1
$\begingroup$

Suppose I have $f(x) = (x^2-3)(x^2-2)$. The roots are $\pm\sqrt{3},\pm\sqrt{2}$. So the splitting field of $f$ over $\mathbb{Q}$, which is a Galois extension, is the smallest subfield of $\mathbb{C}$ containing all the roots. Specifically, $\mathbb{Q}(\pm\sqrt{3},\pm\sqrt{2}) = \mathbb{Q}(\sqrt{3},\sqrt{2}).$

But, a group is Galois if all homomorphism of the roots are also roots of the polynomial. For my $f(x)= (x^2-3)(x^2-2)$, there are clearly automorphisms of the roots that do not work. So, if $\sigma(\pm \sqrt{2})= \pm \sqrt{3}$, then $\sigma(\alpha)$ are not roots of the equation, since we get $((\sqrt{2})^2-3)((-\sqrt{2})^2-3)(3^2-2)((-\sqrt{3})^2-2) \neq 0$.

Does this mean $f$ doesn't have a Galois group?

$\endgroup$
1
  • 1
    $\begingroup$ The $\sigma$ you define is not an automorphism. For any automorphism $\sigma$, and any value $\alpha$ in the field it should hold that $\sigma(\alpha^2) = \sigma(\alpha)^2$. $\endgroup$
    – BHT
    Nov 21, 2019 at 17:57

1 Answer 1

2
$\begingroup$

If $f(x)$ is a polynomial in $F[x]$ (assume separable), and $K$ is the splitting field of the polynomial, then it is true that every element of $\mathrm{Gal}(f)$ (the Galois group of the splitting field $K$ over $F$) must permute the roots of $f$. In fact, if $g$ is a factor of $f$ over $F$, then the roots of $f$ that are roots of $g$ must be permuted amongst themselves.

So if $\deg(f)=n$, then $\mathrm{Gal}(f)$ is isomorphic to a subgroup of $S_n$.

But it not true that every permutation of the roots necessarily defines an element of $\mathrm{Gal}(f)$. Not even if $f$ is irreducible: there are cubics whose Galois group is cyclic of order $3$, so $\mathrm{Gal}(f)$ is not isomorphic to $S_3$.

Your error here is thinking that every permutation of the set $\{\sqrt{2},-\sqrt{2},\sqrt{3},-\sqrt{3}\}$ must correspond to an element of $\mathrm{Gal}(f)$. It does not. In fact, the Galois group is isomorphic to the Klein $4$-group, generated by the map that sends $\sqrt{2}\mapsto-\sqrt{2}$ and fixed $\sqrt{3}$; and the map that sends $\sqrt{3}\mapsto-\sqrt{3}$ and fixes $\sqrt{2}$.

$\endgroup$
5
  • $\begingroup$ I see. So, if $f=x^3-3$, then the only root is $3^{1/3}$. The possible field homomorphisms of $\mathbb{Q}(3^{1/3})$ send $3^{1/3}$ to itself or to its negative. Since $-3^{1/3}$ is not a root of $f$, then does that mean that $\mathbb{Q}(3^{1/3})$ is not a Galois exension of $\mathbb{Q}$? $\endgroup$
    – Peter_Pan
    Nov 21, 2019 at 22:48
  • 1
    $\begingroup$ @Jess: I am deeply troubled that you would even consider sending $3^{1/3}$ to $-3^{1/3}$. Why? Analogy to the square roots? You shouldn't. The three roots of $x^3-3$ are $3^{1/3}$, $\omega 3^{1/3}$, and $\omega^2 3^{1/3}$, where $\omega$ is a complex primitive cubic root of unity. You should not even consider sending $3^{1/3}$ to its negative. And, yes, $\mathbb{Q}(3^{1/3})$ is not a Galois extension: its automorphism group over $\mathbb{Q}$ is trivial, so it is its own fixed field. Remember that $K$ is Galois over $F$ if and only if the fixed field of $\mathrm{Aut}(K/F)$ is $F$. $\endgroup$ Nov 21, 2019 at 22:52
  • 1
    $\begingroup$ The real question when looking at $\mathbb{Q}(3^{1/3})$ and possible automorphisms is: "What can I send $3^{1/3}$ to?" Well, it has to be sent to a root of $x^3-3$ (and not to just similar-looking numbers like $-3^{1/3}$). There is only one real root of $x^3-3$, so there is only one possible thing to send it to in $\mathbb{Q}(3^{1/3})$. So the only automorphism is the identity. $\endgroup$ Nov 21, 2019 at 22:53
  • $\begingroup$ Yes I was thinking as an analogy to square roots, and I see why that doesn't make the slightest sense to do. Is the condition that "$K$ is Galois over $F$ if and only if the fixed field of Aut$(K/F)$ is $F$" identical to the condtin that "a number field $K$ is a Galois extension of $F$ iff #Aut$(K) = [K:F]$? $\endgroup$
    – Peter_Pan
    Nov 21, 2019 at 22:56
  • 1
    $\begingroup$ @Jess: The two conditions turn out to be equivalent, yes. $\endgroup$ Nov 21, 2019 at 23:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .