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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 $\mathbb{Q}$?

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  • $\begingroup$ Excellent question! This is the archetypical example of the way in which abstraction in mathematics (which, as a categorist, I can hardly oppose) loses touch with the motivating situation by failing to explain the reverse journey. I would love to see a good explanation here. $\endgroup$
    – Paul Taylor
    Commented Apr 21, 2015 at 17:11
  • $\begingroup$ The identity morphism is the only automorphism of $\mathbb{Q}(x)$. If you want to consider nontrivial automorphisms of a field containing $x$ that do not fix $x$, you must consider, at a minimum, $\mathbb{Q}(x,z)$. $\endgroup$
    – user14972
    Commented Apr 21, 2015 at 19:18

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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.

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  • $\begingroup$ 1 OK. This extension is not Galois since $\mathrm{Aut}[\mathbb{Q}[x]/\mathbb{Q}]= 6 > [\mathbb{Q}[x]:\mathbb{Q}]=3$ $$.$$ 2 The action of the Galois group of $x^3 - x - 1$ cannot be extended to all of $\mathbb{C}$ but maybe it can be done using Riemann surfaces? The Riemann surface for $y^2 = x^2 + ax + b$ is always a sphere $\hat{\mathbb{C}}$ but for $y^2 = x^3 + ax + b$ it is hyperelliptic curve. Maybe the Galois action can be extended to that object. $\endgroup$
    – cactus314
    Commented Apr 21, 2015 at 18:08
  • $\begingroup$ The action of the Galois group can be extended to all of $\mathbb{C}$. It just won't be a continuous action. $\endgroup$
    – Pace Nielsen
    Commented Apr 21, 2015 at 18:25
  • $\begingroup$ There is no natural way to extend the automorphisms to $\mathbf C$. There are lots of extensions and no clean formula for them. What is it you really want to do, and why? Your comment is similar to (but more subtle than) the fact that for a vector space $V$ and subspace $W$, any linear map $W \rightarrow W$ can be extended to a linear map $V \rightarrow V$. But there are tons of ways of doing this, so unless you have a specific goal in mind it's not clear that the extension has any worthwhile features. $\endgroup$
    – KCd
    Commented Apr 21, 2015 at 19:11
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    $\begingroup$ In your first comment, it is wrong that ${\rm Aut}(\mathbf Q(x)/\mathbf Q)$ has order $6$. In fact its size is $1$. Or maybe you are not writing what you meant to write. $\endgroup$
    – KCd
    Commented Apr 21, 2015 at 19:13
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    $\begingroup$ I realized where you got the example from, but that doesn't tell me why you said ${\rm Aut}(\mathbf Q(x)/\mathbf Q)$ has size $6$. The term "Galois group of a polynomial" over some field $K$ means the Galois group of its splitting field over $K$, not of the field generated over $K$ by just one root. If the Galois group of a cubic is isomorphic to $S_3$, that does not mean adjoining one root gives an extension of fields with automorphism group of size $6$, but rather that adjoining all the roots gives an extension with automorphism group of size $6$. $\endgroup$
    – KCd
    Commented Apr 21, 2015 at 22:10
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Let $\theta = \sqrt{-23}$, and let $K = \mathbf{Q}(x,\theta)$. One can describe the action of $S_3$ on $K$ as follows. The element $\sigma = (123) \in S_3$ fixes $\theta$ and sends $x$ to

$$ \frac{1}{23} \left(2 \theta + \left(\frac{-23 + 9 \theta}{2} \right) x - 3 \theta x^2\right).$$

Here the $S_3$ action is determined by ordering the roots as $(x, \sigma x, \sigma^2 x)$. With this choice, the element $\tau = (23)$ fixes $x$ and sends $\theta$ to $- \theta$.

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You define your splitting field as $\mathbb{Q}[X,Z]/(X^3-X-1,Z^2+XZ+X^2-1)$

The three roots of your equation in the splitting field are $x,z$ and $-z-x$. Now for any element $ \sigma \in S_3$ you know how to construct a field automorphism: the image of $x$ has to be either $x$,$z$ or $(-x-z)$ depending on the value of $\sigma(1)$ and the image of $z$ has to be either $x$,$z$ or $(-x-z)$ depending on the value of $\sigma(2)$.

so for example $(12)$ exchange $x$ and $z$ and $(13)$ send $x$ to $-x-z$ and preserve $z$.

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    $\begingroup$ If by rational fonction you mean with rational coefficient then there cannot be : such a function would preserve the subfied generated by $x$ and this fields does not contains $z$. $\endgroup$
    – Simon Henry
    Commented Apr 21, 2015 at 16:11
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    $\begingroup$ This is completely explicit : if you have any element of the field it can be written as a polynome in $x$ and $z$ and for any $\sigma \in \S_3$ you just replace all the $x$ and $z$ by their image under $\sigma$, exactly as you do in the two other examples you mentioned. $\endgroup$
    – Simon Henry
    Commented Apr 21, 2015 at 16:13
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    $\begingroup$ @john: Except in very rare circumstances, field automorphisms do not act like plugging a field element into a rational function. Instead, in many common situations it's exactly the other way around: you want to view your field elements as rational functions, and then the Galois action is to evaluate them. e.g. here, any permutation of $\{ x, z, -x-z \}$ corresponds to a Galois action, e.g. the action $r(x,y) \mapsto r(z, -x-z)$. $\endgroup$
    – user14972
    Commented Apr 21, 2015 at 19:22
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    $\begingroup$ The other two roots can't be written as rational functions of $x$ with rational coefficients, however the discriminant of $f(x)$ is $-23$, thus they can be written as polynomials in $x$ over $\mathbb{Q}(\sqrt{-23})$. $\endgroup$
    – Jarek Kuben
    Commented Apr 21, 2015 at 20:37
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    $\begingroup$ ... with the caveat that only the $A_3 \subseteq S_3$ subgroup of permutations can be written in terms of substitution $x \mapsto p(x)$. (that is, you only get the Galois group of $\mathbb{Q}(x,z) / \mathbb{Q}(\sqrt{-23}))$. $\endgroup$
    – user14972
    Commented Apr 21, 2015 at 20:48

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