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Let $P(X)\in F[X]$ be a polynomial of degree $n\ge 1$ such that the splitting field $K$ satisfies $[K:F]=n!$. Let $\alpha\in K$ be a root of $P$.

(1) Show that $\text{Aut}_F(F(\alpha))=1$ if $n\ne 2$.

(2) Show that $K$ is the Galois closure of $F(\alpha)/F$.

I feel awkward about this problem because I do not know what to do with $[K:F]=n!$. Any hints are appreciated, thanks in advance.

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Ryze
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    Looks like this must be a separable case so the splitting field is Galois over $F$. In that case the Galois group can be identified with a group of permutations of roots of $P(X)$. As there are $n$ of those, it is a subgroup of $S_n$, together with given data leaving us with only a single alternative. Your first claim has been covered already on our site. Gimme a minute. The second claim also follows easily. – Jyrki Lahtonen Oct 07 '19 at 08:27
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    Combine this with the fact that $S_{n-1}=Gal(K/F(\alpha))$ is its own normalizer in $S_n$. – Jyrki Lahtonen Oct 07 '19 at 08:43

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Note that the decomposition field of a polynomial of degree $n$ has degree at most $n!$ ( easy induction on $n$). Now, since $p! q! < (p+q)!$ for $p$, $q\ge 1$, we conclude that if the decomposition field of a polynomial of degree $n$ is of degree $n!$, then the polynomial is irreducible.

Therefore, in our problem, polynomial $P$ is irreducible, so the extension $F(\alpha)/F$ is of degree $n$, hence the extension $K/F(\alpha)$ is of degree $(n-1)!$. In $F(\alpha)[X]$ we have $P(X)=(X-\alpha)P_1(X)$. Now, $K$ is the decomposition field of $P_1(X)$ (over $F(\alpha)$). Therefore, $P_1(X)$ is irreducible in $F(\alpha)[X]$. If $n>2$, this implies $P_1(X)$ has no linear factors, so no roots in $F(\alpha)$. This now implies point (1).

Now (2) is true since $P(X)$ is irreducible.

orangeskid
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