# Simple modules of the group algebra $kC_p$ over different fields $k$

I want to find the simple modules for the group algebra $A=kC_7$, where $k$ is a field and $C_7$ is the cyclic group of order $7$.

When $k = \mathbb{C},$ by Maschke's Theorem $A$ must be semisimple. Then by Artin-Wedderburn Theorem, $A$ is isomorphic to a direct product of matrix rings, i.e. $$A = M_{n_1}(\mathbb{C})\times \cdots \times M_{n_s}(\mathbb{C})$$ But $A$ is a finite dimensional commutative algebra over $\mathbb{C}$, so every simple $A$-modules must be $1$-dimensional over $\mathbb{C}.$ Now $A$ is semisimple, so it has exactly $s$ simple modules up to isomorphism such that $\dim_\mathbb{C}M_i = n_i.$ So $n_i = 1$ for all $i$, i.e. $M_{n_i}(\mathbb{C}) \cong \mathbb{C}.$ Comparing dimensions we have that $A \cong \mathbb{C}^7.$ I believe there is nothing wrong with this?

What happens when $k = \mathbb{R}?$ I suppose there should be the trivial module $\langle x \rangle$ where $v_g.x = x$ for all $v_g \in \mathbb{R}C_7.$ I also know that $\mathbb{R}C_7$ is semisimple. But other than that, I have no idea how to proceed, because most of the things I learned only works for algebraically closed fields. (Perhaps I could try to find the complement of the trivial module, but I'm not too sure how to do that..) Would the situation change if I let $k = \mathbb{Q}?$

Finally, suppose $k = \mathbb{F}_7,$ so $A = \mathbb{F}_7C_7.$ Then $A \cong k[x]/(x^7-1).$ But $x^7-1 = (x-1)^7$ in $k$, so the only maximal ideal of $k[x]/(x^7-1)$ is $(x-1)$ (well, mod the quotient, of course). This corresponds to $v_g-v_e$ in $A$, where $e$ is the identity of $C_7$ and $e \neq g \in C_7$. So the only simple module would be $A/(v_g-v_e),$ but what would this actually look like inside $A$? Furthermore, is there a way to figure out the simple modules without resorting to quotienting by $k[x]?$

There is a good way to calculate a complete set of primitive orthogonal idempotents of the group algebra $kC_p$. I suppose you know what is the complete set of primitive orthogonal idempotents of a group algebra. You can see the book Elements of the representation theory of associative algebras written by I. Assem, D. Simson, A. Skowronski. On page 32, by Corollary 5.17, you can find all the simple modules by the primitive orthogonal idempotents of a group algebra.

I know we can get $p$ primitive orthogonal idempotents of the group algebra $kC_p$ (which is related to the irreducible complex character of $C_p$), if $k=\mathbb{C}$. So $\mathbb{C}C_7$ has $7$ simple module. You can check that $e_i\mathbb{C}C_7\ncong e_j\mathbb{C}C_7$, for all $i\neq j$, where $\{e_1,e_2,...e_7\}$ is a complete set of primitive orthogonal idempotents. Then $\mathbb{C}C_7$ is a basic algebra and so its simple module is one-dimensional. If we let $k=\mathbb{R}$ or $\mathbb{Q}$,or other fields, the number of primitive orthogonal idempotents may less than $p$ by directly calculate. You can try it.

I'll start with the case of $\mathbb{C}[C_7]$. As you've noticed, this is ismorphic to $\mathbb{C}[x] / (x^7 - 1)$. Since $x^7 - 1$ splits into linear factors over $\mathbb{C}$, by the chinese remainder theorem the ring $\mathbb{C}[x] / (x^7 - 1)$ is isomorphic to a direct product of $7$ copies of $\mathbb{C}$, and so you can see that any simple module must be one-dimensional.

If we now try to do the same thing with $\mathbb{R}[C_7]$, the polynomial $x^7 - 1$ splits into one linear factor, and three quadratic factors. Hence $\mathbb{R}[C_7]$ is isomorphic to a direct product of one copy of $\mathbb{R}$ and three copies of $\mathbb{C}$ (each considered as a 2-dimensional $\mathbb{R}$-algebra). So here, simple modules may be two-dimensional, if they are modules over one of the $\mathbb{C}$ factors: and easy examples of these are mapping a generator $x \in C_7$ to a $2 \times 2$ rotation matrix $X$ such that $X^7 = 1$.

If you want to know more about how to classify real representations using complex representations, there are some good answers on this site, for example this one. The basic idea is to take a real representation, think of it as a complex representation, and then look at how it decomposes over $\mathbb{C}$. You'll also notice that the polynomial $x^7 - 1$ splits into one linear factor and one degree six factor over $\mathbb{Q}$, and so you can expect some Galois theory to be involved: see this answer

Finally, in the finite field case things get more complicated, and I can't give a good answer here. You may want to try searching for "modular representations of cyclic groups".

• For the cyclic group of order 7 over a field of characteristic 7 there are no difficulties, really: the group algebra is isomorphic to k[x]/(x^7), which is a Nakayama algebra May 26, 2017 at 3:10