If we have $F_n\geq\ldots\geq F_1$ fields such that $F_{i+1}$ is a separable splitting field over $F_i$ for all $i=1,\ldots,n-1$ is it true that $F_n$ is a splitting field over $F_1$? I think I can prove it for the finite case, but am unsure about the infinite case. For the finite case here's what I had. For simplicity we'll assume characteristic $0$, so everything is separable.

$\textbf{Proof:}$ If $n=1$ there is nothing to show. Suppose $n>1$. Let $$\sigma:F_{n-1}\to F_{n-1}$$ be an automorphism leaving $F_1$ fixed. Since $F_n$ is a splitting field over $F_{n-1}$ it follows that we can extend $\sigma$ in $[F_n:F_{n-1}]$ ways to an automorphisms of $F_n.$ If $F_{n-1}$ were a splitting field over $F_1,$ then there would be $[F_{n-1}:F_{n-1}]$ such $\sigma,$ hence there would be at least $$[F_{n}:F_{n-1}][F_{n-1}:F_1]=[F_n:F_{n-1}]$$ automorphisms of $F_n$ leaving $F_1$ fixed, and $F_n$ would be a splitting field.

If $F_{n-1}$ is not a splitting field over $F_{1}$, then applying the contrapositive of the above to $F_{n-1}$ it would follow that $F_{n-2}$ were not a splitting field over $F_1,$ so $n-2>1.$ By infinite descent $F_{n-1}$ is a splitting field over $F_1,$ and so $F_n$ is a splitting field over $F_1.\blacksquare$

The above argument hinges on the fact that $F_n$ has $[F_{n}:F_1]<\infty$ isomorphisms into a subfield of $\overline{F_n}$ leaving $F_1$ fixed, hence since we've shown that many automorphisms every isomorphisms of that type is an automorphism. This argument cannot apply to the infinite case, for even if we have $\infty$ automorphisms we cannot apply this counting argument. Is there a more general way to make it work for the infinite case?

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    $\begingroup$ Did you look at $\Bbb{Q}(\sqrt[4]{2})/\Bbb{Q}(\sqrt{2})/\Bbb{Q}$ $\endgroup$ – reuns Apr 17 at 12:09
  • $\begingroup$ @reuns Thank you, and I see the mistake now, or I believe I do. The map sending $\sqrt{2}\to-\sqrt{2}$ can be extended to an isomorphism of $\mathbb{Q}(\sqrt[4]{2}),$ but because $\mathbb{Q}(\sqrt{2})$ is not fixed we cannot conclude that this extension is an isomorphsim. And actually it is not, because it requires we send $\sqrt[4]{2}\to\pm i\sqrt[4]{2}.$ Thank you :) Are there additional conditions we can impose to make the claim I was falsely asserting true? $\endgroup$ – Melody Apr 17 at 12:29
  • $\begingroup$ Meant cannot conclude it is an automorphism, too late to edit and correct now. $\endgroup$ – Melody Apr 17 at 12:37
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    $\begingroup$ I don't know any easy to way to check that $E/K$ is Galois when $E/F,F/K$ are Galois, let $E =F(\alpha) \cong F[x]/(h(x))$, it requires searching for a root of $h^\sigma(x)$ in $E$ for each $\sigma \in Gal(F/K)$ (in my example $h(x) = x^2-\sqrt{2}, h^\sigma(x)=x^2+\sqrt{2}$) $\endgroup$ – reuns Apr 17 at 12:43
  • $\begingroup$ @Melody If $E$ is the splitting field of $p(x)\in K[X]$ over $K$ and $K$ is the splitting field of $q(x)\in F[X]$ over $F$, then $E$ will be contained in the splitting field of $q(x)\prod_{h_i\in Gal(K:F)} \bar h_i(p(x))$ over $F$ where each $\bar h_i$ is an extension of $h_i$ to $K[X]$. In particular, that splitting field will be the smallest one over $F$ that contains $E$ whenever $p(x)$ has at least one coefficient that can generate $K$ from $F$. It looks like generally $E$ won't be a splitting field over $F$. The easiest additional condition you could impose would be to have $p(x)\in F[X]$ $\endgroup$ – Cardioid_Ass_22 Apr 17 at 13:33

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