It can be shown the Galois group of $x^3 - x - 1$ is $S_3$. What is the orbit or $x$ under this permutation action? In other words, we can find three things, elements of the extension field that are literally permutated by Galois group?

Certainly we can name the roots $x=x_1, x_2, x_3$ and say they are permuted. So I guess I'm asking to express $x_2$ and $x_3$ as polynomials in $x_1=x$. The group $S_3$ is acting on the $\mathbb{Q}$-vector space:

$$K = \mathbb{Q}[x]/(x^3 - x - 1) \simeq \mathbb{Q}\cdot 1 + \mathbb{Q}\cdot x + \mathbb{Q}\cdot x^2 $$

and therefore we are getting a representation of the permutation group.

Edit The Galois closure $L$ is itself a quadratic extension of $K$, with $[L:\mathbb{Q}]=[L:K][K:\mathbb{Q}]=6$.

There is a way to save all of this. $S_3$ acts on $L = \mathbb{Q}[x]/p(x)$ for some irreducible sextic polynomial $p(x)$ with $\deg p = 6$.

In remains to find $p(x)$ and then... what is the orbit of $x$ under $S_3$ in the splitting field of $K$ ?

Another example: the cubic $x^3 - 3x - 1$ has Galois group is $A_3$ (which has only $|A_3|=3$ elements) and acts on the extension $\mathbb{Q}[x]/(x^3 - x - 1)$ by two permutations: $x \mapsto x^2 - x - 2$ and $x \mapsto -x^2 + 2$, and we get a representation.

  • $\begingroup$ This action fixes elements of $\mathbb Q$. Thus it sends $1.x $ to $1.x$. You probably want to ask another thing or I missed something. $\endgroup$ – mesel Jan 14 '18 at 17:43
  • $\begingroup$ By the so called normal basis theorem the action of a Galois group of a polynomial on its splitting field always gives a representation that is isomorphic to the regular representation. In other words, the comment by Lord Shark generalizes. There are certain "obvious" invariant $\Bbb{Q}$-subspaces such as the space of zero trace elements. I'm ashamed I don't know much about this. I recall having used it in the case of a finite field (when the group is necessarily cyclic) sometimes... $\endgroup$ – Jyrki Lahtonen Jan 14 '18 at 17:50
  • $\begingroup$ OTOH, when you get the spitting field by adjoining single root, then you can think of orbits in the way you described. For example, if you take the polynomial $x^3+x^2-2x-1$, then the Galois group over $\Bbb{Q}$ is cyclic of order three. The orbit of the coset of $x$ is then stable under the mapping $\alpha\mapsto \alpha^2-2$, and consists of $x,x^2-2$ and $(x^2-2)^2-2=1-x-x^2$. Those zeros sum to $-1$, of course. So, by linearity, $1\mapsto 1$, $x\mapsto x^2-2$, $x^2=(x^2-2)+2\mapsto 3-x-x^2$. $\endgroup$ – Jyrki Lahtonen Jan 14 '18 at 18:00
  • $\begingroup$ @JyrkiLahtonen it's interesting you say the invariant was obvious and then immediately that you don't know. it suggests there's a certain complexity of Galois theory that we're just not acknowledging. $\endgroup$ – cactus314 Jan 14 '18 at 18:17
  • $\begingroup$ I meant that I don't know of a general way of finding the components of that left regular representation in terms of e.g. the intermediate fields. $\endgroup$ – Jyrki Lahtonen Jan 14 '18 at 19:19

No, the Galois group is not acting on $K=\Bbb Q[x]/(x^3-x-1)$ since that isn't a Galois extension of $\Bbb Q$. Rather it acts on the Galois closure which is a degree $6$ extension. One cannot express $x_2$ as a polynomial in $x_1=x$, but of course $x_3=-x_1-x_2$.

  • $\begingroup$ i stand corrected? what is the Galois closure? Certainly there is an $S_3$ action there... and also what other intermediate field... certainly there's a quadratic? and another cubic extension. $\endgroup$ – cactus314 Jan 14 '18 at 18:19
  • $\begingroup$ The Galois group is $S_3$, so if you need a census of the subfields of $L$ consider the subgroups of $S_3$. $\endgroup$ – Lord Shark the Unknown Jan 14 '18 at 18:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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