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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.

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  • $\begingroup$ 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
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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 $x_3=-x_1-x_2$.

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  • $\begingroup$ 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

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