^{op}$".) – 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 $$D_1\cong D_2$$.

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