# Galois group of $x^3 - 2$ over $\mathbb Q$

I know the Galois group is $S_3$. And obviously we can swap the imaginary cube roots. I just can't figure out a convincing, "constructive" argument to show that I can swap the "real" cube root with one of the imaginary cube roots.

I know that if you have a 3-cycle and a 2-cycle operating on three elements, you get $S_3$. I have a general idea that based on the order of the group there's supposed to be at least a 3-cycle. But this doesn't feel very "constructive" to me.

I wonder if I've made myself understood in terms of what kind of argument I'd like to see?

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Lots of good answers here. Another way to see the Galois group is to note that an automorphism of this splitting field over $\mathbf Q$ is completely determined by where the three roots of $X^3 - 2$ are sent; more precisely, the Galois group embeds in $S_3$. Since the extension has degree $6$, you're done! – Dylan Moreland Aug 11 '11 at 2:36
@Gerry thanks for editing my question. I don't know how I expect to learn Galois theory if I can't learn to use Latex. – Marty Green Aug 11 '11 at 4:08

A brute-force way to see it is easy enough. The roots of $X^3 - 2$ are $\sqrt[3]{2},\; \omega\sqrt[3]{2},\; \omega^2\sqrt[3]{2}$, so the splitting field of your polynomial is $K = \mathbb{Q}(\sqrt[3]{2},\omega)$. You are asking why it's legitimate to send $\sqrt[3]{2}$ to $\omega\sqrt[3]{2}$. Well, let's try it. Let $\sigma: K \to K$ be a map that sends $\sqrt[3]{2}$ to $\omega\sqrt[3]{2}$. How are we going to extend this to an element of the Galois group? We need $\sigma$ to be an automorphism that fixes $\mathbb{Q}$.

Our field $K$ is a $\mathbb{Q}$-vector space generated by six basis elements: $1,\; \sqrt[3]{2},\; \sqrt[3]{4},\; \omega,\; \omega\sqrt[3]{2},\; \omega\sqrt[3]{4}$. If we can define how $\sigma$ acts on those, we can extend linearly to the whole space: that is, we can extend $\sigma$ such that $\sigma(a+b) = \sigma(a) + \sigma(b)$ for all $a$ and $b$. We also know that, if we define $\sigma$ sensibly, we should get that $\sigma(ab) = \sigma(a) \sigma(b)$. By "sensibly" here, I mean that $\sigma$ should send each element of $K$ to one of its conjugates - that is, it should send $x$ to any $y$ that satisfies the same minimal polynomial as $x$. We also know that it's enough to make sure $\sigma$ is multiplicative on the basis, and in fact we only need to define it on $\sqrt[3]{2}$ and $\omega$, since the rest will then follow automatically by multiplicativity.

So we have a few options from what we've deduced so far. $\sigma$ can send $\sqrt[3]{2}$ to any of $\sqrt[3]{2},\; \omega\sqrt[3]{2},\; \omega^2\sqrt[3]{2}$, and it can send $\omega$ to either of $\omega$ and $\omega^2$. So let's say $\sigma(\sqrt[3]{2}) = \omega\sqrt[3]{2}$ (as we wanted) and $\sigma(\omega) = \omega$ (no good reason for this choice, but we had to make one). Where do all the other basis elements go? Can you convince yourself now that $\sigma$ is an element of the Galois group?

(What would have happened if we'd chosen $\sigma(\omega) = \omega^2$?)

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There is much to be learned from the many insightful answers and comments to my question, but you are the only one who took seriously my request for a constructive demonstration. Well done and thanks! – Marty Green Aug 11 '11 at 4:06
The fact that the splitting field has degree $6$ requires some justification. (Not much justification: you just need to exhibit subfields of degrees $2$ and $3$.) @Marty: in our defense, you weren't very clear about what "constructive" meant. – Qiaochu Yuan Aug 11 '11 at 4:28
@Qiaochu: I doubt that there is anyone here providing more sophisticated and insightful answers than you, so please don't be put off when I say that your methods are the exact opposite of what I think of as constructive. – Marty Green Aug 11 '11 at 4:51

What you're probably missing is that, although the real cube root of 2 stands out, all three complex cube roots of 2 are algebraically the same, that is, they cannot be told apart, and so can be permuted at will. Hence $S_3$.

