Working my way through this resource on abstract algebra, and a few questions pop up from time to time.

Here is a statement regarding fields and Galois groups I find puzzling:

Proposition 23.17: Let be $E$ be a splitting field over $F$ of a separable polynomial. Then $E_{G(E/F)} = F$.

Here, $G(E/F) : = \{\sigma \in \mathrm{Aut}(E) : \sigma(\alpha) = \alpha, \; \forall \alpha \in F\}$ is the Galois group of $E$ over $F$, and $E_{G(E/F)} = \{ \alpha \in E : \sigma(\alpha) = \alpha, \; \forall \sigma \in G(E/F)\}$ is the fixed field of $E$ w.r.t $G(E/F)$.

To me, this intuitively feels like a tautology - i.e. using brackets for comprehension, it is the statement that "the sub-field of $E$ consisting of elements fixed by [the group whose members leave $F$ invariant] is $F$".

There is a short proof provided, which does not make this clear to me:

Let $G=G(E/F)$. Clearly, $F \subset E_G \subset E$. Also, $E$ must be a splitting field of $E_G$ and $G(E/F) = G(E/E_G)$. By Theorem 23.7 [which is about a splitting field $E$ for a polynomial in $F[x]$ which is separable]

$|G| = [E:E_G] = [E:F]$.

Therefore $[E_G:F]=1$. Consequently $E_G =F$.

What makes the Proposition a (relatively) non-trivial statement? The key piece seems to be that we have a splitting field whose minimal polynomial over $F$ is separable in $E$, but I can't see a way through.

(This is slightly before the fundamental theorem on Galois theory is proven, so it would be helpful to see this without using it.)


1 Answer 1


What is clear is that every element in $F$ is fixed by every automorphism of $E$ that fixes $F$.

However, to go the opposite way, i.e. to claim that every element of $E$ that is fixed by [every automorphism of $E$ that fixes $F$] must belong to $F$, is not trivial.

This non-trivial direction is equivalent to (by taking contrapositive) the assertion that if I give you a random element $\alpha$ of $E$ that is not in $F$, you have to give me an automorphism of $E$ that fixes $F$ but not $\alpha$.

An example where this is not true is $\Bbb Q(\sqrt[3]2)$ as an extension of $\Bbb Q$; here we do not have enough automorphisms, since any automorphism must send $\sqrt[3]2$ to a cube root of $2$, but the only cube root of $2$ in $\Bbb Q(\sqrt[3]2)$ is $\sqrt[3]2$ itself; i.e. every automorphism is the identity.

So in this case, "the group of automorphisms that fix $\Bbb Q$" consists just of the identity, and so the sub-field of $\Bbb Q(\sqrt[3]2)$ fixed by "the group of automorphisms that fix $\Bbb Q$" is unfortunately just $\Bbb Q(\sqrt[3]2)$ itself, instead of $\Bbb Q$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .