October 24th
Today I learned the proof of Wedderburn's Theorem showing that finite domains $D$ are all fields. Namely, taking not-necessarily-commutative ring axioms plus the integral domain axiom, we must have a field if finite. Quickly, every nonzero element $a\in D\setminus\{0\}$ has an inverse by considering $\langle a\rangle,$ which by pigeonhole must eventually overlap with itself. As soon as $a^x=a^y$ with $x \gt y,$ then integral domain lets us cancel to say\[a\cdot a^{x-y-1}=1,\]providing us with our inverse.
The hard part, then, is showing that $D$ is commutative. The proof is actually quite nice. There's an algebraic step, which is basically writing down the class equation, and then there's an analytic step, which is basically bounding cyclotomic polynomials.
Because we're interested in commutativity information, we'll use the conjugacy class equation, under multiplication. Namely, make $D^\times$ act on itself by conjugation. Orbits of individual elements are conjugacy classes (by definition), so we get to write\[|D^\times|=\sum_{\mathcal C\text{ class}}|\mathcal C|.\]The elements with conjugacy class size $1$ are the center $Z(D^\times),$ so\[|D^\times|=|Z(D^\times)|+\sum_{\mathcal C\text{ nontrivial class}}|\mathcal C|.\]Considering the orbit-stabilizer theorem, we write this as\[|D^\times|=|Z(D^\times)|+\sum_{[x]\text{ nontrivial class}}|D^\times x|.\]But the number of elements in the orbit of $x$ is the index of $D^\times$ of the subgroup of $D^\times$ fixing $x$ (by the Orbit-Stabilizer Theorem), and the set of elements fixing $x$ is its centralizer. So we get to write\[|D^\times|=|Z(D^\times)|+\sum_{[x]\text{ nontrivial class}}\frac{|D^\times|}{|C(x)|}.\]
We can actually put some values on each of the quantities of this class equation. Namely, using strong induction on the size of our domain (and that $2$-element unitary rings are fields as our base), we can assume that all finite domains smaller than $D$ are fields. Now, $Z(D)$ already commutes with itself, so it's a finite field of size $q.$ It follows that we can view $D$ and $C(x)$ for any $x\in D^\times\setminus Z(D^\times)$ both as $Z(D)$-vector spaces, of size $q^n$ and $q^d$ for positive integers $n$ and $d.$ Plugging this all in to the conjugacy class equation (and remembering to convert to the multiplicative groups) gives\[q^n-1=(q-1)+\sum_{[x]\text{ nontrivial class}}\frac{q^n-1}{q^d-1}.\]Note that $C(x)\subsetneq D^\times$ implies that $d \lt n.$ In particular, $C(x)=D^\times$ is equivalent to $x\in Z(D).$
That is the conjugacy class equation step. However, note that the cyclotomic polynomial $\Phi_n(x)$ divides $x^n-1$ and even each $\frac{x^n-1}{x^d-1}$ because $d \lt n.$ Throwing this at our class equation tells us\[\Phi_n(q)~\left|~\left(q^n-1\right)-\sum_{[x]\text{ nontrivial class}}\frac{q^n-1}{q^d-1}=q-1\right.\]as polynomials in $q$ and hence also for integers $q.$ But if $n \gt 1,$ then\[|\Phi_n(q)|=\prod_{\gcd(k,n)=1}\left|q-\zeta_n^k\right| \gt |q-1|\]because of the placement of $\zeta_n^\bullet$ on the unit circle. This conflicts with $\Phi_n(q)\mid q-1,$ so we must have $n=1$ instead. Thus, $D=Z(D),$ and we're done here.