October 30th
Today I learned a cute but unsurprising connection in some infinite Galois theory. Basically, we can ask about the Galois group\[\op{Gal}\left(\bigcup_{k=0}^\infty\QQ(\zeta_{p^k})/\QQ\right).\]Note that this kind of makes sense because each finite extension $\QQ(\zeta_{p^k})/\QQ$ is Galois, so we're more or less inducting up to infinity.
So what does this Galois group look like? Well, the motivating observation is that, really, we have\[\QQ(\zeta_{p^\infty}):=\bigcup_{k=0}^\infty\QQ(\zeta_{p^k})\cong\varprojlim\QQ(\zeta_{p^\bullet}).\]And we understand $\QQ(\zeta_{p^\bullet})$ really well. So we might hope that\[\op{Gal}(\QQ(\zeta_{p^\infty})/\QQ)\stackrel?=\varprojlim\op{Gal}(\QQ(\zeta_{p^\bullet})/\QQ)=\varprojlim(\ZZ/p^\bullet\ZZ)^\times=\ZZ_p^\times.\]To formalize this intuition, we begin by fixing some $\sigma\in\op{Gal}(\QQ(\zeta_{p^\infty})/\QQ).$ For any prime power $p^\bullet,$ we can restrict $\sigma$ to an automorphism $\sigma_\bullet$ of $\QQ(\zeta_{p^\bullet})/\QQ.$ Do note that we can identify $\sigma_\bullet$ with\[\sigma_\bullet(\zeta_{p^\bullet})=\zeta_{p^\bullet}^{a_\bullet}\]where $a_n\in(\ZZ/p^n\ZZ)^\times.$ This almost lets us map into $\ZZ_p^\times,$ but we have to also note that\[\zeta_{p^n}^{a_{n+1}}=\zeta_{p^{n+1}}^{pa_{n+1}}=\sigma_{n+1}(\zeta_{p^{n+1}}^p)=\sigma(\zeta_{p^n})=\sigma_n(\zeta_{p^n})=\zeta_{p^n}^{a_n}.\]So we know $a_{n+1}\equiv a_n\pmod{p^n},$ which lets us say that the sequence $(a_\bullet)$ generated by $\sigma$ does indeed live in $\ZZ_p^\times.$ Of course, we can map this backwards again. For any sequence $(a_\bullet)\in\ZZ_p^\times,$ we simply force\[\sigma(\zeta_{p^\bullet})=\zeta_{p^\bullet}^{a_\bullet}\]like before, and this is an automorphism of $\QQ(\zeta_{p^\infty}).$ I guess more formally, we'd have to show that $\sigma(\alpha+\beta)=\sigma(\alpha)+\sigma(\beta)$ and so on, but these hold because $\QQ(\alpha,\beta)$ is a subfield of some sufficiently large $\QQ(\zeta_{p^\bullet}),$ and $\sigma$ projects to an automorphism $\sigma_\bullet$ there. The fact that all of these projections agree to a single automorphism $\sigma$ when brought upwards is exactly the $a_{n+1}\equiv a_n$ discussed above.
So we get a nice statement $\op{Gal}(\QQ(\zeta_{p^\infty})/\QQ)\cong\ZZ_p^\times.$