Identifying the Galois Group

[Ganesh Ujwal]

Identifying the Galois Group \(\displaystyle G(\mathbb{Q}[\sqrt{2}, \sqrt[3]{2}]/\mathbb{Q})\)
I am trying to determine the Galois group \(\displaystyle G(\mathbb{Q}[\sqrt{2}, \sqrt[3]{2}]/\mathbb{Q})\). I am fairly confident I have the correct answer, but I need someone to confirm my work since I have just taught myself this material today.


First, note that \(\displaystyle K = \mathbb{Q}[\sqrt{2}, \sqrt[3]{2}]\) is a field extension of degree \(\displaystyle 6\). However, it is not a splitting field of some polynomial with coefficients in \(\displaystyle \mathbb{Q}\). Thus, the order of the Galois group will be strictly less than $6$ since it is not a Galois extension.


Consider the polynomial \(\displaystyle f(x) = (x^2 - 2)(x^3 - 2)\). This polynomial has \(\displaystyle 3\) roots in \(\displaystyle K\), namely \(\displaystyle \sqrt{2}\), \(\displaystyle -\sqrt{2}\), and \(\displaystyle \sqrt[3]{2}\). Any \(\displaystyle \mathbb{Q}\)-automorphism of \(\displaystyle K\) will permute these three roots. Note that for any such automorphism, \(\displaystyle 0 = \phi(0) = \phi(\sqrt{2} - \sqrt{2}) = \phi(\sqrt{2})+\phi(-\sqrt{2})\). Hence, the only two possible automorphisms are the identity and that which swaps \(\displaystyle \sqrt{2}\) with \(\displaystyle -\sqrt{2}\).


We conclude that \(\displaystyle G(K/\mathbb{Q}) \cong \mathbb{Z}_2\).

 

[HallsofIvy]

A more "simple-minded" technique- look at numbers of the form \(\displaystyle a\sqrt{2}+ b\sqrt[3]{2}+ c\) where a, b, and c are rational numbers. Numbers of that form are "closed under addition" but not "closed under multiplication" because a product of two such numbers will have a term of the form \(\displaystyle \sqrt{2}\sqrt[3]{2}= \sqrt[6]{8}\sqrt[6]{9}= \sqrt[6]{72}\). So try numbers of the form \(\displaystyle a\sqrt[6]{72}+ b\sqrt{2}+ c\sqrt[3]{2}+ d\). Again that is "closed under addition". Is it "closed under multiplication"? If not, what needs to be added?

 

[Limax]

A more "simple-minded" technique- look at numbers of the form \(\displaystyle a\sqrt{2}+ b\sqrt[3]{2}+ c\) where a, b, and c are rational numbers. Numbers of that form are "closed under addition" but not "closed under multiplication" because a product of two such numbers will have a term of the form \(\displaystyle \sqrt{2}\sqrt[3]{2}= \sqrt[6]{8}\sqrt[6]{9}= \sqrt[6]{72}\). So try numbers of the form \(\displaystyle a\sqrt[6]{72}+ b\sqrt{2}+ c\sqrt[3]{2}+ d\). Again that is "closed under addition". Is it "closed under multiplication"? If not, what needs to be added?

The correct description is \(\displaystyle \mathbb Q\left(\sqrt2,\sqrt[3]2\right)=\{a+b2^{1/6}+c2^{1/3}+d2^{1/2}+e2^{2/3}+f2^{5/6}\,:\,a,b,c,d,e,f\in\mathbb Q\}\). But we're not actually interested in this. What we want to know are the automorphisms of \(\displaystyle \mathbb Q\left(\sqrt2,\sqrt[3]2\right)\) which fix every element in \(\displaystyle \mathbb Q\): these form a group, called the Galois group.

Ganesh Ujwal's solution is spot on. Bravo. There is a theorem that states that if \(\displaystyle K,L\) are fields and \(\displaystyle L:K\) is a Galois extension, then the order of its Galois group is equal to the degree of the extension. It would be all too easy to apply this theorem blindly. However, as Ganesh pointed out, \(\displaystyle \mathbb Q\left(\sqrt2,\sqrt[3]2\right):\mathbb Q\) is not a normal extension, hence not Galois, so the theorem does not hold. Thus the Galois group has to be computed another way. Any \(\displaystyle \mathbb Q\)-automorphism \(\displaystyle \phi\) of \(\displaystyle \mathbb Q\left(\sqrt2,\sqrt[3]2\right)\) permutes \(\displaystyle \sqrt2,-\sqrt2,\sqrt[3]2\) but it can't interchange \(\displaystyle \sqrt2\) and \(\displaystyle \sqrt[3]2\) – for example, if \(\displaystyle \phi(\sqrt2)=\sqrt[3]2\) then we would have \(\displaystyle 2=\phi(2) = \phi\left(\left(\sqrt2\right)^2\right) = \left[\phi(\sqrt2)\right]^2 = \left(\sqrt[3]2\right)^2\) which is absurd. Hence there are only two automorphisms, the identity and \(\displaystyle \sqrt2\mapsto-\sqrt2\).