# $[L:K]=n!\ \Longrightarrow \ f$ is irreducible and $\text{Gal}(L/K)\cong S_n.$

How do I go about proving the following theorem ?

Let $f\in K[T]$ have degree $n$ and splitting field $L/K$. Then we have $$[L:K]=n!\ \Longrightarrow \ f\text{ is irreducible and Gal}(L/K)\cong S_n.$$

I managed to prove that $f$ has to be irreducible, but I'm stuck showing that $\text{Gal}(f/K)\cong S_n$.

(Some things I know, that might be useful in the proof: The Galois group of $f$ has to be, thanks to irreducibility, a transitive subgroup of $S_n$; $S_n$ is transitive.
I already browsed through Dummit's book and M. Artins book on algebra but couldn't find anything that would help me.)

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How many subgroups of $S_n$ of order $n!$ are there? – Hagen von Eitzen Jan 8 at 19:14
@HagenvonEitzen Well...one...but how does that help me ? – user47580 Jan 8 at 20:47

Let $f$ be a polynomial of degree $n$ over a field $K$ and $E$ its splitting field. Assume that $E$ is a Galois extension and let $\alpha_1,\dots,\alpha_n$ be the roots of $f$. Any $K$-automorphism of $E$ is determined by its action on $\alpha_1,\dots,\alpha_n$ since $\alpha_1,\dots,\alpha_n$ generate $E$ over $K$. We also know that for any $K$-automorphism $\sigma$ of $E$ that $\sigma(\alpha_i)=\alpha_j$ for some $j$. In particular we see that $\mathrm{Gal}(K/E)$ provides a faithful group action on $\alpha_1,\dots,\alpha_n$. This means that $\mathrm{Gal}(K/E)$ embeds into $S_n$.
Now concerning your problem. First observe that if $f$ had repeated roots $L/K$ could not have degree $n!$, so $L/K$ is a separable extension. We have that $L/K$ is normal because it is a splitting field so it is a Galois extension. So it makes sense to talk about $\mathrm{Gal}(K/L)$. By the argument in the previous paragraph we know that $\mathrm{Gal}(K/L)$ embeds in $S_n$, but $\mathrm{Gal}(K/L)$ has order $n!$ so it must be that the embedding is in fact an isomorphism.
 Could you please be a little more explicit ? – user47580 Jan 8 at 21:09 @user47580 I've rewritten the answer to include more details. Let me know if there are any specific parts you don't follow. – JSchlather Jan 8 at 21:24 Hey, thanks a lot. After thinking about it a bit, a single thing is still unclear to me though: How can we rigorously infer from $f$ having repeated roots in $L/K$ that its degree has to be $n!$ ? – user47580 Jan 8 at 22:52 Suppose that $f$ has a repeated root $\alpha$. Then $f(x)=(x-\alpha)^2g(x)$ over $K(\alpha)$. The splitting field $E$ of $g(x)$ over $K(\alpha)$ is also the splitting field of $f$. Now $[E:K(\alpha)]\leq (n-2)!$ and $[K(\alpha):K]\leq n$ so $[E:K]\leq n (n-2)! show 1 more comment (I think) I see two possible points of confusion you're having, so I'll try to straighten them out. Firstly, in your edit, you've added that you know that the Galois group must be a subgroup of$S_n$. Now in Hagen von Eitzen's comment, he asks how many subgroups of$S_n$(i.e., potnetial Galois groups) have order$n!$(as required in your problem). So how many potential Galois groups are there, and what are they? Secondly, in your comments to Jacob Schlather's (great) answer, you're asking good questions, but I get the feeling you've missed the point of the first paragraph in his answer. What he's saying is, because the elements of the Galois group must permute the roots of$f$, and$f$has$n$(not necessarily distinct) roots, each permutation is a permutation on$n$(not necessarily distinct) letters. Now the key is that if those$n$letters are not distinct, then there cannot be$n!$permutations. Perhaps it's helpful to think about it this way: not counting repetition, there are$n!$permutations of all the roots, so if some of the roots were repeated, say$r_1$and$r_2$, then every permutation swapping those two, would be the same as another permutation that has the same effect on the rest of the roots, but doesn't swap$r_1$and$r_2$, and so you have fewer than$n!$distinct permutations (i.e., automorphisms!). Thus, as$L/K$is Galois, we know$|\operatorname{Gal}(L/K)| = [L:K] = n!$, and so if any of the roots were repeated, there would be fewer than$n!$(permutations) automorphisms of$L$fixing$K\$, contradicting the hypotheses.