# Group rings of elementary Abelian p-groups over finite fields

I am interested in the following situation.

Let $p, q$ be two distinct primes.
Let $G$ be an elementary Abelian p-group, say $G = G_1 \oplus \cdots \oplus G_\ell$ for cyclic groups $G_i$ of order $p$, $\ell \geq 1$.

Now consider the group ring $\mathbb F_q[G]$. I am interested in the structure of this ring. What I already found about this is the following:
(1) $\mathbb F_q[G]$ is a semisimple, commutative ring due to Maschke's Theorem and it can be written as a direct sum of finite fields. Moreover, these fields have to be of characteristic q.
(2) Since $G$ is finite and Abelian, $\mathbb F_q[G]$ is a finite commutative ring, and can thus be decomposed into local rings $$\mathbb F_q[G] = e_0\mathbb F_q[G] \oplus \cdots \oplus e_k\mathbb F_q[G],$$ for certain idempotent, non-trivial, pairwise orthogonal elements $e_i$. First question: I guess (1) implies that these local rings are finite fields?

My main question is the following: Is there any way in which the size of the finite fields in the decomposition of $\mathbb F_q[G]$ can be bounded, in terms of $p,q$ only and independent of $\ell$? How do these finite fields look like? They are of the form $\mathbb F_q[X]/f(X)$, but where does the polynomial $f$ come from in this particular case? Or in other words, is there a simple way to express the elements $e_i$ in the decomposition into local rings? In which way does the decomposition of $G$ into cyclic $p$-groups occur in the decomposition of $\mathbb F_q[G]$?

I am very grateful for hints and ideas about any of these points. Thanks a lot!

• Yes, indeed, thanks a lot. q is meant to be the characteristic of the finite field which is different from the prime exponent p of the elementary Abelian group G. Changed it. – yalu Dec 15 '17 at 15:20
• Because $\Bbb{F}_q[C_p]\simeq \Bbb{F}_q[x]/\langle x^p-1\rangle$ the polynomials $f(x)$ that pop up naturally are the irreducible factors of $x^p-1$ in $\Bbb{F}_q[x]$. There will be the obvious linear factor $x-1$, and the other factors all have the same degree $m$ that is also the order of $q$ modulo $p$ (=the degree of the minimal polynomial of primitive $p$th root of unity over $\Bbb{F}_q$. – Jyrki Lahtonen Dec 15 '17 at 16:03

I would approach this question as follows.

First consider the case of a cyclic group: $G=C_p$. In my answer to this old question I explain how you can write $\Bbb{F}_q[C_p]$ as a product of extension fields of $\Bbb{F}_q$. A key parameter is the order $m$ of $q$ in the multiplicative group $\Bbb{Z}_p^*$. Because the cyclic group is here of a prime order we can actually be a bit more precise, namely $$\Bbb{F}_q[C_p]\simeq \Bbb{F}_q\oplus \Bbb{F}_{q^m}^{(p-1)/m}.$$ This is because all the non-trivial $q$-cyclotomic cosets modulo $p$ will have size $m$.

Your group is a product of finitely many copies of $C_p$. In this case we can use the fact that $$\Bbb{F}_q[G\times H]=\Bbb{F}_q[G]\otimes_{\Bbb{F}_q}\Bbb{F}_q[H]$$ the group algebra of a direct product is the tensor product of the group algebras of the factor groups.

Tensor product distributes over direct sums, so all we need to do is to figure the tensor products of the fields that appear in the above decomposition. Tensoring with $\Bbb{F}_q$ obviously won't change anything, so we only need to consider the tensor product of two degree $m$ extensions.

Claim. As $\Bbb{F}_q$-algebras we have an isomorphism $$\Bbb{F}_{q^m}\otimes_{\Bbb{F}_q}\Bbb{F}_{q^m}\simeq \Bbb{F}_{q^m}^m.$$ That is, we just get the $m$-fold Cartesian power of the extension field.

Proof. The field extension is simple so we can write $\Bbb{F}_{q^m}=\Bbb{F}_q(\alpha)$, where $\alpha$ is a zero of some irreducible polynomial $p(x)\in\Bbb{F}_q[x]$ of degree $m$. We have the short exact sequence $$0\to \Bbb{F}_q[x]\to\Bbb{F}_q[x]\to\Bbb{F}_{q^m}\to0,\qquad(*)$$ where the first map is multiplication of polynomials by $p(x)$, and the second map is the evaluation homomorphism $f(x)\mapsto f(\alpha)$. We are to tensor $(*)$ with $\Bbb{F}_{q^m}$ over $\Bbb{F}_q$. The extension field is obviously a flat module. Equally obviously when we tensor the polynomial algebra we just extend the scalars. Therefore we get the exact sequence $$0\to \Bbb{F}_{q^m}[x]\to\Bbb{F}_{q^m}[x]\to\Bbb{F}_{q^m}\otimes\Bbb{F}_{q^m}\to0$$ and the isomorphism $$\Bbb{F}_{q^m}\otimes\Bbb{F}_{q^m}\simeq \Bbb{F}_{q^m}[x]/\langle p(x)\rangle.$$ The extension $\Bbb{F}_{q^m}/\Bbb{F}_q$ is Galois, so $p(x)$ splits into a product of distinct linear factors in the ring $\Bbb{F}_{q^m}[x]$, $$p(x)=(x-\alpha)(x-\alpha^q)\cdots(x-\alpha^{q^{m-1}}).$$ By the Chinese Remainder Theorem $$\Bbb{F}_{q^m}[x]/\langle p(x)\rangle\simeq\bigoplus_{i=0}^{m-1} \Bbb{F}_{q^m}[x]/\langle x-\alpha^{q^i}\rangle.$$ But all the summands in the last form are obviously isomorphic to $\Bbb{F}_{q^m}$. The claim follows.

So your group algebra is the product of a single copy of $\Bbb{F}_q$ and then a bunch of copies of $\Bbb{F}_{q^m}$.

• Well, that wound up being a very ring-theoretic explanation! I thought something pithy from representation theory would emerge. – rschwieb Dec 15 '17 at 16:02
• @rschwieb I feel your pain. This ended up a bit boring :-/ – Jyrki Lahtonen Dec 15 '17 at 16:06
• Actually the theory of characters leads to the same destination. We need an extension field containg the required roots of unity for the algebra to split. Once we have that we get all the non-trivial irreducible reps of $G$ by composing with a suitable outer automorphism. So the same extension field will appear many times over, each replica combining $m$ $\Bbb{F}_q$-conjugates of a given character $G\to\Bbb{F}_{q^m}^*$. – Jyrki Lahtonen Dec 15 '17 at 16:14
• Dear Jyrki, thanks a lot for your thorough explanation! This is very helpful and, in particular, shows what I expected, namely that the order of the fields occurring in the decomposition can be bounded in terms of p and q! Thanks again! – yalu Dec 18 '17 at 9:41

Since $G$ is finite and Abelian, $\mathbb F_q[G]$ is a finite commutative ring, and can thus be decomposed into local rings

This is true, but it doesn't give you any new information. The decomposition into a finite product of fields is already a finite product of local rings.

Is there any way in which the size of the finite fields in the decomposition of $\mathbb F_q[G]$ can be bounded, in terms of $p,q$ only and independent of $\ell$?

This recent question will be helpful, since all of the fields in the decomposition are in fact representatives of the simple modules for the group ring.

Every field $F$ appearing in the decomposition must satisfy $\dim_{\mathbb F_{q}}(F)<|G|$, and the size of $F$ is, obviously, $|F|=|\mathbb F_q|^{\dim_{\mathbb F_{q}}(F)}$

In which way does the decomposition of $G$ into cyclic $p$-groups occur in the decomposition of $\mathbb F_q[G]$?

I guess you are aware of the fact that for any subgroup $H$ of $G$, $\frac{1}{|H|}\sum_{h\in H}h$ is an idempotent of the group ring. So, knowing about the subgroup structure of elementary abelian $p$-groups is going to give you idempotents.

• Unfortunately, I think I'm probably missing some key representation theoretic insights... I have not yet had the opportunity to absorb a good group representation theory course. – rschwieb Dec 15 '17 at 15:27
• Thanks a lot for your answer. But this does only give us a bound on the size of the fields which depends on $\ell$. Your other comment about the idempotents is interesting. If you denote the idempotents for subgroups H by $e_H$, then you get a family of idempotents $e_i$ by considering the subgroups without the $i$th Summand. Does this mean that all fields in the Decomposition are the same? – yalu Dec 15 '17 at 15:36
• @yalu and $q$, because $|G|$ is some power of $q$. – rschwieb Dec 15 '17 at 15:41
• @yalu Unfortunately, I'm not sure how to deduce what is possible for the fields. You would think that given the uniform nature of an elementary abelian $q$ group, the fields would somehow be the same too. I guess Jyrki will be able to shed more light on that. I'm also not confident about how to choose subgroups which decompose how you want. Maybe the cyclic summands of $G$ are the ones to pay attention to. – rschwieb Dec 15 '17 at 15:42