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I am having trouble understanding the difference between a proposition and theorem in chapter 14 (Galois theory) of Dummit and Foote.

Prop. 5 (p.562). Let $E$ be a splitting field over $F$ of the polynomial $f(x)\in F[x]$. Then $|\operatorname{Aut}(E/F)|\leq [E:F]$.

Thm. 9 (p.570) Let $G=\{\sigma_{1}=1,\sigma_{2},\cdots,\sigma_{n} \}$ be a subgroup of automorphisms of a field $K$ and let $F$ be the fixed field. Then $[K:F]=n=|G|$.

My question: to me it seems that $G=Aut(K/F)$, then by Prop. 5 it seems we get the inequality $G=\operatorname{Aut}(K/F)\leq [K:F]$. If this is the case, I do not understand why we get equality in Thm. 9 and not in Prop. 5.

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One thing to note is that in the situation of Prop 5 it does not have to be true that $F$ is the fixed field of $\operatorname{Aut}(E/F)$. Indeed, this fails if an irreducible factor of $f$ is not separable. The standard example is the spitting field of $f(X) = X^p - t$ over the function field $\mathbf F_p(t)$: the extension you get has degree $p$, but no nontrivial automorphisms. – Dylan Moreland Apr 6 '12 at 4:26
The Galois correspondence mappings are only bijective given seperability and normality. Another example like @Dylan's would be $f=x^3-2$ over $\mathbb{Q}(t)$ adjoining the real root, $\alpha$, for which the automorphism group of – bgins Apr 6 '12 at 6:45

This question was pretty thoroughly answered in the comments, so I'm just collecting some thoughts as a cw post here.

In proposition $5$ there's no reason that the fixed field of $\mathrm{Aut}(E/F)$ is $F$. For instance in a situation like $E=\mathbb Q(\sqrt[3]{2})$ and $F=\mathbb Q$ there are no non-trivial automorphisms so the fixed field is $E$, but the degree of the extension is $3$.

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