Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Can someone please tell me how to determine all $\mathbb{Q}$-automorphisms of some splitting field $F$ of $X^3-3 \in \mathbb{Q}[X]$ and to determine $[F:\mathbb{Q}]$ ?

I think $[F:\mathbb{Q}]=3$, since I can construct a specific splitting field containing the three (complex) roots of the polynomial $X^3-3 $, but I don't know how to prove that it equals $3$. And for the automorphisms, I don't even know how to approach it.

share|cite|improve this question
the reason why you can't prove $[F:\mathbb{Q}]=3$ is that $[F:\mathbb{Q}]=6$ – user8268 Nov 6 '12 at 22:12
yes, once you have in hand one of the roots of the polynomial, you need to write down explicitly what the other roots are. Then you need to notice what other irrationality needs to be adjoined to the setup to get all three in any field. You'll see that there is a unique quadratic extension of $\mathbb Q$ there, and three different cubic extensions. This is an exercise that everyone must do for her/himself. – Lubin Nov 7 '12 at 2:27
up vote 4 down vote accepted

Let $F$ be a field. Let $F^\times$ be the multiplicative group of $F$. Let $(a, b) \in F^\times \times F$. We denote by $\lambda(a, b)$ the permutation $x \rightarrow ax + b$ on $F$. Let $(a, b), (c, d) \in F^\times \times F$. Since $a(cx + d) + b = acx + ad + b$, $\lambda(a, b)\circ\lambda(c, d) = \lambda(ac, ad + b)$. Hence $\{\lambda(a, b) | (a, b) \in F^\times \times F\}$ is a permutation group on $F$. We call this group the affine general linear group of degree $1$ on $F$ and denote it by $AGL(1, F)$. Since $\lambda(a, b)(0) = b$ and $\lambda(a, b)(1) - b = a$, $(a, b)$ is uniquely determined by $\lambda(a, b)$. Hence $|AGL(1, F)| = |F^\times \times F|$.

Proposition Let $p, q$ be prime numbers which may or may not be equal. Let $a$ be an integer which is divisible by $q$, but not not divisible by $q^2$. $X^p - a$ is irreducible in $\mathbb{Q}[X]$ by the Eisenstein's criterion. Let $K$ be the splitting field of $X^p - a$ in $\mathbb{C}$. Then the Galois group $G$ of $K/\mathbb{Q}$ is isomorphic to $AGL(1, F)$, where $F = \mathbb{Z}/p\mathbb{Z}$.

Proof: Let $\zeta$ be a primitive $p$-th root of unity. Let $\alpha$ be a root of $X^p - a$ in $\mathbb{C}$. $\alpha,\alpha\zeta,\dots,\alpha\zeta^{p-1}$ are distinct roots of $X^p - a$. Hence $K = \mathbb{Q}(\alpha,\alpha\zeta,\dots,\alpha\zeta^{p-1})$. Since $\zeta = \alpha\zeta/\alpha \in K$, $K = \mathbb{Q}(\alpha, \zeta)$.

Let $\sigma \in G$. Since $\sigma(\zeta)^p = 1$, there exists an integer $b$ such that $\sigma(\zeta) = \zeta^b$. Clearly $b$ (mod $p$) is uniquely determined by $\sigma$ and $b$ (mod $p) \in F^\times$. On the other hand, there exists an integer $c$ such that $\sigma(\alpha) = \alpha\zeta^c$. Clearly $c$ (mod $p$) is uniquely determined by $\sigma$. We denote $\phi(\sigma) = \lambda(b$ mod $p、c$ mod $p) \in AGL(1, F)$. Hence we get a map $\phi\colon G \rightarrow AGL(1, F)$. Let $\sigma, \tau \in G$. Suppose $\phi(\sigma) = \lambda(b$ mod $p, c$ mod $p$) and $\phi(\tau) = \lambda(d$ mod $p, e$ mod $p$). Then $\sigma\tau(\zeta) = \zeta^{bd}$ and $\sigma\tau(\alpha) = \sigma(\alpha\zeta^e) = \alpha\zeta^c\zeta^{be} = \alpha\zeta^{be + c}$. Hence $\phi(\sigma\tau) = \phi(\sigma)\phi(\tau)$. Hence $\phi$ is a homomorphism. It is easy to see that $\phi$ is injective.

It remains to prove that $\phi$ is surjective. Since $[\mathbb{Q}(\zeta) \colon \mathbb{Q}] = p - 1$, $[K \colon \mathbb{Q}(\zeta)] > 1$. Let $H$ be the Galois group of $K/\mathbb{Q}(\zeta)$. Let $\tau \in H$. There exists an integer $c$ such that $\tau(\alpha) = \alpha\zeta^c$. We define a map $\psi\colon H \rightarrow F$ by $\psi(\tau) = c$ mod $p$. It is easy to see that $\psi$ is an injective homomorphism. Hence $|F| = p$ is divisible by $|H|$. Since $|H| > 1$, $|H| = p$, Hence $[K \colon \mathbb{Q}] = p(p-1)$. Hence $\phi$ is surjective. QED

share|cite|improve this answer
Seems to me that OP is a real beginner, and that your explanation will be too highfalutin for her/him. – Lubin Nov 7 '12 at 2:29
@Lubin It is a proof rather than an explanation. I believe anyone who knows basics of Galois theory and group theory can understand it. – Makoto Kato Nov 7 '12 at 3:18
Makoto, you may be right. But, reading between the lines, I bet OP doesn't know what a semi-direct product is, and won't derive much benefit from your efforts, at least, not this year. – Gerry Myerson Nov 7 '12 at 5:42
@GerryMyerson If there are people(including the OP or otherwise) who can understand the proof and learn something from it, that would be my pleasure. – Makoto Kato Nov 7 '12 at 21:32
@MakotoKato +1 for your answer, but it would be good if you added a section that addressed beginners. For example, state all the automorphisms of the extension (in the usual way that we show that the Galois group of the polynomial above is $S_3$). Even for me reading your post takes some time to fully digest what is going on, what about for a beginner? – user38268 Dec 15 '12 at 14:04

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.