# Tower of Splitting Fields

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?

• Did you look at $\Bbb{Q}(\sqrt[4]{2})/\Bbb{Q}(\sqrt{2})/\Bbb{Q}$ – reuns Apr 17 at 12:09
• @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? – Melody Apr 17 at 12:29
• Meant cannot conclude it is an automorphism, too late to edit and correct now. – Melody Apr 17 at 12:37
• 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 $h(x) = x^2-\sqrt{2}, h^\sigma(x)=x^2+\sqrt{2}$) – reuns Apr 17 at 12:43
• @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 $p(x)\in F[X]$ – Cardioid_Ass_22 Apr 17 at 13:33