# How to determine irreducible modules of $\text{C}_5$ over $\mathbb{Z}_2$?

Since $2\nmid 5$ and $5=1^2+2^2$, we can apply Maschke's theorem: $\mathbb{Z}_2\text{C}_5=V_1\oplus(V_2\oplus V_3)$ with $V_1$ the trivial module and $V_2,V_3$ 2-dimension irreducible modules. But I have trouble in finding $V_2,V_3$.

You cannot apply Mashke's theorem in this way as $\mathbb{Z}_2$ is not algebraically closed. In fact, you have $$\mathbb{Z}_2C_5\cong \mathbb{Z}_2[x]/(x^5-1)\cong\mathbb{Z}_2[x]/(x+1)\oplus\mathbb{Z}_2[x]/(x^4+x^3+x^2+x+1)$$ where the second isomorphism is the Chinese remainder theorem. We can define an action of $C_5=\langle c\rangle$ on the right hand side or the isomorphisms above by letting $c$ act by multiplication by $x$.
The first summand, $\mathbb{Z}_2[x]/(x+1)$, is the trivial module.
The summand $\mathbb{Z}_2[x]/(x^4+x^3+x^2+x+1)$ is also irreducible and 4-dimensional over $\mathbb{Z_2}$ (it is easy to check that $x^4+x^3+x^2+x+1$ is irreducible over $\mathbb{Z}_2$ as it has no linear or quadratic factors). The matrix for the action of $c$ on this module (in the basis $1,x,x^2,x^3$) is $$c\mapsto\begin{pmatrix}0&0&0&1\\1&0&0&1\\0&1&0&1\\0&0&1&1\end{pmatrix}.$$
• In your expression of the Chinese remainder theorem, the second part should be $\mathbb{Z}_2[x]/(x^4+x^3+x^2+x+1)$ or $\mathbb{Z}_2/(x^4+x^3+x^2+x+1)$ as you wrote above? Commented Mar 30, 2016 at 11:08