Determine the Galois Group of $(x^2-2)(x^2-3)(x^2-5)$

Solution 1:

Note that any automorphism $\sigma$ must send $\alpha\in \mathbb Q(\sqrt2,\sqrt3,\sqrt5)$ to another root of the minimal polynomial of $\alpha$. In particular, $\sigma\sqrt2=\pm \sqrt2,\sigma\sqrt3=\pm\sqrt3,$ and $\sigma\sqrt5=\pm\sqrt5$. Since $\{\sqrt2,\sqrt3,\sqrt5\}$ is a generating set for $\mathbb Q(\sqrt2,\sqrt3,\sqrt5)$ as a field over $\mathbb Q$, we see that $\sigma$ is uniquely determined by $\sigma\sqrt2,\sigma\sqrt3,$ and $\sigma\sqrt5$. Thus the only possible automorphisms are $$\begin{align} a+b\sqrt2+c\sqrt3+d\sqrt5&\mapsto a+b\sqrt2+c\sqrt3+d\sqrt5\\ a+b\sqrt2+c\sqrt3+d\sqrt5&\mapsto a-b\sqrt2+c\sqrt3+d\sqrt5\\ &\vdots\\ a+b\sqrt2+c\sqrt3+d\sqrt5&\mapsto a-b\sqrt2-c\sqrt3-d\sqrt5\\ \end{align}$$ and in order to show that these are all automorphisms, it suffices to show that there are exactly $8$ automorphisms of $\mathbb Q(\sqrt2,\sqrt3,\sqrt5)$. Since $\mathbb Q(\sqrt2,\sqrt3,\sqrt5)$ is the splitting field of a polynomial, we know that it is Galois, so the number of automorphisms it has is equal to its degree. Recall that $$[\mathbb Q(\sqrt2,\sqrt3,\sqrt5):\mathbb Q]=[\mathbb Q(\sqrt2,\sqrt3,\sqrt5):\mathbb Q(\sqrt2,\sqrt3)][\mathbb Q(\sqrt2,\sqrt3):\mathbb Q(\sqrt2)][\mathbb Q(\sqrt2):\mathbb Q]$$ and so it suffices to show that each degree on the RHS is $2$. Clearly they are at most $2$, as each extension is obtained by adjoining a root of a quadratic polynomial. Thus it suffices to show that each extension on the RHS is nontrivial, i.e. that $\sqrt2\notin \mathbb Q,\sqrt3\notin \mathbb Q(\sqrt2)$ and $\sqrt5\notin \mathbb Q(\sqrt2,\sqrt3)$. The first is a famous theorem. Since $\mathbb Q(\sqrt2)$ has basis $\{1,\sqrt2\}$, if $\sqrt3\in \mathbb Q(\sqrt2)$ we would have $\sqrt3=a+b\sqrt2$ with $a,b\in\mathbb Q$, so $3=a^2+2b^2+2ab\sqrt2$. Since $\sqrt2$ is irrational we must have $2ab=0$, so $a=0$ or $b=0$, thus we need only observe that $3$ and $3/2$ are not squares in $\mathbb Q$. The same technique (with some additional effort) works to show that $\sqrt5\notin \mathbb Q (\sqrt2,\sqrt3)$, observing that $\{1,\sqrt2,\sqrt3,\sqrt6\}$ is a basis for $\mathbb Q(\sqrt2,\sqrt3)$.

Once you see that these are the automorphisms, it should be relatively easy to see what their fixed fields are. For example, the map $$a+b\sqrt2+c\sqrt3+d\sqrt5\mapsto a-b\sqrt2+c\sqrt3+d\sqrt5$$ has fixed field $\mathbb Q(\sqrt3,\sqrt5)$ while the map $$a+b\sqrt2+c\sqrt3+d\sqrt5\mapsto a-b\sqrt2-c\sqrt3+d\sqrt5$$ has fixed field $\mathbb Q(\sqrt6,\sqrt5)$ (why $\sqrt6$?). These fixed fields are all the maximal subfields of $\mathbb Q(\sqrt2,\sqrt3,\sqrt5)$, and the remaining intersections are pairwise intersections of these subfields (since the only other nontrivial subgroups of the Galois group are generated by pairs of automorphisms), which are easy to determine.

Solution 2:

So for your first question, the splitting field will be generated by the roots of the polynomial you are trying to split. Sometimes, some of the roots might have relations and make some other roots redundant, and also, there will be times that it is "nicer" to find some other generators for your polynomial (for instance the splitting field for $x^4-10$ would be $\mathbb{Q}(\omega, \sqrt[4]{10})$, where $\omega$ is the primitive fourth root of unity).

Back to your problem, take your polynomial and factor it into irreducible monic polynomials. In your case you would get that $(x^2-2)(x^2-3)(x^2-5)$ is already factored into irreducible parts. Note that $x^2-2$ is the minimal polynomial of $\sqrt{2}$, so $\sqrt{2}$ can only be mapped to $\pm\sqrt{2}$, similarly for $\sqrt{3}$ and $\sqrt{5}$. This reduces the possibilities for your mappings a lot. You have that you can send $\sqrt{2}$ to $\pm\sqrt{3}$, $\sqrt{3}$ to $\pm\sqrt{3}$, and $\sqrt{5}$ to $\pm\sqrt{5}$. Each combination of possibilities gives you an automorphism, so your Galois Group will be of order $2^3=8$, also note that every map is of order $2$, so the only group with order $8$ and every element of order $2$ is $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$.

From here is it straight forward to compute the fixed fields. Say you have the automorphism that negates $\sqrt{2}$ and $\sqrt{3}$ but fixes $\sqrt{5}$, then the fixed field would be just $\mathbb{Q}(\sqrt{5})$ (do you see why?), and it's the same for every automorphism.