The polynomial $f(x) = x^6 - 7x^5 + 21x^4 - 41x^3 + 63x^2 - 63x + 27$ defines a Galois extension $H$ of $\mathbb{Q}$. The Galois group of the extension $H/\mathbb{Q}$ is dihedral, and depending on your notation is $D_6$ or $D_3$.

(Aside: in fact I generated the polynomial by defining $g_1(\tau) = \eta(\tau/3)/\eta(\tau)$, where $\eta(\tau)$ is the Dedekind eta function, and letting $w = g_1(\tau)^2$ where $\tau = (-1 + \sqrt{-23})/2$ and taking the minimum polynomial of $w$ over $\mathbb{Q}$. In fact $w$ generates the Hilbert class field $H$ of $\mathbb{Q}(\sqrt{-23})$ over $\mathbb{Q}$.)

Suppose that the roots of $f(x)$ are written $r_1, r_2, \ldots, r_6$. Also suppose that they are ordered such that $r_1, r_2$ and $r_3, r_4$ and $r_5, r_6$ are complex conjugate pairs.

An automorphism in $G =$ Gal$(H/\mathbb{Q})$ defines a permutation on the roots.

My question is this: is there a $\sigma \in G$ which as a permutation on the roots is given by $(r_1 r_2)(r_3 r_4)(r_5 r_6)$.

  • $\begingroup$ I fed it into a graphing program, and it looks like the polynomial is positive for all real $x$, so, no real roots. $\endgroup$ Feb 29, 2016 at 23:15
  • $\begingroup$ Yes, the roots are approximately: [1.2849201454990266329556999790597843244 - 0.24186999128647610163946329705918097326*I, 1.2849201454990266329556999790597843244 + 0.24186999128647610163946329705918097326*I, 2.2548776662466927600495088963585286919 - 0.42445224584391909433345158075420027900*I, 2.2548776662466927600495088963585286919 + 0.42445224584391909433345158075420027900*I, -0.039797811745719393005208875418313016315 - 1.7315935245259645748258041542679657077*I, -0.039797811745719393005208875418313016315 + 1.7315935245259645748258041542679657077*I] So no real roots. $\endgroup$
    – Bill
    Feb 29, 2016 at 23:47

3 Answers 3


The complex conjugate is such automorphism.


I present here two different answers. The first is how I see it. The second is how another mathematicians sees it, even after I explained my answer and followed it with a four hour discussion. I wonder if anyone else is convinced by the second answer?

Answer 1: Let $\tau$ be complex conjugation. It is well known and easy to see that $\tau(H) = H$. Clearly $\tau$ fixes $\mathbb{Q}$ and thus it is a $\mathbb{Q}$-automorphism of $H$. By definition it is therefore an element of Gal$(H/\mathbb{Q})$ and by definition it has the required action on the roots of $f$.

Answer 2: If complex conjugation were a Galois automorphism of $H$ then $H$ would have a totally real subfield. As it does not (admitted), it is not a Galois automorphism.

$H$ is the field $\mathbb{Q}[x]/(f)$. Given an ordering of the roots $r_1, \ldots, r_6$ there is a map $\phi : H \to \mathbb{C}^6$ which is defined by $x \pmod{f} \mapsto (r_1, r_2, \ldots, r_6)$.

An automorphism in $G =$ Gal$(H/\mathbb{Q})$ induces a permutation on the roots. However, there is no automorphism in $G$ which maps $(r_1, r_2, r_3, r_4, r_5, r_6)$ to $(r_2, r_1, r_4, r_3, r_6, r_5)$. Instead an automorphism that permutes the roots so that $r_1$ is sent to $r_2$ and vice versa will send $(r_1, r_2, r_3, r_4, r_5, r_6)$ to something like $(r_2, r_1, r_6, r_5, r_4, r_3)$.

A demonstration follows using the favourite computer algebra system of this mathematician, which confirms his claim.

Are there two fundamentally different definitions of the terms used in my question in the literature? Specifically, what is meant by a permutation of the roots given by a Galois automorphism?

  • 1
    $\begingroup$ What makes you admit that $H$ has no totally real subfield? $\endgroup$ Feb 29, 2016 at 23:17
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    $\begingroup$ No. The value $r_1 + r_2$ is a root of $x^3 - 7x^2 + 18x - 17$. The value $r_3 + r_4$ is a root of $x^3 - 7x^2 + 15x - 17$ and $r_5 + r_6$ is a root of $x^3 - 7x^2 + 12x + 1$. (Given the ordering of roots given by the numerical approximations above, anyway.) $\endgroup$
    – Bill
    Feb 29, 2016 at 23:56
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    $\begingroup$ A field fixed by complex conjugation must be real, but it does not need to be totally real. For example $\mathbb Q(\sqrt[3]2)\subset \mathbb Q(\sqrt[3]2,\zeta_3)$ has two complex embeddings into $\mathbb C$ even though the field itself is real. $\endgroup$
    – Mathmo123
    Mar 1, 2016 at 8:53
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    $\begingroup$ $\mathbb Q(\sqrt[3]2)$ is a real number field - by which I mean an abstract number field that has been embedded into $\mathbb R$. $\mathbb Q[X]/(X^3-2)$ is an abstract field which can be embedded into $\mathbb C$ in three ways, one of which real. Here there is an ambiguity because the extension is not Galois. In the case where a number field $K$ is Galois, any embedding of $K$ into $\mathbb C$ will be an automorphism of $K$. Either way, if a number field is Galois and has complex roots, then its Galois group contains an element corresponding to complex conjugation. There is no ambiguity there. $\endgroup$
    – Mathmo123
    Mar 1, 2016 at 9:23
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    $\begingroup$ An embedding of $K = \mathbb{Q}[x]/(x^3-2)$ into $\mathbb{C}$ isn't an automorphism of $K$. An automorphism is an isomorphism from a field to itself. An embedding of $K$ into $\mathbb{C}$ is nothing more than an injection, or if you prefer, an isomorphism onto its image. In my example in the question, $\mathbb{Q}[x]/(f)$ is a Galois number field. Yet there is no single automorphism in the Galois group which corresponds to complex conjugation. The automorphism is different depending on which embedding you choose. The situation would be as you say if the number field were abelian, but it isn't. $\endgroup$
    – Bill
    Mar 1, 2016 at 10:11

Your friend is right likely as confused as I was.

