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Let $F/K$ be a finite extension and suppose $F$ is the splitting field of a separable polynomial over $K$. Show that if Gal(F/K) is cyclic then for each divisor $d$ of $[F : K]$ there exists exactly one subfield $E$ of $K \subseteq E \subseteq F$ such that $[E : K]=d$.

Is this actually true? what I get is that one can find a subfield $E$ of $F$ such that $[F : E] = d$.

My work:

First, since $F/K$ is Galois then $[F : K] = |Gal(F/K)|$ so that if $d$ divides $[F : K]$ then d divides $|Gal(F/K|)$. By assumption the latter group is cyclic so for each divisor we can find a unique subgroup $H$ of $Gal(F/K)$ such that $|H| = d$.

But now the fundamental theorem implies $[F : F^{H}] = |H| = d$ so that $E=F^{H}$. I get then that $[F : E] = d$ and not $[E : K]=d$. What am I doing wrong?

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Let $n=[F:K]$ and do your argument for $n/d$. – Gerry Myerson May 20 '11 at 2:49
@Gerry Myerson: doh, i'm stupid, can you please post your reply as an answer so I can accept it? – user6495 May 20 '11 at 3:00
up vote 4 down vote accepted

Posted as an answer by request: Let $n=[F:K]$, and do your argument for $n/d$.

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thanks, can't believe I didn't see that. – user6495 May 20 '11 at 4:17
@user6495, I've been known to miss things that are even more obvious - we all have our moments.... – Gerry Myerson May 20 '11 at 5:52

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