# Galois group for splitting field of $\sqrt[3]{2+\sqrt2}$ and $\sqrt{2+\sqrt[3]2}$

Let $$\mathbf F = \mathbb Q(\omega)$$, where $$\omega = e^{2\pi i \over > 3}$$, determine galois group for splitting field of $$\sqrt[3]{2+\sqrt2}$$ and $$\sqrt{2+\sqrt[3]2}$$ over F

My solution is:

Let $$\alpha = \sqrt[3]{2+\sqrt2}$$ and $$\alpha' = \sqrt[3]{2-\sqrt2}$$ and $$\alpha_1 ... \alpha_6 = \alpha, \alpha', \omega\alpha, \omega\alpha', \omega^2\alpha, \omega^2\alpha'$$

I can easily determine the splliting fiel $$\mathbf K = \mathbf F(\alpha, \alpha', \omega\alpha, \omega\alpha', \omega^2\alpha, \omega^2\alpha')$$ and $$\alpha$$ has order 6 over F.

The order of Galois group is 6 or 18 which is determine by whehter $$\alpha' \subset \mathbf F(\alpha)$$ There is only two of 6 group: $$S_3$$ and $$C_6$$ For order 18 group as transitive subgroups of $$S_6$$ is $$S_3 \times C_3$$ As I do not know which is the Galois group, I choose subgroup (12)(34)(56) which belong all of them, which is fix $$\alpha^3 \subset \mathbf Q(\sqrt2)$$ and $$\alpha\alpha' \subset \mathbf Q(\sqrt[3]2)$$ The fix field $$\mathbf Q(\sqrt2, \sqrt[3]2)$$ has six order over $$\mathbf Q$$ so the Galois group must >12 = 6 * 2 so it is the $$S_3 \times C_3$$

Is ths analysis correct?

I want to use this same to determine Let $$\beta = \sqrt{2+\sqrt[3]2}, \beta' = \sqrt{2+\omega\sqrt[3]2}, \beta'' = \sqrt{2+\omega^2\sqrt[3]2}$$ and $$\beta_1 ... \beta_6 = \beta, \beta', \beta'', -\beta, -\beta', -\beta''$$ which Galois group has order 6,12 and 24.

Is there any better solution, this should be a lot if information about the group of order 6,12,18 and 24 and transitive subgroups of $$S_6$$

• For $\alpha$ the splitting field is of degree $18$ over $F$, splitting field being $F(\alpha, \sqrt [3]{2})$ so the Galois group is of order $18$ as well. Aug 23, 2023 at 4:27
• For $\alpha=\root3\of{2+\sqrt2}$ we are looking at the splitting field of the minimal polynomial $f(x)=x^6-4x^3+2$, irreducible by Eisenstein. I think $\alpha'\notin\Bbb{Q}(\alpha)$, as $\alpha\alpha'=\root3\of2$. This suggests to me that $[\Bbb{Q}(\alpha,\alpha'):\Bbb{Q}]=18$,and, consequently, the splitting field of $f(x)$ is a degree $36$ extension of $\Bbb{Q}$. Aug 23, 2023 at 9:15
• This is corroborated by an application of Dedekind's theorem (using factorizations of $f(x)$ modulo various primes). There are elements of orders two and three with different cycle types in the Galois group (over the rationals), so the Sylow groups have orders at least $4$ and $9$ respectively. On the other hand, you identified the zeros of $f(x)$, so it is obvious that the extension degree cannot be higher than $36$. Aug 23, 2023 at 9:17
• So I agree with @ParamanandSingh. Over $F$ the splitting field has degree $18$. Aug 23, 2023 at 9:23
• @ParamanandSingh,@JyrkiLahtonen, Here are some question, that how could we make sure that $\sqrt [3]{2} \notin F(\alpha)$, and Even we know the order of group, how to decide which it is? Aug 23, 2023 at 9:32

