3

The question is:

Let $\overline{K}$ be an algebraic closure of $K$, $\sigma\in G(\overline{K}:K)$ and $E=\{x\in\overline{K}:\sigma(x)=x\}$. Prove that every finite extension $L\mid E$ is a Galois extension and that $G(L:E)$ is cyclic.

The accepted answer of this old question is the next hint:

"If $x\in\overline{K}$, the set $\{\sigma^i (x):i\in\mathbb{N}\}$ is finite. It follows that a suitable polynomial of the form:

$$\prod_{i=0}^{n-1}(t-\sigma^i (x))\in E[t]$$

Using this you can hopefully prove that if $N$ is a normal closure of $L\mid E$, then $G(N:E)=\langle\sigma\rangle$."

Now, I do not know why a suitable polynomial of the form $\prod_{i=0}^{n-1}(t-\sigma^{i}(x))\in E[t]$, and assuming that, how could you prove $G(N:E)=\langle\sigma\rangle$?

A different way to solve the problem would be appreciated too.

Community
  • 1
Srinivasa Granujan
  • 1,110

1 Answers1

5

Since $x\in\overline{K}$, there exists $p\in K[t]$ such that $p(x)=0$, and then $p(\sigma(x))=\sigma(p(x))=0$. Hence the set $\{\sigma^i(x):i\in\mathbb{N}\}$ is contained in the set of roots of $p$ and so it is finite. Since $\sigma$ is an automorphism, there exists $n$ such that $\sigma^n(x)=x$ and $\sigma^i(x)\neq\sigma^j(x)$ for $i,j< n$, $i\neq j$.

Now consider the polynomial $$q(t)=\prod_{i=0}^{n-1}(t-\sigma^i(x))\in \overline{K}[t]$$ Since $\sigma^n(x)=x$, $$\sigma(q(t))=\prod_{i=0}^{n-1}(t-\sigma^{i+1}(x))=q(t)$$ and hence $q(t)\in E[t]$. Since $\sigma^i(x)\neq\sigma^j(x)$ for $i,j< n$ and $i\neq j$, $q(t)$ is separable and we have shown that every algebraic extension of $E$ is separable. If $L/E$ is finite, take $N/E$ a normal closure. Since, by hypothesis, $N^\sigma=E$, we have $<\sigma>=\operatorname{Gal}(N/N^\sigma)=\operatorname{Gal}(N/E)$ by the fundamental theorem of Galois theory. Since $\operatorname{Gal}(N/E)$ is abelian, all subextensions of $N/E$ are Galois: in particular $L/E$ is normal and $L=N$.

Giulio Bresciani
  • 2,918
  • Thank you for your answer (even though this comment is 3 years late!). What do you mean by your notation N^(sigma)=E? In fact could you explain to me the line "Since, by hypothesis, ...=Gal(N/N^(sigma))=Gal(N/E)"? (I do actually understand the theorem and proof of the fundamental theorem of Galois theory) – Pianoman1234 Jul 14 '17 at 23:46
  • Also are you able to show that Gal(N/E) is cyclic, or just any method to show that Gal(L/E) is cyclic (which is the question the OP asked)? – Pianoman1234 Jul 14 '17 at 23:50
  • 1
    Well, $\sigma$ acts on $N$ because $N/E$ is normal, $N^{\sigma}$ is the standard notation for the fixed field of this action. Now the rest should be clear: $N/N^\sigma$ is Galois with group $<\sigma>$ by standard Galois theory, and $N^\sigma=E$ because $E$ is defined as the fixed field of $\sigma$. Hence $Gal(N/E=N^\sigma)=<\sigma>$ is abelian (and cyclic). This implies that all of its subgroups are normal, and hence all subextensions of $N/E$ are normal, in particular $L$, which implies $L=N$ has Galois group $Gal(L=N/E)=<\sigma>$. As you see, there was an answer to the original question. – Giulio Bresciani Jul 15 '17 at 05:57
  • Ah i see now thank you. Could you just explain how "N/N^(sigma) is Galois with group by standard Galois theory"? – Pianoman1234 Jul 15 '17 at 10:16
  • 1
    $N/N^\sigma$ is normal by construction and is separable because of what is written in the answer above, hence it is Galois. You obviously have an injective map $<\sigma>\to{\rm Gal}(N/N^\sigma)$, since $\sigma$ is an automorphism of $N/N^\sigma$. The fact that it is surjective is exactly the fundamental theorem of Galois theory. – Giulio Bresciani Jul 15 '17 at 11:16
  • What do you mean by: "You obviously have an injective map onto Gal(N/N^sigma)". I understand all of your proof up to that point, and I understand how is a subgroup of Gal(N/N^sigma) just not how =Gal(N/N^sigma). Could you just spell it out really simply for me this bit: "we have =Gal(N/N^sigma)...by the fundamental theorem of Galois theory". – Pianoman1234 Jul 15 '17 at 23:23