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I am getting quite confused about the distinction between algebra homomorphisms and $A$-module homomorphisms, where $A$ is an algebra. If $A=\mathbb CG$, the group algebra, then I have a result in my representation theory course which says that:

If $G$ is abelian, then all irreducible $\mathbb C G$-modules are 1-dimensional.

The proof is that there are $n=|G|$ conjugacy classes, so $n$ irreducible modules and since $\mathbb C G \cong \bigoplus_i S_i^{\oplus \dim(S_i)} $ as $\mathbb CG$-modules ($\mathbb CG$ the regular $\mathbb CG$-module) with the $S_i$ irreducible, a dimension count says that all the $S_i$ are 1-dimensional.

Then, later, we prove that:

If $G$ is abelian, then $\mathbb CG\cong\mathbb C^{\oplus n}$ as algebras

The proof of this is a corollary of the fact that any semi-simple algebra is isomorphic to a direct sum of matrix algebras over $\mathbb C$.

These results seem very similar, since a corollary to the first result says that $\mathbb CG\cong\bigoplus_i S_i$ as $\mathbb CG$-modules and then each $S_i$ is isomophic to $\mathbb C$ as a vector space. I'm wondering:

Does one of these results prove the other, or are isomorphisms of $\mathbb CG$ modules and of algebras are just completely separate things and should be treated as such.

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  • $\begingroup$ @Engloutie For me an algebra hom $f$ is such that $f(ab)=f(a)f(b)$ plus some other requirements. But for an $A$-module hom, $g$, $g(am)=a\cdot g(m)$. Why is this the same? $\endgroup$
    – James
    Commented May 6, 2015 at 15:17
  • $\begingroup$ Sorry, it doesn't make sense the way I said it. I'm deleting it to avoid confusion! $\endgroup$
    – Thomas Poguntke
    Commented May 6, 2015 at 15:28

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An algebra is basically a module with a concept of multiplication. So think for example of $\mathbb R^2$. That is a vector space (module) over $\mathbb R$. But if you think of it as the complex plane $\mathbb C$ you can endow $\mathbb R^2$ with a meaningful multiplication. That makes it a module (over $\mathbb R$) but also something more, it has more structure, such a thing is called an algebra. If you prove isomorphism as algebras, you have to show the function respects the multiplication. And in that case you have shown isomorphism as modules because an algebra is a module, plus more.

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  • $\begingroup$ I may be using a slightly different definition of a module and algebra as you are as I don't completely understand. It would help if you could show me how to form a $A$-module isomorphism $g$ from some algebra isomorphism $f:A \to B$ where $A$ and $B$ are algebras. $\endgroup$
    – James
    Commented May 6, 2015 at 15:24
  • $\begingroup$ Is your definition of algebra different from the one on this wiki page? en.wikipedia.org/wiki/Algebra_%28ring_theory%29 $\endgroup$
    – Gregory Grant
    Commented May 6, 2015 at 15:25
  • $\begingroup$ Because if so then an $A$-algebra isomorphism is automatically an $A$-module isomorphism. $\endgroup$
    – Gregory Grant
    Commented May 6, 2015 at 15:26
  • $\begingroup$ That's basically what I tried and failed to say above. The $A$-module structure on $B$ is given by $a.b = f(a)b$. $\endgroup$
    – Thomas Poguntke
    Commented May 6, 2015 at 15:31
  • $\begingroup$ I think it's the same definition, but in lectures an algebra was defined first and then a module over an algebra was defined. To build this up, is it that a module over a ring is defined, then that is used to define an algebra and then you go a further stage up and get a module over an algebra? That seems really confusing. For me an algebra hom $f$ is such that $f(ab)=f(a)f(b)$ plus some other requirements. But for an $A$-module hom, $g$, $g(am)=a\cdot g(m)$. Why is this the same? (Here $a, b \in A$ and $m$ is in the module.) $\endgroup$
    – James
    Commented May 6, 2015 at 15:32

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