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!




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 $\varepsilon\colon \mathbb{C}[D_8]\to \mathbb{C}$.

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

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