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Let $K$ be a field and let $C_3$ be cyclic group of three elements. I am reading part of this book that seems to use the fact that if $\text{char}(K)=2$ then $KC_3$ is a semisimple ring and so simple modules over $KC_3$ have dimension 1 over $K$ or equivalently simple representations have degree 1. Is this generally true, if $KG$ is semisimple (possibly do we need to be Artinian too?) that any simple $KG$-module is 1 dimensional as a vector space over $K$?

Anonmath101
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  • The group ring $\mathbb C[S_3]$ is semisimple but has an irreducible $2$-dimensional module. (the standard representation of $S_3$.) – Kenta S Feb 27 '23 at 19:40
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    The essential fact used here is that $G$ is abelian, so (irreducible) $G$-representations are $1$-dimensional. – Kenta S Feb 27 '23 at 19:41
  • Ah, the book mentioned the fact it’s semisimple, but it’s not important here? – Anonmath101 Feb 27 '23 at 19:42
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    It is not true that simple modules over $KC_3$ have dimension $1$ over $K$, if $K$ is an arbitrary field of characteristic $2$. (In particular, it is not true for $K=\mathbb{F}_2$.) – Eric Wofsey Feb 27 '23 at 19:59
  • @KentaS Were you thinking of the case when $K$ is an algebraically closed field? – rschwieb Mar 07 '23 at 18:33

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I am reading part of this book that seems to use the fact that if $char()=2$ then $_3$ is a semisimple ring and so simple modules over $_3$ have dimension $1$ over $$ or equivalently simple representations have degree $1$.

In any case, $KC_3\cong K[x]/(x^3-1)$. This is going to wind up differently depending on the field $K$ because $x^3-1$ will factor differently.

Take for example $K=F_2$, the field of two elements. Since $(x-1)(x^2+x+1)$ is a factorization into irreducibles, we have that $KC_3\cong F_2\times F_4$ using the Chinese remainder theorem. It has two simple modules, one being $1$ dimensional and the other $2$ dimensional over $K$.

If $K=F_3$, it factors into $(x-1)^3$, and $KC_3$ is not semisimple at all, it's local. But indeed it does have one single simple module and that one has $K$ dimension $1$.

If $K=F_4$, $x^2+x+1$ actually factors into linear factors so that $KC_3=F_4\times F_4\times F_4$. (Note this is characteristic 2 as well.) If $\alpha$ is a primitive element, then $\alpha\neq \alpha^{-1}$ and $\alpha+\alpha^{-1}=1$, so $(x+\alpha)(x+\alpha^{-1})=x^2+(\alpha+\alpha^{-1})x+\alpha\alpha^{-1}=x^2+x+1$.

You get dimensions $1$ and $2$ for simple modules in the case of $K=F_5$, and for $K=F_7$ there are only $1$ dimensional simple modules again. (The common reason for $2,5,7$ can be found here.) This explains all the $F_pC_3$ cases.

One thing to note is that there will always be at least one simple module that is $1$ dimensional for $KG$ because the quotient by the augmentation ideal is $1$ dimensional. But as demonstrated above it is not even consistent for a single characteristic.

possibly do we need to be Artinian too?

Most people mean "semisimple artinian" when they say semisimple, nowadays. At any rate it is superfluous here because $KC_3$ is $3$ dimensional, and so obviously Artinian. Maschke's theorem tells us the ring will be semisimple precisely when $|G|$ is a unit in $K$, and that's why it applies in characteristic $2$ for $C_3$.

rschwieb
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  • Note that the quotient by the augmentation ideal is equivalently just the trivial $G$-representation of $K$! – Kenta S Mar 07 '23 at 20:20