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Let $f(x)\in\mathbb Q[X]$ be a polynomial of degree $n$, and let $K$ be a splitting field of $f$ over $\mathbb Q$. Suppose that $\mathrm{Gal}(K/\mathbb Q)$ is the symmetric group $S_n$ with $n>2$.

(a) Show that $f$ is irreducible over $\mathbb Q$.

(b) If $\alpha$ is a root of $f$, show that the only automorphism of $\mathbb Q(\alpha)$ is the identity.

I am beginning to learn Galois Theory, but I feel this question is a little hard for me. Can someone tell me how to solve this problem?

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    For a simple case, suppose $f=gh$ with $g$ and $h$ irreducible, of degrees $p$ and $q$, respectively, with $p+q=n$, $p\ne0\ne q$. Then the splitting field of $g$ would be of degree $\le p!$, the splitting field of $h$ of degree $\le q!$, and their compositum of degree $\le p!q! – 2017-02-06

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a) If $f$ is reducible then the action of the Galois group on the set of roots is not transitive. Indeed the Galois group never maps a root of one factor of $f$ into a root of another factor. The permutations generating the Galois group generate a group that is a proper subgroup of $S_n$, otherwise the action would be transitive.

b) The action of the Galois group is transitive so the set of roots has only one orbit. By the orbit-stabilizer theorem the stabilizer of a root $\alpha$ is the trivial group.

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    The Galois group of $(x-1)^2$ certainly maps a root of one factor in a root of the other factor. So the hypothesis that $n>2$ has to be used. – 2017-02-06
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    This is indeed true. At first sight I wonder if this case can be covered by stating that the polynomial is separable? – 2017-02-06
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    See Lubin's comment. – 2017-02-06
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    I don't see where in this comment the fact $n>2$ is used, I"m totally confused. And what about the case $(x-1)^3$? – 2017-02-06