2
$\begingroup$

Good evening,

I was wondering whether somebody could point me in the right direction here. Why does the existence of a surjective ring homomorphism $\mathbb{C}[D_8]\rightarrow\mathbb{C}:g\in D_8\mapsto 1\in \mathbb{C}$ imply that $\mathbb{C}[D_8]$ has at least one simple module of dimension 1?

It has been given as a hint, but I don't understand where it has come from.

I am writing $\mathbb{C}[D_8]$ as a product of simple $\mathbb{C}$-algebras.

So far I have used Maschke's theorem combined with the Artin-Wedderburn theorem and the fact that the only finite-dimensional division algebra over $\mathbb{C}$ is $\mathbb{C}$ itself and deduced that either

$$\mathbb{C}[D_8]\cong \mathbb{C}\times \mathbb{C}\times \mathbb{C}\times \mathbb{C}\times M_2(\mathbb{C})$$ or $$\mathbb{C}[D_8]\cong M_2(\mathbb{C})\times M_2(\mathbb{C}).$$

(I know $\mathbb{C}[D_8]\ncong \mathbb{C}^8$ because the LHS is not commutative whereas the RHS is).

Now, using the hint would immediately give the former as the correct decomposition. I just don't understand where it has come from. Could anybody point me in the right direction here? I don't like using things that I don't fully understand!

Thanks,

Andy.

$\endgroup$
1
$\begingroup$

There is nothing special about this particular homomorphism. Whenever you have a ring homomorphism $\varphi\colon R\to S$, then any $S$-module $M$ becomes an $R$-module via $r\cdot m:=\varphi(r)m$. In particular, in your situation the $1$-dimensional simple $\mathbb{C}$-module $\mathbb{C}$ gets a $\mathbb{C}[D_8]$-structure via the counit map $\varepsilon\colon \mathbb{C}[D_8]\to \mathbb{C}$.

$\endgroup$
  • $\begingroup$ Thank you, using restriction of scalars hadn't even crossed my mind! $\endgroup$ – andrew1601 Nov 25 '16 at 13:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.