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So I am trying to do the following problem: if $K/F$ is an extension of degree $p$ (a prime), and if $L/F$ is a normal closure of $K/F$, then $[L:K]$ is not divisible by $p$.

This is what I tried. If $F$ is separable, then primitive element theorem holds and we get $K=F(\alpha)$. Let $f(x)$ be the minimal polynomial of $\alpha$ over $F$. We know that $deg(f)=p$. If $K$ contains all the roots of $f$, then it is normal, and we are done. If not, then there is a root $\beta$ that when we add it we generate an extension of at degree at most $p-1$. If we have all the roots then we are done, otherwise we keep adding roots and at each step the maximum degree of the extension we obtain goes down, so we have that $[L:K]$ is a divisor of $(p-1)!$, so it is not divisible by $p$ since $p$ is a prime.

I am not quite sure of how to solve this problem when we dont know that $F$ is not separable. Any thoughts?

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  • $\begingroup$ Remember that 'separable' is a property of field extensions, not fields. The primitive element theorem holds for $K/F$ if $K$ is separable over $F$, not if '$F$ is separable'. $\endgroup$ – Paul VanKoughnett Mar 20 '13 at 3:03
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    $\begingroup$ If you can't get from $F$ to $K$ by adjoining a single element, doesn't the tower law say that the degree of $K/F$ isn't prime? $\endgroup$ – Gerry Myerson Mar 20 '13 at 3:13
  • $\begingroup$ Wow, thanks Dr. Myerson! $\endgroup$ – Kanye West Mar 20 '13 at 3:29
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If $[K:F]=P$, then for any $\alpha\in K\setminus F$, we get $K=F(\alpha)$.

Let $f(x)$ be the minimal polynomial of $\alpha$ over $F$, then the normal closure $L/F$ of $K/F$ is the splitting field of $f(x)$ over $K$. If we let $f(x)=g(x)(x-\alpha)$, then $L$ is the splitting field of $g(x)$ over $K$, but $\deg(g)=p-1$, so $[L:K]$ is less than $(p-1)!$
so it is not divisible by $p$ since $p$ is a prime.

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