Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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
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
up vote 6 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 $F$.

share|cite|improve this answer

I think what you needed as a hypothesis was that the extension is Galois. By this, I mean that the polynomial $f(x)$ splits into distinct linear factors in an extension, so I am just assuming the negation of the problem about having repeated roots that other folks were suggesting. In this case, the transitive action on the roots of $f(x)$ should give irreduciblity, as the action permutes the roots of any irreducible factor of $f(x)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.