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Exercise III.2.11 (Aluffi, Algebra Ch 0): Let $R$ be a division ring consisting of $p^2$ elements, where $p$ is a prime. Prove that $R$ is commutative (and thus $R$ is a field).

Note: I do so without invoking Wedderburn's Theorem, that every finite division ring is a field. Also, within the context of this text, Rings must include unity.

Proof

Assume towards a contradiction that $R$ is not commutative. Then it's center, $C$, is a proper subring of $R$. Now, since $|R| = p^2$, $|C| = p$, as subrings are also subgroups and thus by Lagrange's Theorem $|C|$ must divide $|R|$. Notice that since $0_R, 1_R \in C$, $|C| > 1$.

Now, let $r \in R, r \not \in C$ be arbitrary. Notice that the centralizer of $r$ contains both $r$ and $C$, as r commutes with itself and the center is the intersection of all centralizers. Thus the centralizer must contains more than $p$ elements. Since the centralizer is also a subring, this forces the centralizer to be all of $R$. Lastly, since $r$ was arbitrary it follows the centralizer for every element is all of $R$. Therefore, the center of $R$ is $R$ and thus we have a contradiction and it follows that $R$ is commutative.

Personal Notes: I feel that this proof is a little wordy, but beyond that I'm curious for any advice on if there is a necessity for more clarity or if there are any elements that should be cut from the proof.

BabyDietCoke
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    I think it's a fine proof, and gets to the point. A possibility that occurred to me would be to observe that everything must be $p$-torsion (additively) so that $R$ is also a 2-dimensional vector space over $\Bbb{Z}_p$, hence also a $\Bbb{Z}_p$-algebra. It thus has a vector space basis of the form ${1,x}$ for some $x\in R$. Obviously the basis elements commute, and the claim follows from bilinearity of the product. It is questionable whether this approach saves lines. Depends on how familiar you are with the concept of algebras over a field, I suppose. – Jyrki Lahtonen Feb 17 '20 at 18:44

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