10
$\begingroup$

We have the following lemma (used in the proof of Abel's Theorem):

If $\text{Char}(F)=0$, $E/F$ is a radical extension, and $K/E$ is the Galois closure of $E/F$, then $K/F$ is also a radical extension.

Proof: We have $$ F=:E_0\subset E_1\subset E_2\subset\cdots\subset E_n:=E $$ such that $E_{i+1}=E_i(\alpha_i)$ where $\alpha_i^m\in E_i$.

Define $K_{i+1}:=K_i(\sigma(\alpha_i),\sigma\in\text{Gal}(K/F))$ and $K_0:=E_0=F$. Note that $E_i\subset K_i$ for all $i$. Since $ (\sigma(\alpha_i)))^m=\sigma(\alpha_i^m)$ where $\alpha_i^m\in E_i$, we have $(\sigma(\alpha_i)))^m\in K_i$ and therefore $K_{i+1}/K_i$ is a radical extension.

But $K_n=K$. Indeed, since $K$ is the Galois closure of $E/F$ and we have the tower of fields $K/K_n/E/F$, it suffices to show $K_n/F$ is a Galois extension. But this is the case because $K_{i+1}$ is the splitting field of the minimal polynomial of $\alpha_i$ over $F$ for all $i$.

Hence the result. $\blacksquare$

Questions:

  1. $E/F$ must be finite, right?
  2. Where is used the hypothesis $\text{Char}(F)=0$ ?
  3. To even talk about the Galois closure, we must verify that $E/F$ is a separable extension. Why is it in this case?
  4. Why is $K_{i+1}$ the splitting field of $\alpha_i$ for all $i$ ?
  5. (Why) was it essential to consider a new tower of fields $K_i$ ? I thought that $K$ must be the splitting field of the minimal polynomial of $\alpha_{n-1}$ because of

Proposition: If $E/F$ is a finite separable extension, then its Galois closure exists.

Proof: We know that if $E/F$ is a finite separable extension, then $E=F(\alpha)$ for a certain $\alpha\in E$. Let $m\in F[x]$ be the minimal polynomial of $\alpha$. Then by hypothesis $m$ is separable. Let $E'$ denote the splitting field of $m$ over $F$. We know $E'/F$ is a Galois extension. Also, suppose $E\subset\tilde{E}\subset E'$ is such that $\tilde{E}/F$ is a Galois extension. Since $E'/F$ is finite, we know $\tilde{E}/F$ must be finite and we know that a finite Galois extension is, in particular, normal. Therefore we have $\alpha\in\tilde{E}$ with $m$ splitting over $\tilde{E}$. By unicity of the splitting field, $\tilde{E}=E'$. $\blacksquare$

So, does the proof actually amount to show that this splitting field is part of a radical extension of $F$ ?

Thoughts:

  1. I think so because I only know that the Galois closure exists if $E/F$ is finite.
  2. I think $\text{Char}(F)=0$ tells us that there exists an extension of $F$ in which there is a primitive root of unity $\zeta$. In turn I think this shows that $x^m-\alpha_i^m\in F[x]$ has $m$ different roots, namely $\zeta^k\alpha_i$ for $k=1,2,\ldots,m$. Hence $x^m-\alpha_i^m$ is separable and this implies that the minimal polynomial of $\alpha_i$ over $F$ is separable. Hence this would show that $E/F$ is separable because $E=F(\alpha_0,\alpha_1,\ldots,\alpha_{n-1})$ and we know that an extension generated by separable elements is separable.
  3. See 2.
  4. I understand that every $\sigma(\alpha_i)$ must be a root of the minimal polynomial of $\alpha_i$ over $F$. However I don't see why we get all the roots. I see $\#\text{Gal}(K_{i+1})=[K_{i+1}:F]\geq [E_{i+1}:F]=\deg(m_{\alpha_i})$ where $m_{\alpha_i}\in F[x]$ is the minimal polynomial of $\alpha_i$, but maybe some automorphisms in $\text{Gal}(K/F)$ send $\alpha_i$ to a same image...?
$\endgroup$
  • $\begingroup$ 2 & 3) There are a lot of pieces here, but one point: all characteristic zero field extensions are separable. $\endgroup$ – Slade Oct 15 '14 at 2:53
  • $\begingroup$ 1 & 4 & 5) This is all true even if $E/F$ is infinite, but that's not really important. What's important is that, given some expression in radicals, adding the extra roots those radicals imply only involves more radicals—for example, if we have $\sqrt{5+\sqrt{5}}$, we need to add $\sqrt{5-\sqrt{5}}$. This should be done inductively, so that we only need to worry about one radical at a time. $\endgroup$ – Slade Oct 15 '14 at 2:56
  • $\begingroup$ @Slade 2 & 3) Ah! I now see that if $F$ has characteristic $0$ and $E/F$ is an algebraic field extension, then $E/F$ is a separable extension. That would be because we know that if $f\in F[x]$ with $\text{Car}(F)=0$, then $f$ is separable. To show this we first note that $f$ is separable iff $\gcd(f,f')=1$, where $f'$ is the formal derivative of $f$, then using this we show that if $f$ is irreducible then $f$ is separable iff $f'\neq 0$ and the result is a consequence of this because then the leading coefficient of $f'$, $\deg(f) a_{\deg(f)}\neq 0$. But why must $E/F$ be algebraic? $\endgroup$ – Guest Oct 15 '14 at 3:10
  • $\begingroup$ @Slade 2 & 3) (cont.) Besides this, do you know if a radical extension $E/F$ must be algebraic (even if we drop the hypothesis $\text{Car}(F)=0$)? It looks true, but I can only get this : If $E=F(\alpha)$ where $\alpha^m\in E$ then I see that $\alpha$ is algebraic over $F$ (since it is a root of the polynomial $x^m-\alpha^m\in F[x]$) and therefore we know $E(\alpha)$ must be algebraic, but since using the same argument seems only to show that if $F=:E_0\subset E_1\subset \cdots\subset E_n:=E$ where $E_{i+1}=E_i(\alpha_i)$, then $\alpha_i$ is algebraic over $E_i$ (not necessarily $F$)... $\endgroup$ – Guest Oct 15 '14 at 3:22
  • $\begingroup$ Finite extensions are always algebraic. (though it is true, more generally, that the composition of two algebraic extensions is algebraic—this is a nice exercise) $\endgroup$ – Slade Oct 15 '14 at 5:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.