Ytukyg

I factor the polynomial as $(x-sqrt{3})(x+sqrt{3})(x-sqrt[3]{3})(x-rhosqrt[3]{3})(x-rho^2sqrt[3]{3})$ where $rho$ is the primitive 3rd root of unity. The splitting field will be $mathbb{Q}(sqrt{3}, sqrt[3]{3}, rhosqrt[3]{3}).$ When it comes to writing down the Galois group I run into a problem. I have two automorphisms which fix $mathbb{Q}$,
$$sigma:bigg(sqrt[3]{3} mapsto rhosqrt[3]{3}, rho to rho bigg)$$
$$tau:bigg(sqrt[3]{3} mapsto sqrt[3]{3}, rho to rho^2 bigg).$$

My question is whether these maps should also fix $sqrt{3}?$ Because $rho = (-1+sqrt{-3})/2$ and $rho^2 = (-1-sqrt{-3})/2$, $tau(sqrt{-3}) = -sqrt{-3}.$ Does this necessarily imply that $tau(sqrt{3}) = -sqrt{3}$ and thus permutes the first two roots above? Or does $tau(i) = -i$ and hence fixes $sqrt{3}$?

$tau(isqrt 3)=-tau(isqrt 3)$ doesn't really imply anything about $tau(sqrt 3)$ because there's no way to algebraically express $sqrt 3$ in terms of adding/multiplying/squaring $isqrt 3$. This can be shown more formally by proving that $x^2-3$ has no roots in $Bbb{Q}(isqrt 3)$: Any $x in Bbb{Q}(isqrt 3)$ is in the form of $x=a+bisqrt 3$ because $isqrt 3$ has minimal degree $2$. However, when you square this, you get $x^2=a^2-3b^2+2absqrt 3i$. If we want to solve $x^2=3$, we need a zero imaginary part, which happens only if $a=0$ or $b=0$. If $b=0$, then $x$ is rational, so $x^2 neq 3$. If $a=0$, then $x$ is imaginary, so $x^2<0rightarrow x^2neq 3$. Thus, $x^2neq 3$ for all $xin Bbb{Q}(isqrt 3)$.

Therefore, $tau(sqrt 3)$ is not determined by $tau(isqrt 3)$, so it could or could not permute the roots, depending on what you choose.

The splitting field over $mathbf Q$ of $(X^2-3)(X^3-3)$ is obviously $L=mathbf Q (sqrt 3, sqrt [3]3, omega)$, where $omega$ is a primitive cubic root of unity. Then the description of $mathcal G=Gal(L/mathbf Q)$ is just a question of organizing the relevant subgroups/quotients of $mathcal G$. The most convenient way seems to introduce $K=mathbf Q (sqrt [3]3, omega)$ and write $L=K(sqrt 3)$. The group $Gal(K/mathbf Q)cong S_3$ admits a unique quotient of order $2$, hence a unique quadratic subfield which is $mathbf Q (sqrt {-3})$, and so $K$ and $k=mathbf Q(sqrt 3)$ are linearly disjoint over $mathbf Q$. It follows that $mathcal G$ is a direct product, more pecisely the direct product of the subgroups $Gal(L/K) cong Gal(k/mathbf Q) cong C_2$ and $Gal (L/k)cong Gal(K/mathbf Q)cong S_3$. It remains only to make the generators explicit.
The generator $sigma$ of $Gal(L/K)$ is determined by $sigma (sqrt 3)=-sqrt 3$ (the other field generators $sqrt [3]3, omega$ being automatically invariant), whereas those of $Gal (L/k)$ are the transposition $tau:omega to omega^2$ and the $3$-cycle $gamma: sqrt [3]3 to omega sqrt [3]3$ (the other non-mentioned field generators being automatically invariant).