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Let $F$ be a field, $f(x)$ a non-constant polynomial, $E$ the splitting field of $f$ over $F$, $G=\mathrm{Aut}_F\;E$. How can I prove that $G$ acts transitively on the roots of $f$ if and only if $f$ is irreducible?

(if we suppose that $f$ doesn't have linear factor and has degree at least 2, then we can take 2 different roots that are not in $F$, say $\alpha,\beta$, then the automorphism that switches these 2 roots and fixes the others and fixes also $F$, is in $G$, but I'm not using the irreducibility, so where is my mistake?

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How do you know that there is an automorphism that switches any two roots and fixes the others? (Consider $f = (x^2 - 2)(x^2 - 3)$ over $\mathbb{Q}$.) –  Qiaochu Yuan May 4 '12 at 15:07
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The claim is false: e.g. take $F=\mathbb{Q}$, $f(x)=x^2$. The splitting field is just $F$, there's only one root so the automorphism group trivially acts transitively, but $f$ is reducible. –  Chris Eagle May 4 '12 at 15:09
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In addition, the claim that there is an automorphism switching any two roots and fixing the others is false even for irreducible $f$; take, for example, any cubic $f$ with Galois group $C_3$ (in characteristic not equal to $2$ and $3$ this is equivalent to the discriminant being square in the base field). –  Qiaochu Yuan May 4 '12 at 15:27
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up vote 3 down vote accepted

As noted, the claim is false if $f(x)$ is a perfect power of an irreducible polynomial. One direction always holds.

To prove that if $f(x)$ is irreducible then the action is transitive, you can use the following result as a lemma:

Theorem. Let $F$ and $L$ be fields, and let $\sigma\colon F\to L$ be a field isomorphism. Let $g(x)\in F[x]$ be a nonzero polynomial, let $\sigma g(x)\in L[x]$ be the corresponding polynomial. If $K$ is a splitting field for $g(x)$ over $F$, and $M$ is a splitting field of $h(x)$ over $L$, then $\sigma$ extends to an isomorphism $\tau\colon K\to M$ such that $\tau|_{F}=\sigma$.

With this theorem in hand, proceed as follows: let $u,v$ be two roots of $f(x)$ in $K$. Then there exists an isomorphism $\sigma\colon F(u)\cong F(v)$ that is the identity on $F$ and maps $u$ to $v$ (since $F(u)\cong F[x]/(f(x)) \cong F(v)$). Now view $K$ as a splitting field for $f(x)$ over both $F(u)$ and $F(v)$ to obtain an extension of $\sigma$ to all of $K$. This gives you an automorphism of $K$ that fixes $K$ and maps $u$ to $v$, proving that $\mathrm{Aut}_F(K)$ acts transitive on the roots.

Note however that it may be impossible to find an automorphism that has a particular cycle structure on the roots; for instance, your automorphism may just permute the roots cyclically, as in the case of a splitting field of an irreducible polynomial of degree $3$ with three real roots over $\mathbb{Q}$.

If $f(x)$ is not irreducible and not a power of an irreducible polynomial, let $g_1(x)$ and $g_2(x)$ be two distinct irreducible factors of $f(x)$ in $F[x]$; if $u$ is a root of $g_1(x)$, then for every $\sigma\in\mathrm{Aut}_F(K)$, we have $\sigma(u)$ is a root of $g_1(x)$; hence it is never a root of $g_2(x)$, since $g_2(x)\neq g_1(x)$ and distinct irreducibles have distinct roots in the splitting field; so there are roots of $f(x)$ (namely, those of $g_2(x)$) that are not in the $\mathrm{Aut}_F(K)$-orbit of $u$, proving that the action is not transitive.

The proof of the theorem above is by induction on $[K:F]$. If $[K:F]=1$, then $\tau=\sigma$ works. If $[K:F]\gt 1$, let $h(x)$ be an irreducible factor of $g(x)$ of degree greater than $1$ (which must exist, otherwise $g$ splits in $F$), and let $\sigma h$ be the corresponding factor of $\sigma g$. Let $u\in K$ be a root of $h$, let $v\in M$ be a root of $\sigma h$. Then $\sigma$ extends to an isomorphism $\rho\colon F(u)\cong L(v)$ that maps $u$ to $v$, since $F(u)\cong F[x]/(h(x)) \cong L[x]/(\sigma h(x)) \cong L(v)$. Then, inductively, $\rho$ extends to an isomorphism $\tau\colon K\to M$ that restricts to $\rho$ on $F(u)$ and hence to $\sigma$ on $F$.

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