December 7th
Today I learned that all fields complete with respect to an archimedean valuation are either $\RR$ or $\CC.$ The proof reminds of the reasons why I don't like analysis very much, so some technical steps will be jumped over. Indeed, fix $K$ a field complete with respect to some archimedean valuation named $|\bullet|.$ To start, note that $K$ must have characteristic $0$ because otherwise\[\{|k|:k\in\ZZ\}\]is finite and therefore bounded, breaking $|\bullet|$ being archimedean. Taking quotients, we also get a copy of $\QQ$ in $K$ (explicitly, $m/n\mapsto (m\cdot1)/(n\cdot1)$), and because $K$ is complete under $|\bullet|,$ we also get all Cauchy sequences of $\QQ,$ which means that $K$ has a copy of $\RR.$
We'd like to work in $\CC$ because $\CC$ has some nicer structure, so we look at either $K$ or $K(i)$ depending on if $T^2+1$ is irreducible in $K$ or not. There is some technicality in that we want to be sure that $K(i)$ is still complete with respect to an extension of $|\bullet|.$ We extend this by writing\[|a+bi|_{K(i)}=\sqrt{\left|a^2+b^2\right|_K}.\]This function detects $0$ because $|a+bi|=0$ would imply $a^2+b^2=0,$ for which $b\ne0$ would imply $(a/b)^2=-1,$ implying $T^2+1$ reducible. Brahmagupta's identity tells us that this function is multiplicative. The hard part, of course, is verifying the triangle inequality. I'm going to omit this because the argument isn't very much fun.
The fun part of the argument is doing what we can when $\CC$ is inside of $K.$ We want to show $K\subseteq\CC$ to complete the proof. Well, suppose for the sake of contradiction that $\alpha\in K\setminus\CC.$ To detect contradiction, we focus on the function $z\mapsto|z-\alpha|.$ This function is continuous, for reasons we will gloss over. Further,\[|z-\alpha|\ge|z|-|\alpha|,\]so as $|z|\to\infty,$ we have that $|z-\alpha|\to\infty$ as well. It follows that the function has an achieved minimum. Namely, over any closed disk, the function surely achieves its minimum somewhere by continuity. Sending the size of the closed disk to infinity, the only way for there not to be an absolute minimum is for the function to periodically get smaller and smaller, which violates $|z-\alpha|\to\infty.$
We're going to show that the set of points that the function achieves its minimum is open, which will readily finish the proof. Suppose that the function achieves its minimum at $a,$ and we will show that there is a neighborhood of points around $a$ which also minimize $|z-\alpha|.$ Fix $\beta=a-\alpha$ so that $\alpha\notin\CC$ implies $|\beta| \gt 0.$ Our neighborhood is going to have radius $|\beta|,$ so fix any $\varepsilon$ with $0 \lt |\varepsilon| \lt |\beta|.$ For some $n\in\ZZ^+$ to be sent to infinity later, note\[\frac{\beta^n-\varepsilon^n}{\beta-\varepsilon}=\prod_{k=0}^{n-1}\left(\beta-\varepsilon\zeta_n^k\right).\]Surely $|\beta-\varepsilon\zeta_n^\bullet|\ge|\beta|$ because $\beta=a-\alpha$ minimizes $|z-\alpha|.$ But then we may rewrite the above as\[\frac{|\beta-\varepsilon|}{|\beta|}\le\frac{\left|\beta^n-\varepsilon^n\right|}{|\beta|^n}=\left|1-\left(\frac\varepsilon\beta\right)^n\right|.\]As $n\to\infty,$ we are forced into $|\beta-\varepsilon|\le|\beta|,$ so combining this with $|\beta-\varepsilon\zeta_n^\bullet|\ge|\beta|,$ we indeed see that $\beta+\varepsilon$ is also minimizing $|z-\alpha|.$
However, the solution set to a continuous function equal to a value is closed, so the fact that it's shown to also be open (and nonempty) implies that it must be $\CC.$ (Visually, we can imagine extending the open neighborhood around our minimum by $a$ by successive $|\beta|$ increments to cover all of $\CC.$) In particular, the function\[z\mapsto|z-\alpha|_K\]is constant over all of $\CC.$ However, this violates the archimedean property, as we showed earlier that $|z|\to\infty$ must imply $|z-\alpha|\to\infty$ as well. This contradiction completes the proof.