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Consider the following:

Let $E/K$ be a separable field extension of degree $p$ ($p$ a prime). Suppose $f\in K[x]$ is an irreducbile polynomial which has more than one root in $E$. Show $f$ splits in $E[x]$.

I've tried a couple different ideas, but I haven't been able to make anything out of it. Here is what I know for certain, though: Any of the roots in $E$ must be a primitive element for $E$. Since $E/K$ is separable, we conclude $f$ must be separable and therefore its splitting field $F$ (containing $E$) is a Galois extension of $K$.

If I could show $Aut(F/E)$ is normal in $Gal(F/K)$ I'd be done, for then $E/K$ is normal. I know $Aut(F/E)$ has index $p$ in $Gal(F/K)$. This doesn't seem to get me anywhere though since I don't know much about $Gal(F/K)$.

I am having a difficult time finding a way to use the "has more than one root in E" hypothesis.

Can anybody give me a nudge in the right direction?

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    Ah, shoot. Take $F$ to be a splitting field of $f$ containing $E$. I'll edit it.2012-12-06

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This is a nice Galois Theory question because it illustrates how you can use group theory to get information about the structure of the fields.

Lemma: Let $G$ be a group and $H$ a subgroup of index $p$. If $H$ is not normal in $G$ then $H$ has $p$ conjugate subgroups.

Proof: Use orbit-stabilizer theorem.

Now what does this tell you about your case when you pull back to the fixed fields?

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    @JohnM That's it exactly.2012-12-06