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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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Also, Are you using the $Gal$ notation even if the extension is not noraml hence not Galois ? –  Belgi Dec 6 '12 at 18:05
    
I am using the following definition of a separable extension: An algebraic extension $E/K$ is called separable if every element of $E$ is separable over $K$. We say that an element $\alpha\in E$ is separable over $K$ if its minimal polynomial has no multiple roots (in any splitting field). And yes, I apologize for using the Gal notation. I should have wrote $Aut(F/E)$ in this case. This is a symptom of jumping back and forth between a few different texts. –  John Myers Dec 6 '12 at 18:11
    
Another question: What is $F$ ? its not in the question but it is in the attempt... –  Belgi Dec 6 '12 at 18:12
    
Ah, shoot. Take $F$ to be a splitting field of $f$ containing $E$. I'll edit it. –  John Myers Dec 6 '12 at 18:13

1 Answer 1

up vote 1 down vote accepted

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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But we are not given that the extension is Galois, how do you apply any knowledge in group theory here ? –  Belgi Dec 6 '12 at 18:36
    
Did you look at how John M was attempting to solve the question? I'm building on his work. –  JSchlather Dec 6 '12 at 18:40
    
Great. Put $H=Aut(F/E)$ and $G=Gal(F/K)$. First, as mentioned above, if $H$ is normal in $G$, then we're done. If $H$ is not normal, then it has precisely $p$ conjugate subgroups in $G$ (by the lemma). But the conjugate subgroup $\sigma H \sigma^{-1}$ in $G$ ($\sigma \in G$) corresponds to the conjugate subfield $\sigma E$ in $F$. However, by hypothesis $E$ contains at least two roots of $f$ and hence $E$ has at most $p-1$ conjugate subfields. Thus the first case must hold, i.e., $H$ must be normal in $G$. –  John Myers Dec 6 '12 at 19:43
    
@JohnM That's it exactly. –  JSchlather Dec 6 '12 at 20:02

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