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I found an answer to my bigger question here, but I'm curious about my attempted proof in the case where $|G|=p$. I'm nearly certain this does not work, but I can still learn something from it. Do I have the right idea at all?

Let $p$ be a prime, $P=\langle c \rangle$ a group of order $p$, and $F$ a field of characteristic $p$. I want to show that the only irreducible representation of $P$ over $F$ is the trivial representation. I am told to do this "using the fact that $F$ contains all $p$th roots of $1$ (namely $1$ itself)."

I am not sure about how to use the $p$th roots (just $1 \in F$), but here's what I did:

Let $A$ be a left $FP$-module, where $FP$ is the group ring of $P$ over $F$. Also, let $P$ be a left $F$-module. Take an element $0 \ne g \in A$ and consider the elements $g, (1c)g, (1c^2)g, ..., (1c^{p-1})g$. Then these elements form a nontrivial proper(?) submodule of $A$ since for an arbitrary $a_1c^{j_1}+\cdots + a_pc^{j_p} \in FP$, we have $(a_1c^{j_1}+\cdots + a_pc^{j_p})(1c^ig)=\cdots=(a_1c^{i+j_1} + \cdots + a_pc^{i+j_p})g = c^kg$ for some $c^k \in P$, since $P$ is an $F$-module. It follows that $A$ is reducible, and we have proved the contrapositive.

Thanks.

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Alex Petzke
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1 Answers1

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The issue in your argument is that a cyclic submodule need not be a proper one.

Any module contains a cyclic submodule (pick an element and form the submodule it generates), and any cyclic $R$-module is isomorphic to a quotient $R/I$. Use $F[C_p]\cong F[X]/(X^p-1)$ and the factorization of $X^p-1$ in characteristic $p$ to determine all ideals and hence all quotients of $F[C_p]$ and determine which one has no proper submodule (equivalently, which quotient is a field).

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