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I'm struggling to understand the proof that the normal closure of a radical extension of fields is also a radical extension, which is crucial since it allows us to work with radical and normal extensions (and thus, in characteristic 0, with radical Galois extensions).

The proof I'm reading is lemma 4.17 in Rotman's Advanced Modern Algebra, pp. 211-212, which I attach at the end of this message.

The definition Rotman uses is the following: an extension $F\subset F(u)$ is pure if there exists $m \geq 1$ such that $u^m\in F$. An extension $F\subset K$ is radical if there exists a tower (which I like to call radical tower) $$F=K_0\subset K_1\subset \dots \subset K_t=K$$ such that $K_i\subset K_{i+1}$ is pure for all $i$.

What I'm struggling to prove is: if $F\subset K$ is a radical extension, and $N$ is the normal closure of the extension, then $F\subset N$ is a radical extension.

Since I find Rotman's proof confusing (and not convincing), I have rewritten the proof. It might seem longer in appearance, but the first part of the proof (the one that troubles me) was unsatisfying. Let $m_{\alpha,F}$ denote the minimal polynomial for an element $\alpha$ over $F$.

What I ask is:

1) Is my rewriting of the proof correct?

2) Is this what's behind the argument of Rotman, or is there another, more simple way of reading it?

Rotman's proof seems shorter, but I don't understand when he says "it follows that $E=k(\sigma(u_1),\dots,\sigma(u_t):\sigma \in G)$": what are those $u_i$?

Here's my rewriting:

Let $F\subset K$ be a radical extension. There exist $u_1,\dots,u_t$ such that $$F\subset F(u_1)\subset F(u_1,u_2)\subset \dots \subset F(u_1,\dots,u_t) = K$$ is a radical tower. Let $E$ be the normal closure of $F\subset K$. I claim that$$E=F\left(\{\sigma(u_i): \sigma \in \text{Gal}^E_F, i=1,\dots,t\}\right)$$

Indeed, $$E=\text{Split}_F\left(m_{u_1,F} \dots m_{u_t,F}\right)=F(\{\text{roots of }m_{u_1,F} \dots m_{u_t,F}\})$$

Since the Galois group of a polynomial acts transitively on its roots, $\{\text{roots of }m_{u_i,F}\}=\{\sigma(u_i):\sigma \in \text{Gal}_F(m_{u_i,F})\}$. Let us see that is also equals $\{\sigma(u_i):\sigma \in \text{Gal}^E_F\}$.

$(\subset)$: We can extend every $\sigma\in \text{Gal}_F(m_{u_i,F})$ to $\tilde{\sigma}\in \text{Gal}_F^E$ by the theorem of extension of a polynomial to its splitting field, by taking the polynomial $m_{u_1,F} \dots m_{u_t,F}\in F[X]$.

$(\supset)$: If $\sigma\in \text{Gal}^E_F$, then $\sigma$ restricts to an $F$-isomorphism $F(u_i)\to F(\sigma(u_i))$, then (corollary 1.9 p.236 of Hungerford) $u_i$ and $\sigma(u_i)$ must be roots of the same irreducible polynomial in $F[X]$. This proves that $\sigma(u_i)$ is a root of $m_{u_i,F}$.

Let $B_i=F\left(\{\sigma(u_j):1\leq j\leq i,\sigma \in\text{Gal}^E_F\}\right)$. What we just proved is that $E=B_t$. To see that $F\subset E= B_t$ es radical, we use induction on $t$, and this part of Rotman's proof is satisfying to me.

Below is Rotman's proof.

enter image description here

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I think he should have said that $u_1, \ldots, u_t$ are elements of $B$ such that for each $i = 1, \ldots, t - 1$ some power of $u_{i + 1}$ is contained in $F(u_1, \ldots, u_i)$. Your proof looks good. I think you just needed to write out the details for yourself, which is good! –  Dylan Moreland Jul 31 '11 at 22:26
    
@Dylan: What puzzles me is that if all the details I wrote out are necessary, then I think the author should have been a bit more verbose on the proof... –  Bruno Stonek Jul 31 '11 at 22:36
    
I must say I'm a bit surprised by the lack of answers to this question. It shouldn't take too long to answer for somebody comfortable with Galois theory. –  Bruno Stonek Aug 3 '11 at 13:45
    
What would you like to see in such an answer? It looks like both you and Rotman have correct proofs, and for me it's hard to imagine a shorter argument! –  Dylan Moreland Aug 3 '11 at 13:49
    
The one problem I have with what Rotman is saying here (and I could be missing something) is that you might have repeated $p_i$ in part (i), e.g. for $\mathbf{Q}(\sqrt[4]{2}, i\sqrt[4]{2})$. –  Dylan Moreland Aug 3 '11 at 13:58
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1 Answer 1

up vote 2 down vote accepted

I think that both you and Rotman have the same argument in mind. Here's another way of looking at this.

Let $B$ be a finite extension of $k$, and let $\sigma_1, \ldots, \sigma_s$ be the distinct embeddings over $k$ of $B$ into an algebraic closure $B^a$ of $B$. I claim that a normal closure of $B/k$ is the compositum $E = (\sigma_1B) \cdots (\sigma_sB)$. If $B$ is given as $k(\alpha_1, \ldots, \alpha_n)$ then $\sigma_iB = k(\alpha_1^{\sigma_i}, \ldots, a_n^{\sigma_i})$, and we have rules like $k(\alpha)k(\beta) = k(\alpha, \beta)$. Each of the $\sigma_i$ will extend to an element of $\operatorname{Gal}(E/k)$, and your result follows.

Of course, you need to prove that $E$ really is normal over $k$, and minimal among such extensions containing $B$. But that's not too hard, and conceptually I think this is nice.

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