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I've thought about this problem a bit, but I may be heading in the wrong direction. I know that if $f$ is a polynomial of prime degree over $\mathbb{Q}$ with precisely two nonreal zeros, then the Galois group of $f$ is isomorphic to $\mathbb{S}_p$. It follows that this polynomial's Galois group is isomorphic to $\mathbb{S}_3$, from which we can can conclude that the Galois group is solvable. Thus $f$ is solvable by radicals, meaning there exists some field $M$ containing $\Sigma$, the splitting field of $f$, such that $M:\mathbb{Q}$ is a radical extension. From here, I don't know how I can draw any conclusions about $\Sigma$ itself. Any help is appreciated.

Radical Extension: An Extension $L:K$ in $\mathbb{C}$ is radical if $L=K(\alpha_1,...,\alpha_{m})$ where for each $j=1,...,m$, there exists an integer $n_j$ such that $\alpha_j^{n_j}\in K(\alpha_1,...,\alpha_{j-1})$ for $j\geq 2$.

It is not necessarily true that everything expressible by the radicals contained in $M$ will be contained in $\Sigma$. So if we have $m = \sqrt{6}+\sqrt[4]{1+\sqrt[3]{4}}$, then letting $\alpha=\sqrt{6}$, $\beta=\sqrt[3]{4}$, and $\gamma=\sqrt[4]{1+\beta}$, we may create the radical extension $\mathbb{Q}(\alpha,\beta,\gamma)$. We also have $\mathbb{Q}(m)\subset \mathbb{Q}(\alpha,\beta,\gamma)$, but $\mathbb{Q}(m)$ is not a radical extension (at least that's my understanding - I just started learning this material so it's quite possible that I'm wrong).

Edit: I appreciate the answer already given, but it uses a fair amount of machinery I have yet to develop (quadratic subfields, Kummer field extensions, etc). Is there a more elementary route to take in attacking this problem?

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What does it mean to be a radical extension? I assume you want this term to mean that the extension is generated by radicals, which means that every element of the extension can be written using elements of $\mathbb{Q}$ and field operations. Since every element of $\Sigma$ is an element of $M$, this follows for free. –  Brett Frankel Nov 19 '12 at 17:18
    
If you read for example planetmath.org/RadicalExtension.html you can see that what you call a radical extension is what they call a radical tower. –  Phira Nov 19 '12 at 18:01
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1 Answer 1

Let $f$ be an irreducible cubic, whose roots generate non-normal cubic extensions, and let $L$ be its splitting field, which has degree $6$ and Galois group $S_3$. Let $K$ denote a cubic subfield of $L$. It is not difficult to show that $K$ is radical if and only if the discriminant of $f$ (or $K$) has the form $D = -3m^2$. If $L$ is radical in your sense, then either $K$ must be radical, or $L$ must be radical over its quadratic subfield $k$. Since $L/k$ is cyclic, $L$ can be normal only if $k = {\mathbb Q}(\sqrt{-3})$ and $L/k$ is Kummer. But since $L = K(\sqrt{{\rm disc}\ K})$, we have $k = {\mathbb Q}(\sqrt{-3})$ if and only if the discriminant of $f$ has the form $-3m^2$. Thus $L$ is radical in your sense if and only if $K$ is radical, which is the case if and only if the discriminant of $f$ has the form $-3m^2$.

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When you write, "$L$ can be normal only if...," do you mean, "$L$ can be radical only if..."? After all, $L$ is normal --- it's a splitting field. –  Gerry Myerson Nov 20 '12 at 3:29
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