September 13th
Today I learned that the Cayley-Hamilton Theorem can sometimes be used as a lemma (!) instead of just always a theorem, from one of Tom's riddles. Ready yourself; this is exceedingly cool. The statement is that for a finite extension $K/\QQ,$ given the existence of an embedding of rings\[\varphi:K\hookrightarrow\QQ^{n\times n}\]for some $n\in\NN,$ then $[K:\QQ]\le n.$ In particular, $K\cong\varphi(K).$ Intuitively, we feel this result as a size constraint: as an embedding of groups, an idiot degree bound says that $[K:\QQ]\le n^2.$ Somehow the extra ring structure is sharpening this bound in a natural way.
To set up the proof, we recall that every finite extension is generated by a single element, so fix $K=\QQ(\alpha).$ This is nice because it tells us that $\varphi$ is entirely determined by $\varphi(\alpha)=:\bf A,$ so we may focus on $\bf A$ alone. In particular, we can use the Cayley-Hamilton Theorem (!) we know that $\bf A$ is a root of\[p(x)=\det({\bf A}-x{\bf I})\]extended to $\QQ^{n\times n}.$ But $\bf A$ is an $n\times n$ matrix as is, so $\deg p=n.$ Then, using our isomorphism $K\cong\varphi(K),$ we see that $\alpha=\varphi^{-1}(\bf A)$ is also a root of $p(x).$
To convince you that this argument is cool, we remark that this implies that no matter what our embedding $\varphi$ looks like, the fact $p(x)\in\QQ[x]$ implies that all Galois conjugates of $\alpha$ are eigenvalues of $\bf A.$ To reiterate, we know nothing about what $\varphi$ looks like a priori, but its structure is still very strangely forced by its input field $K.$
Anyways, it remains to finish the argument. We know that $\alpha$ is a root of $p$ of degree $\deg p=n,$ so $\alpha^n$ can be written as a $\QQ$-linear combination of lesser powers of $\alpha.$ But then the basis $\left\{1,\alpha,\alpha^2,\ldots\right\}$ of $K$ has no more than $n$ elements, from which $[K:\QQ]\le n$ follows.