Let $\mathbb Q\subset\mathbb Q(\sqrt{3}+\sqrt[3]{5})$. I want to find the Galois group of the given field extension.

It would be easy for me if I could find a basis of the given field extension but how does one find a basis of that field extension ? Are there any better ways of finding the Galois group in this case ?

  • $\begingroup$ May be you should first find degree of the minimal polynomial... $\endgroup$ – kk lm Feb 5 '16 at 13:53
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    $\begingroup$ $\mathbb Q(\sqrt{3}+\sqrt{5}^3)$ is most probably equal to $\mathbb Q(\sqrt{3},\sqrt{5}^3)$ $\endgroup$ – lhf Feb 5 '16 at 14:05
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    $\begingroup$ What is $\sqrt{5}^3$ really? Whatever it is, you can better edit, please. $\endgroup$ – Piquito Feb 5 '16 at 14:07
  • $\begingroup$ What a deception!!! I have solved for $\mathbb Q(\sqrt 3+5\sqrt 5)$. I leave this question. $\endgroup$ – Piquito Feb 5 '16 at 15:05
  • $\begingroup$ Why is the extension Galois? Answering this question may also help. $\endgroup$ – johnnycrab Feb 5 '16 at 18:25

As user lhf points out in the comments, $\mathbb{Q}(\sqrt3 + \sqrt[3]{5})$ is most probably equal to $\mathbb{Q}(\sqrt3, \sqrt[3]{5})$. (Why?)

Now $M:= \mathbb{Q}(\sqrt3, \sqrt[3]{5})/\mathbb{Q}$ is not normal, because $[M:\mathbb{Q}] = 6$, thus the minimal polynomial $X^3-5$ of $\sqrt[3]{5}$ over $\mathbb{Q}$ is also irreducible over $\mathbb{Q}(\sqrt3)$ and is not decomposable into linear factors over $M$.
$X^3-5=(X-\sqrt[3]{5}) \cdot (X-\zeta\sqrt[3]{5}) \cdot (X-\zeta^2\sqrt[3]{5})$ where $\zeta = \exp(\frac{2\pi i}{3}) = -\frac{1}{2} + i\frac{\sqrt3}{2}$.

Now $L:= M(i) = \mathbb{Q}(\sqrt3, \sqrt[3]{5},i)$ is a normal closure of $M$ over $\mathbb{Q}$ and $L/\mathbb{Q}$ is a Galois extension.

$[L:\mathbb{Q}] = 12 = \#Gal(L/\mathbb{Q})$ Now consider $$f(X) = (X^2-3)\cdot(X^3-5)\cdot(X^2+1) \in \mathbb{Q}[X]$$ A $\sigma \in Gal(L/\mathbb{Q})$ is uniquely defined by the images $\sigma(\sqrt3)$, $\sigma(\sqrt[3]{5})$ and $\sigma(i)$. If you keep in mind that there are 12 distinct $\mathbb{Q}$-automorphisms and you know how $\sigma$ maps roots of an irreducible factor of a polynomial, it should be very easy to find the Galois group.

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Knowing that $\mathbb{Q}(\sqrt{3}+ \sqrt[3]{5})$ equals $\mathbb{Q}(\sqrt{3},\sqrt[3]{5})$ it is easy to see that

$\mathrm{Gal}(\mathbb{Q}(\sqrt{3}, \sqrt[3]{5})/ \mathbb{Q})\cong \mathbb{Z}_2$.

Because, if $\sigma\in \mathrm{Gal}(\mathbb{Q}(\sqrt{3}, \sqrt[3]{5})/ \mathbb{Q})$, then $\sigma(\sqrt{3})$ has to be a root of $x^2-3$ and $\sigma(\sqrt[3]{5})$ has to be a root of $x^3-5$. Since $\sigma(\sqrt{3})$ and $\sigma(\sqrt[3]{5})$ must be elements of $\mathbb{Q}(\sqrt{3}, \sqrt[3]{5})$, the only possibilities are $\sigma(\sqrt{3})=\pm \sqrt{3}$ and $\sigma(\sqrt[3]{5})=\sqrt[3]{5}$. (The other two roots of $x^3-5$ are not real.) The Galois group consists of two elements only.

(The existence of $\sigma$, follows from the existence of the corresponding $\sigma: E \rightarrow E$, see below.)

How to prove that $\mathbb{Q}(\sqrt{3}+ \sqrt[3]{5})$ equals $\mathbb{Q}(\sqrt{3},\sqrt[3]{5})$?

It is enough to show that $[\mathbb{Q}(\sqrt{3}+ \sqrt[3]{5}): \mathbb{Q}]=6$, since $\mathbb{Q}(\sqrt{3}+ \sqrt[3]{5})\subseteq \mathbb{Q}(\sqrt{3},\sqrt[3]{5})$ and $[\mathbb{Q}(\sqrt{3}, \sqrt[3]{5}): \mathbb{Q}]=6$.

There are many ways to prove this claim. The most elegant is via Galois correspondence:

The field $\mathbb{Q}(\sqrt{3}+ \sqrt[3]{5})$ is contained in the field $E=\mathbb{Q}(\sqrt{3}, \sqrt[3]{5}, \zeta)$, $\zeta$ primitive cubic root of $1$. Τhe field $Ε$ is splitting field of $(x^2-3)(x^3-5)(x^2+x+1)\in \mathbb{Q}[x]$. The Galois group $G:=G_{\mathbb{Q}}(E)$ has $12$ elements (notice $[E:\mathbb{Q}]=12$) and is generated by $\sigma, \tau, \rho$, which are fully determined by their action on $\{\sqrt{3}, \sqrt[3]{5}, \zeta\}$. This action is as follows:

$$ \sigma(\sqrt{3})=-\sqrt{3}, \sigma(\sqrt[3]{5})=\sqrt[3]{5}, \sigma(\zeta)=\zeta $$ $$ \rho(\sqrt{3})=\sqrt{3}, \rho(\sqrt[3]{5})=\zeta\sqrt[3]{5}, \rho(\zeta)=\zeta $$ $$ \tau(\sqrt{3})=\sqrt{3}, \tau(\sqrt[3]{5})=\sqrt[3]{5}, \tau(\zeta)=\zeta^2 $$ The subgroup $H$ of $G$ which fixes the elements of $\mathbb{Q}(\sqrt{3}+ \sqrt[3]{5})$ has only two elements $\mathrm{id}$ and $\tau$. The Galois correspondence informs that $2=|H|=[E: \mathbb{Q}(\sqrt{3}+ \sqrt[3]{5})]$. Hence $[\mathbb{Q}(\sqrt{3}+ \sqrt[3]{5}): \mathbb{Q}]=6$.

Additional info:

Knowing that $[\mathbb{Q}(\sqrt{3}+ \sqrt[3]{5}): \mathbb{Q}]=6$, we get that the degree $\mathrm{irr}(\sqrt{3}+ \sqrt[3]{5}, \mathbb{Q})$ is $6$. One can calculate (tedious) that the monic polynomial $p(x):=x^6- 9x^4-10x^3+27x^2-90x -2$ has $\sqrt{3}+ \sqrt[3]{5}$ as root. Hence $p(x)=\mathrm{irr}(\sqrt{3}+ \sqrt[3]{5}, \mathbb{Q})$. (Notice that the Eisenstein criterion does not help here!.)

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