Let $$K=\mathbb{Q} (\sqrt{2})$$. Every algebraic integer of $$K$$ is of the form $$p+q\sqrt {2}$$ with $$p, q$$ as integers. Let us observe that the numbers $$a=\sqrt[3]{2+\sqrt {2}},b=\sqrt [3]{2-\sqrt {2}},c=ab=\sqrt [3]{2},d=b/a=\sqrt [3]{3-2\sqrt{2}}$$ are all algebraic integers (as they are cube roots of algebraic integers in $$K$$). We show that none of them lie in $$K$$. If $$a\in K$$ then $$2+\sqrt{2}=(p+q\sqrt{2})^3$$ which implies $$2-\sqrt{2}=(p-q\sqrt {2})^3$$ and then multiplying these equations we get $$2=(p^2-2q^2)^3$$ which is absurd as $$2$$ is not a cube of an integer. Similarly we can show that $$b\notin K, c\notin K$$.

If $$d\in K$$ then $$3-2\sqrt{2}=(p+q\sqrt {2})^3$$ which implies $$3=p^3+6pq^2,3p^2q+2q^3=-2$$ Then $$p\mid 3,q\mid 2$$ and thus $$p=\pm 1,\pm 3,q=\pm 1,\pm 2$$ and none of the combinations satisfy the equations so that $$d\notin K$$.

Since $$a, b, c, d$$ are cube roots of algebraic integers in $$K$$ each of them is of degree $$3$$ over $$K$$. We now show that $$b\notin K(a)$$. To that end let us use the trace map $$\text{tr} \, :K(a, b) \to K$$ and let $$[K(a, b) :K] =n$$.

If $$b\in K(a)$$ then $$b=p+qa+ra^2$$ for some $$p, q, r\in K$$. The numbers $$a, b, c, d$$ as well as their squares are real radicals over $$K$$ and hence their trace is $$0$$. Applying the trace map on this equation we get $$0=np+q\cdot 0+r\cdot 0$$ ie $$p=0$$ and then $$c=ab=qa^2+ra^3$$ Applying trace map again we get $$0=q\cdot 0+nra^3$$ as $$ra^3\in K$$ so that $$r=0$$. We then have $$b=qa$$ or $$d=q\in K$$ which is already proved to be false. This contradiction shows that $$b\notin K(a)$$. It then follows that $$c\notin K(a), d\notin K(a)$$.

Since the numbers $$b, c, d$$ are cube roots of some numbers lying in $$K$$ (and therefore in $$K(a)$$) it follows that each of them is of degree $$3$$ over $$K(a)$$. Let $$M=K(a, b)$$ and then $$[M:K] =9$$ and then $$[M:\mathbb{Q}] =18$$ and since $$M\subseteq \mathbb{R}$$ it follows that $$[M(\omega) :M] =2$$ ie $$[M(\omega) :\mathbb{Q}] =36$$.

In other words $$\mathbb{Q} (\sqrt {2},\omega,a,b)$$ is of degree $$36$$ over rationals and then $$F(\sqrt{2},a,b)$$ is of degree $$18$$ over $$F$$. Since $$\sqrt{2}\in F(a)$$ we can write the same field as $$F(a, b)$$.

This field $$F(a, b)$$ contains all roots of the polynomial $$f(x) =(x^3-2)^2-2$$ and is the splitting of $$f(x)$$ over $$F$$ and is of degree $$18$$ over $$F$$. Since $$f(x)$$ is irreducible over $$F$$ its Galois group must be transitive and as you mention in your post the desired group is $$S_3\times C_3$$. However it would be nice to know the individual automorphisms of $$F(a, b)$$ fixing $$F$$.

• Interesting use of the trace. I think it is correct, but I don't recall having seen the technique before! Then again, my recollection is no longer what it was at some point in my life :-) Aug 24, 2023 at 8:53
• @JyrkiLahtonen: I got the technique from discussion in comments (by orangeskid) below this answer: math.stackexchange.com/a/4251890/72031 Aug 24, 2023 at 11:46