I am learning about Galois groups and I'm somewhat confused by an argument from my book. My book restricts its attention to finite separable extensions $E=F(\alpha)$ of a field $F$. The authors argue as follows:
Let $p(x)$ be the minimal polynomial of $\alpha$ over $F$. Then $[F(\alpha):F]=$ degree of $p(x)=n$, say. Also, by Lemma 4.2, Chapter 15, we get that the order of the group $G(E/F)$ is $\leq n$ .
The lemma they are referring to is the following:
4.2 Lemma Let $F$ be a field, and let $\sigma: F \rightarrow L$ be an embedding of $F$ into an algebraically closed field $L$. Let $E=F(\alpha)$ be an algebraic extension of $F$. Then $\sigma$ can be extended to an embedding $\eta:E \rightarrow L$, and the number of such extensions is equal to the number of distinct roots of the minimal polynomial of $\alpha.$
Now, what I don't quite get is the following: Since the authors say that $F(\alpha)$ is a separable extension of $F$, $\alpha$ must be separable over $F$, i.e. the minimal polynomial $p(x)$ is a separable polynomial. Given that $p(x)$ is irreducible, this implies, by the definition of separable, that $p(x)$ has only simple roots. Thus, by the above Lemma, the function $\sigma$ can be extended to an embedding $\eta:E\rightarrow L$ in exactly $n=\deg(p)$ ways. Then, why do we conclude $|G(E/F)|$ is $\leq n$ and not $|G(E/F)|=n=[E:F]$ by this reasoning?
I have tried to unwind my confusion by comparing my above reasoning to the (finite, algebraic, non splitting field) separable extension $\mathbb{Q}(\sqrt[3]{2})$ of $\mathbb{Q}$. This extension is of degree $3$, but $|G(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q})|=1$. Is my point of confusion that the extensions $\eta$ of $\sigma$ may not have the same images $\eta(E)$ ? I know that is has to with the fact that the other two roots of $x^3-2$ are complex, but I don't quite get it anyway..
Best Answer
If $E=F(\alpha)$ is separable where $\alpha$ has degree $n$, and $E\subseteq L$, where $L$ is algebraically closed, then $E$ has exactly $n$ $F$-embeddings into $L$. But such an embedding $\sigma$ only gives an automorphism of $E$ if $\sigma(E)=E$, and this does not always happen. So there are $\le n$ automorphisms and possibly $<n$.
Take $F=\Bbb Q$, $\alpha=\sqrt[3]2$ and $L= \Bbb C$. Then one of these embeddings has $\sigma(\alpha)=\exp(2\pi i/3)\alpha$ and certainly then $\sigma(E)\ne E$. Indeed $\sigma(E)\not\subseteq\Bbb R$.