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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 $S_6$

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  • $\begingroup$ 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. $\endgroup$
    – Paramanand Singh
    Aug 23, 2023 at 4:27
  • $\begingroup$ 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 $\Bbb{Q}$. $\endgroup$ Aug 23, 2023 at 9:15
  • $\begingroup$ 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 $36$. $\endgroup$ Aug 23, 2023 at 9:17
  • $\begingroup$ So I agree with @ParamanandSingh. Over $F$ the splitting field has degree $18$. $\endgroup$ Aug 23, 2023 at 9:23
  • $\begingroup$ @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? $\endgroup$
    – Wuming Liu
    Aug 23, 2023 at 9:32

1 Answer 1

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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 $F$.

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  • $\begingroup$ 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 :-) $\endgroup$ Aug 24, 2023 at 8:53
  • $\begingroup$ @JyrkiLahtonen: I got the technique from discussion in comments (by orangeskid) below this answer: math.stackexchange.com/a/4251890/72031 $\endgroup$
    – Paramanand Singh
    Aug 24, 2023 at 11:46

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