I have always been frustrated with how indirect discussions of Galois Theory are in Algebra textbooks. Even in fine treatments such as Miles Reid.
Are there any good examples where we can draw the Galois action explicitly as substitutions??
One possibility are the cyclotomic polynomials:
$$ x^4 + x^3 + x^2 + x + 1 = 0 $$
Then for any $1 \leq a < 5$ we can make subsition $x \mapsto x^a$ and this polynomial is fixed.
This gives me some intuition that the Galois action should behave something like the exponential operator and we often write $x \mapsto x^\sigma$ for $\sigma \in \mathrm{Gal}[p(x)]$
Iterating square roots is another examples such as $x = \sqrt{2} + \sqrt{3}$ and we should recover and explicit $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ action. Even here I am not sure I can get the analogy to the exponential function.
What can we do for the example $f(x) = x^3 – x – 1$ which I took from Keith Conrad [1]
The Galois group is $S_3$. How can we get the conjugates $x$?
Have two equations for the roots $x + a+b = 0$ and $xab = 1$ which I can re-write:
$$ a + b = – x \hspace{0.25in}\text{and}\hspace{0.25in} ab = \frac{1}{x} = x^2 – 1$$
Then the quadratic equation is $(z-a)(z-b) = z^2 + xz + ( x^2 – 1) = 0$ which is reducible in $\mathbb{Q}(x)$.
With this satisfactory result, which explicity substitutions achieve the transpositions $S_3 = \langle (12), (13)\rangle$?
It's not totally clear what I mean by "explicit". The crossed out section seems to reproduce Theorem 2.6 in this @KCd's note except I forget to adjoin the square root of the discriminant.
Since $[\mathbb{Q}(x):\mathbb{Q}] = 3$, I would like to compute the Galois action as a matrix in the basis $\{1, x, x^2\}$.
¿How to we write the element that performs (hopefully, I wrote the roots of $x^3 – x – 1$ correctly):
$$ x \mapsto x \hspace{0.25in}\text{and}\hspace{0.25in} z = \frac{-x + \sqrt{4-3x^2}}{2}\mapsto \frac{-x – \sqrt{4-3x^2}}{2}$$
and similar permutations?
It seems that $4 – 3 x^2$ should be a perfect square in $\mathbb{Q}(x)$ since I have defined it as the splitting field.
Since $x^3 – x – 1$ splits in the 6th degree extension $\mathbb{Q}(x, \sqrt{-23})$ how to compute the action of the Galois group $S_3$ on the basis $\{ 1, x, x^2 \} \times \{ 1, \sqrt{-23}\}$ as a $6 \times 6$ matrix with entries in $\mathbb{Q}$?
Best Answer
John, in cyclotomic extensions of $\mathbf Q$ the Galois group acts like a power map only on the roots of unity, not on general elements, so don't think about the Galois group as "exponential." For comparison, complex conjugation is an element of ${\rm Gal}(\mathbf C/\mathbf R)$ and it acts on $i$ by sending it to $-i$, but that doesn't mean the Galois group is negation on all of $\mathbf C$; negation is not a multiplication operation in characteristic $0$.
You are asking for a formula for the conjugates of a cubic polynomial, when you know one root, using rational functions in that root with coefficients in $\mathbf Q$. But the problem with your example of $x^3 - x - 1$ is that adjoining one root to $\mathbf Q$ doesn't give you the other two. If $r$ is one root of that polynomial then the field $\mathbf Q(r)$ doesn't contain either of the other roots. Indeed, if for concreteness we let $r \approx 1.3247$ be the unique real root then $x^3 - x - 1 = (x-r)(x^2 + rx + r^2-1)$ and the discriminant of the second factor is $4-3r^2 \approx -1.26$, so it has no roots in $\mathbf R$, let alone in $\mathbf Q(r)$. So your dream of rational function formulas is impossible. If you are content with formulas for the Galois group as permutations of the roots, then this example has a Galois group of order $6$ that's isomorphic to $S_3$, so the answer is: all 6 permutations of the three roots extend to automorphisms of the splitting field over $\mathbf Q$.
The same thing happens any time the splitting field of a polynomial is not generated by one root of the polynomial: there will be roots that don't have rational formulas in terms of one root. Your example of cyclotomic extensions is misleading in this respect, since there adjoining a single primitive $n$th root of unity does give you all the other $n$th roots of unity as its powers. That is not how life works in general.
Galois himself never suggested that his theory is supposed to be easily computable. He wrote "If now you give me an equation that you have chosen at will, and about which you want to know if it is or is not solvable by radicals, I cannot do any more than indicate the means for answering your question, without wanting to charge either myself or any other person with doing it. In a word, the calculations are impractical."
If you want to see some examples of Galois groups written as groups of permutations, look here.