# Galois Groups vs $\mathbb{Q}(\alpha_i)$

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?

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

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 $$\sqrt{2}$$.

• 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 $\mathbb{Q}$? Nov 21, 2019 at 22:48
• @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 $F$. Nov 21, 2019 at 22:52
• 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. Nov 21, 2019 at 22:53
• 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$(K) = [K:F]$? Nov 21, 2019 at 22:56
• @Jess: The two conditions turn out to be equivalent, yes. Nov 21, 2019 at 23:00