Maximal centralizer in full matrix ring

I will be so thankful if someone can help me with the following question.

Is it possible to obtain all maximal centralizers in the full matrix ring, $M_n(F)$, for an arbitrary finite field $F$? Here, by maximal centralizer I mean: $C=C_R(x)$ is maximal if $C\subseteq C_R(y)$, then $C=C_R(y)$ or $y\in Z(R)$.

In Akbari et al., Linear Alg. App. 390 (2004) 345-355, lemma 3 determines all centralizers with maximum dimension. So some of the maximal centralizers are determined. Is it true that the set of centralizers with maximum dimension and the set of maximal centralizers are equal?

Hamid.

• You probably only care about maximal centralizers that are proper subrings (otherwise $M_n(F)$ is the unique maximal centralizer). Nov 13 '13 at 4:08
• The double-centralizer theorem shows that solving your problem amounts to determining the minimal non-trivial subalgebras of $M_n(F)$ with one generator. Nov 13 '13 at 10:17

Let $x \in M_n(F)$. If the characteristic polynomial of $x$ has distinct prime factors in the ring $F[t]$, then there is some idempotent that commutes with everything that commutes with $x$, hence, if the centralizer is maximal, then it is the centralizer of that idempotent. The centralizer of an idempotent is clearly maximal. So this gives one kind of maximal subgroup.

In the remaining kind, the characteristic polynomial has the form $f(t)^k$ for some irreducible polynomial $f(t)$. Here we split into two cases - either $x$ is semisimple or it isn't. If $x$ is semisimple, its centralizer is a representation $M_k(F')$ where $F'$ is an extension of $F$. The elements which commute with this ring are just the elements of $F'$, and the only ones that have a larger centralizer are the subextensions. So this algebra is maximal if and only if $F'$ is an extension of prime degree. This gives a second kind of maximal subgroup.

If $x$ is not semisimple, we can take the Jordan decomposition $x=x_{ss}+x_n$, and the centralizer of $x$ is contained in the centralizer of $x_n$, but $x_n$ is not in the center of $M_n(k)$, so we may assume $x=x_n$. Then by taking a power we may assume $x^2=0$. In this case, we can check by looking explicitly at the Jordan block that anything that commutes with the centralizer of $x$ must be a linear combination of $1$ and $x$, and so the centralizer of $x$ is maximal.

So there are three cases:

The algebra is the centralizer of an idempotent, the product of two matrix algebras, as in Amitanshu's example.

The algebra is the algebra of matrices over a prime-degree extension of $F$.

The algebra is the centralizer of a nilpotent $x$ satsifying $x^2=0$.

• Will, if the minimal polynomial of $x$ is $f(t)^r$ for some $r>1$, then $F[f(x)]$ is a proper subalgebra of $F[x]$. Nov 13 '13 at 10:15
• @Will, Thanks for your answer. Is there an explicit representation for these cases, specially case 2. Nov 13 '13 at 14:23
• @ClémentdeSeguinsPazzis for your first two comments, Hamid says $F$ is finite, thus perfect with cyclic Galois group. For your third comment, this subalgebra consists of scalar matrices in the semisimple case. In the non-semisimple case, taking that algebra is basically the same as taking $x_n$. But doing that sounds good if the field is not perfect. Nov 13 '13 at 16:04
• @Hamid: Yes, as long as you choose a basis for $F'$ over $F$. Then just make $k$ copies of that basis, and just take all the matrices such that each block (submatrix sending a copy to a copy) is an element of $F'$. Nov 13 '13 at 16:05
• This is published in "Extremal matrix centralizers" (sciencedirect.com/science/article/pii/S0024379512008658) by Dolinar et al. Oct 26 '18 at 13:17

$$\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{pmatrix}$$
It's centralizer algebra has dimension $8$, which is less than the bound $(n-1)^2 + 1 = 10$. But its centralizer is maximal, I think.