Contrast this with the seemingly analogous situation of, say, the 6-th roots of unity. Here the roots are not the same algebraically, because $x^6-1$ is not irreducible.

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I'm trying to apply your argument to, say, x^5 - 2 = 0. Can the fifth roots of two be permuted at will, giving S_5 for the Galois group? – Marty Green Aug 11 '11 at 1:05
Sorry, no, my argument does not generalize in that way. The Galois group of $x^5 - 2$ is the Frobenius group of order 20 (see opensourcemath.org/gap/small_groups.html and mathreference.com/fld-slv,xmp.html). So, perhaps my argument is no argument after all... – lhf Aug 11 '11 at 1:29
@Marty Green: Can take $\sqrt[5]{2}$ to any other root, can take any primitive fifth root of unity to any other, that's all, so order $20$. – André Nicolas Aug 11 '11 at 1:34
The problem with $x^5-2$ is that the 4-th roots of unity cannot be permuted at will, even after removing the root 1. – lhf Aug 11 '11 at 1:36
@Marty The argument works, but "two elements can be permuted at will" should be interpreted as the Galois group acts transitively on the roots of $x^5 - 2 = 0$, rather than Galois group being $S_5$. – Soarer Aug 11 '11 at 1:44

I don't know what kind of argument you'd like to see, but the standard way to compute the Galois group of an irreducible cubic (characteristic not $2$) is to look at the sign of its discriminant $\Delta = \prod_{i \neq j} (r_i - r_j)$. It has a square root $\prod_{i < j} (r_i - r_j)$ in a splitting field of the polynomial which is multiplied by $1$ when an even permutation is applied to the roots and multiplied by $-1$ when an odd permutation is applied to the roots.

By the fundamental theorem of Galois theory, it follows that the Galois group of an irreducible polynomial of degree $n$ is contained in $A_n$ if and only if the discriminant is a square. In your example, the discriminant of the cubic $x^3 + px + q$ is given by $-4p^3 - 27q^2 = -108$ which is not a square, so the Galois group is not contained in $A_3 \cong C_3$, hence must be $S_3$.

Another way is to use Dedekind's theorem that if an irreducible polynomial $f$ factors into distinct irreducible factors $f = \prod f_i \bmod p$, then the Galois group of $f$ has a permutation with cycle type $(\deg f_1, \deg f_2, ...)$. Thus to show that there is a $3$-cycle in the Galois group of $x^3 - 2$ it suffices to find a prime with respect to which this polynomial is irreducible, and this is easy since for cubics irreducibility is equivalent to not having a root. $p = 7$ works.

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The question implicitly assumes the base field is $Q$. The Galois group of a polynomial only exists with reference to a specified field containing the coefficients. It is not enough to assume that $X^3 - 2$ is irreducible, because the Galois group is cyclic of order 3 (that is, a proper subgroup of S_3 rather than the whole thing) if the base field contains primitive cube roots of 1, and $S_3$ otherwise.

If the base field contains no cube roots of 1 or 2, then a splitting field of $X^3 -2$ is obtained by adjoining all the cube roots of 1 and 2. If the adjoined roots are taken from a larger "pre-existing" field such as the complex numbers, an abstract choice of splitting field is equivalent to selecting one nontrivial cube root of 1 (two choices) and one cube root of 2 (three choices). There are six possibilities and they differ by elements of $S_3$.

The group structure is not the direct product of the cyclic subgroups of order 2 and 3, because the operation of changing one's choice of $\sqrt[3]{1}$ will also re-arrange the cube roots of 2, exchanging the two unselected $\sqrt[3]{2}$'s.

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I recently found an answer to my own question that no one else has brought up, so I thought I'd post it. The thing that always bothered me about swapping the real and imaginary cube roots of two was the assymetry of it all. What makes the positive imaginary root distinguishable from the negative imaginary root?

For me at least, the assymetry goes away when you realize that the same permutation that exchanges the real cube root of two with the positive imaginary root ALSO exchanges the real cube root of FOUR with the negative imaginary root.

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One can prove a more general result : Let $f(X)$ be an irreducible polynomial with rational coefficients and of degree $p$ prime. If $f$ has precisely two nonreal roots in the complex numbers, then the Galois group of $f$ is $S_p$

So taking here $p=3$ we're done

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