I have a small exercise I would like to get help with:
Let $K$ be a field and $K^{ab}$ be the composite of all finite abelian extensions of $K$.
Prove $K^{ab}/K$ is abelian and that if $L/K$ is abelian then $L$ can be embedded into $K^{ab}$ (note that $L$ can be infinite)
What I tried:
I think if we let $\{F_{j}\}_{j\in J}$ be all finite abelian extensions of $K$ then we can define a map $$ Gal(K^{ab}/K)\to\Pi_{j\in J}Gal(F_{j}/K) $$
by taking $\sigma\in Gal(K^{ab}/K)$ to $\sigma'$ that acts as $\sigma|_{Fj}$ on each $F_{j}$, since this map is injective if follows that $Gal(K^{ab}/K)$ can be embedded into $\Pi_{j\in J}Gal(F_{j}/K)$ which is abelian.
So we get that $Gal(K^{ab}/K)$ is isomorphic to a subgroup of a product of abelian groups and thus abelian.
Is that correct ?
Can someone help me showing second part of the exercise ? I think it may be possible to use the classification of abelian groups, but I don't see how
I had a much longer answer prepared, but I think that this should suffice. If you want justification of my argument, I’ll expand this answer.
You’re really asking whether the compositum of all finite abelian extensions of $K$ is equal to the compositum of all abelian extensions of $K$, whether finite or not. But all our extensions are algebraic, and every infinite algebraic extension is the compositum of its finite subextensions. So the compositum of all abelian extensions of $K$ is also the compositum of all finite abelian extensions.