Since your extension is Galois, there are $6$ polynomials $P_1,\ldots,P_6$ with rational coefficients such that $f(T)$ factors in $\Bbb Q[X]/(f(X))$ as $\prod (T-P_i(X))$.
Those polynomials will permute the roots, and with composition modulo $f$, they form a group isomorphic to the Galois group, but we have to be careful because their actions on the roots isn't the action of the Galois group itself.

You can actually explicitly recover the polynomials from the roots by looking at what combinations of them are "rational".

We know that the Galois group is isomorphic to $S_3$ so it will have $3$ "transpositions" that will separate the $6$ roots in $3$ pairs.

For example, $\{\{r_1,r_2\},\{r_3,r_6\},\{r_4,r_5\}\}$ is a "rational" object because $(X-(r_1+r_2))(X-(r_3+r_6))(X-(r_4+r_5))$ (and similarly for the products) have rational coefficients, so this partition is invariant by the Galois group. Since its element have only $3$ conjugates, they are fixed by a subgroup $\{id,\tau\}$ of $G$. Hence the Galois group (as a subgroup of the permutation of the complex roots) contains the permutation $\tau = (r_1,r_2)(r_3,r_6)(r_4,r_5)$

Lo and behold, if you interpolate the polynomial that go through the $6$ points in the graph of $\tau$, you obtain
$P_{(1,2)(3,6)(4,5)} = \frac 1 9 (27-21x+14x²-9x^3+4x^4-x^5)$.

You can look for the other two transpositions and you will obtain

$P_{(1,5)(2,6)(3,4)} = \frac 19 (63-63x+41x^2-21x^3+7x^4-x^5)$ $P_{(1,3)(2,4)(5,6)} = \frac 19 (297-483x+350x^2-204x^3+88x^4-16x^5)$

And you obtain the last two elements by making compositions and interpolating (or composing the polynomials then looking modulo $f$)

$P_{(1,4,6)(2,3,5)} = -12+25x-19x^2+11x^3-5x^4+x^5$
$P_{(1,6,4)(2,5,3)} = -24+37x-26x^2+15x^3-6x^4+x^5$

From this polynomial group acting on the roots, we recover the Galois group by taking the transpose of the group law of the polynomials (if you write the composition table, you go from one to the other by transposing it).

Say we fix a root $r_1$ and we decide to call $\sigma_i$ and $P_i$ the automorphism and the polynomial sending $r_1$ to $r_i$.
Then $\sigma_i(r_j) = \sigma_i(P_j(r_1)) = P_j(\sigma_i(r_1)) = P_j(r_i)$.

And $(\sigma_i \circ \sigma_j)(r_1) = \sigma_i (P_j(r_1)) = (P_j \circ P_i) (r_1)$ so the map $\sigma_i \mapsto P_i^{-1}$ is an isomorphism of groups (but not of group actions on the roots !!)

Finally we can check that $\sigma_2(r_3) = P_3(r_2) = r_4$, which is enough to conclude that $\sigma_2$ acts as complex conjugation.

  • $\begingroup$ The Galois group is not abelian, hence I don't think you can't just compose the polynomials modulo f. As I understand it, that only works for abelian Galois groups. $\endgroup$
    – Bill
    Mar 2, 2016 at 19:57
  • $\begingroup$ I don't see why not. Composition is not commutative. $\endgroup$
    – mercio
    Mar 2, 2016 at 19:59
  • $\begingroup$ Similarly I don't think you can apply the polynomial to all the roots to see how it permutes them. If that were the case then for any $\sigma$ in the Galois group I would have $\sigma(P(r_i)) = P(\sigma(r_i))$ which would effectively mean all the elements of the Galois group commuted. $\endgroup$
    – Bill
    Mar 2, 2016 at 20:02
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    $\begingroup$ you have to be careful, because $P_{\sigma_1 \circ \sigma_2} = P_{\sigma_2} \circ P_{\sigma_1}$ (the order of composition is reversed). By definition, if $\sigma(r_1) = P_\sigma (r_1)$, then $\sigma(P(r_1)) = P(P_\sigma(r_1))$ forall polynomial $P$, but ONLY for the root $r_1$. I must say I'm also a bit confused there lol $\endgroup$
    – mercio
    Mar 2, 2016 at 20:47
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    $\begingroup$ oh I forgot that this WAS the question. Say we "focus" on $r_1$ for applying those rules. $\sigma_{(1,2)(3,6)(4,5)}(r_3) = \sigma_{(1,2)(3,6)(4,5)}(P_{(1,3)(2,4)(5,6)} (r_1)) = (P_{(1,3)(2,4)(5,6)} (r_2)) = r_4$, so yes, complex conjugation IS an element of the Galois group after all... $\endgroup$
    – mercio
    Mar 2, 2016 at 21:18

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