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I am trying to understand part of this book I’m reading and it’s trying to find the simple $KC_3$-modules where $K$ is a field of characteristic 2. It says ‘$KC_3$ is semisimple and if $K$ contains a primitive cube root of unity $\omega $, there are 3 1-dimensional simple representations, on which the generator of $C_3$ acts as $1, \omega,$ or $ \omega^2 $.

I am trying to understand how they come to this. Firstly, how is it the case that the simple modules are necessarily 1-dimensional, and secondly, what does $KC_3$ being semisimple have to do with its simple modules?

Anonmath101
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  • Given a simple $R$-module $M$ take $m\ne 0\in M,f:R\to M,f(r)=rm$, $f(R)=M$ so $M\cong R/\ker(f)$. If $R$ is semisimple as $R$-module then $\ker(f)$ has a supplementary submodule $R= \ker(f)\oplus A$ and $M\cong R/\ker(f)\cong A$ is a simple submodule of $R$. – reuns Feb 23 '23 at 16:39
  • How does this apply here? – Anonmath101 Feb 23 '23 at 16:43

2 Answers2

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The fact that simple modules are dimensional boils down to linear algebra: a (finite dimensional) $K$-linear representation of $C_3$ is exactly the same as a (finite dimensional) $K$-linear vector space $V$ and a linear transformation $T: V \longrightarrow V$ such that $T^3 = 1$.

If $K$ contains a primitive $3$rd root of $1$, call it $\omega$, then $X^3 - 1$ splits completely into $(X-1)(X-\omega)(X-\omega^2)$ and is thus diagonalizable. As such, it admits a $T$-invariant one dimensional subspace, and hence if $V$ is simple it must coincide with this one dimensional $C_3$-submodule.

This also proves the claim you are quoting, since on such an eigenspace, $T$ acts by multiplication by one of $1$, $\omega$ or $\omega^2$. Note that $KC_3$ being semi-simple is independent of $K$ being algebraically closed: this is true because the characteristic of $K$ is coprime with the order of $C_3$, so Maschke's Theorem applies.

If $K= \mathbb F_2$, on the other hand, notice that the $KC_3$-module breaks down into the direct product $K \times K[\alpha]$ where $\alpha^2+\alpha+1=0$, and $K[\alpha]$ is a simple two dimensional $KC_3$-module since $t^2+t+1$ has no roots in the base field $K$.

Pedro
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there are 3 1-dimensional simple representations [...] how is it the case that the simple modules are necessarily 1-dimensional

There aren't always. Sometimes there is one $1$ dimensional and one $2$ dimensional, as discussed in Pedro's solution and in your other question.

what does $_3$ being semisimple have to do with its simple modules?

Maybe the most practical thing is that in a semisimple ring, every simple module appears as a direct summand of the ring. This is not always the case (just consider $\mathbb Z$, for example.)

rschwieb
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