Two-sided ideals of a group ring formed by the field with two elements and a simple group

Question

Let $G$ be a simple group. Let $k$ be the field with two elements.

I am interested in the two-sided ideals of $k[G]$.

$k[G]$ at least has the trivial ideals $(0)$ and $(1)$.

There's also a homomorphism $\Delta : k[G] \to k$ be defined as follows:

$\Delta(g_1 + \cdots + g_n) = \text{mod}_2(n)$.

As proof that $\Delta$ is a homomorphism, we consider $-$ and $*$.

$g_1 + \cdots + g_n - (h_1 + \cdots + h_w) = g_1 + \cdots + g_n + h_1 + \cdots + h_w $ . Any element that is duplicated in this sum is equal to 0, but removing a duplicated element preserves the parity of the number of elements in the sum.

For multiplication, first observe that the group elements $G$ additively generate $k[G]$. Next, observe that multiplication of a group element by a group element $g$ is invertible.

From this, we can conclude that $h(g_1 + \cdots + g_n)$ is actually $hg_1 + \cdots + hg_n$, and the elements of the second sum do not contain any duplicates whenever the first sum is duplicate-free.

Exploiting the distributive property, we can see that $(h_1 + \cdots + h_w)(g_1 + \cdots + h_w)$ has parity equal to the parity of $wn$.

So, $\Delta$ is a homomorphism.

The set $\Delta^{-1}(0)$ is a nontrivial ideal of $k[G]$, consisting of precisely the sums of an even number of group elements.

Under what circumstances, if any, can we get an ideal that isn't $(0)$, $(1)$ or $\Delta^{-1}(0)$?

Answer 1

As a warmup, the easiest case to understand explicitly is the case that $G = C_p$ is a cyclic group of odd prime order. Here the Chinese remainder theorem gives

$$\mathbb{F}_2[C_p] \cong \mathbb{F}_2[x]/(x^p - 1) \cong \prod_i \mathbb{F}_2[x]/f_i(x)$$

where $x^p - 1 = \prod_i f_i(x)$ is the irreducible factorization of $x^p - 1$ over $\mathbb{F}_2$. In this case we can completely understand the ideal structure: every ideal is principal, and the ideals are given by the divisors of $x^p - 1$. We always have at least the factorization

$$x^p - 1 = (x - 1)(x^{p-1} + \dots + 1)$$

but $x^{p-1} + \dots + 1$ will not generally be irreducible. As explained e.g. here, the decomposition depends on the order of the Frobenius map $x \mapsto x^2$ acting on the $p^{th}$ roots of unity, or equivalently the multiplicative order of $2 \bmod p$. Writing $d = \text{ord}_2(p)$ for this order, we get that $x^{p-1} + \dots + 1$ is a product of $\frac{p-1}{d}$ irreducible polynomials of degree $d$, which gives

$$\boxed{ \mathbb{F}_2[C_p] \cong \mathbb{F}_2 \times \mathbb{F}_{2^d}^{\frac{p-1}{d}} }.$$

It follows that $\mathbb{F}_2[C_p]$ has exactly $2^{\frac{p-1}{d} + 1}$ ideals; this is the number of subsets of factors (the $2$ base here is unrelated to $\mathbb{F}_2$).

The case $G = C_2$ behaves differently. In this case the group algebra is

$$\mathbb{F}_2[C_2] \cong \mathbb{F}_2[x]/(x^2 - 1) \cong \mathbb{F}_2[x]/(x - 1)^2 \cong \mathbb{F}_2[y]/y^2$$

so we have a nilpotent. The only nontrivial ideal is $(y) = (x - 1)$, which is the augmentation ideal (the kernel of the augmentation map $\Delta$). So we don't get any new ideals in this case.

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Now let's consider the general case that $G$ is finite, simple, and nonabelian. The maximal (two-sided) ideals $m_i$ of $\mathbb{F}_2[G]$ are in natural bijection with the isomorphism classes of irreducible representations $V_i$ of $G$ over $\mathbb{F}_2$; given such a representation we get an action map $\rho_i : \mathbb{F}_2[G] \to \text{End}(V_i)$ whose kernel is a maximal ideal $m_i$, and conversely given a maximal ideal $m_i$ we get a quotient $\mathbb{F}_2[G]/m_i$ which is a simple Artinian $\mathbb{F}_2$-algebra and hence semisimple with a unique simple module over $\mathbb{F}_2$, which is $V_i$.

Under this bijection the augmentation ideal corresponds to the trivial representation. So to exhibit a different maximal ideal it suffices to show that $G$ has at least one nontrivial irreducible representation over $\mathbb{F}_2$. There are probably simpler arguments for this but here's the first one that comes to mind:

$G$ has at least one nontrivial representation over $\mathbb{F}_2$, namely the regular representation $V$ given by the action on $\mathbb{F}_2[G]$ itself. $V$ has a composition series exhibiting it as an iterated extension of irreducible representations (we need to talk about extensions here because $|G|$ is not necessarily odd so $\mathbb{F}_2[G]$ is not necessarily semisimple; in fact $|G|$ must be even by the Feit-Thompson theorem). If all of these irreducibles are the trivial representation (which can happen if $G$ is not simple, e.g. if $G = C_{2^k}$) then $G$ preserves a complete flag in $V$ and hence is conjugate to a subgroup of the Borel subgroup of upper-triangular $n \times n$ matrices over $\mathbb{F}_2$, where $n = \dim V = |G|$. But this group is solvable and hence so is any subgroup of it; taking the contrapositive, if $G$ is simple nonabelian then it cannot be conjugate to such a subgroup, so the composition series of $V$ contains at least one nontrivial irreducible as desired. In conclusion:

If $G$ is a finite simple group, then $\mathbb{F}_2[G]$ has at least one ideal other than $(0), (1)$, or the augmentation ideal unless $G = C_2$.

This argument can be refined to show that if $G$ is finite (but not necessarily simple) then $G$ has at least one nontrivial irreducible representation over $\mathbb{F}_2$ unless $G$ is a $2$-group.