More examples of Galois groups

Let $F=Q$. Now, this time consider the polynomial $(x^2-2)(x^2-3)$. Clearly the roots of the polynomial are $\pm\sqrt{2},\pm\sqrt{3}$. The splitting field for this polynomials is $E(\sqrt{2},\sqrt{3}) = \{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}|a,b,c,d, \in Q\}$. Notice, that $x^2=\sqrt{6}$ is also a root.

Dimension of $E$ is $4$ and easy to see that $E$ is a vector space over $F$.

To derive Galois group, we start listing out the automorphisms of the extended field $E$ that fix $F$.

Thus we are seeking $\sigma$ that takes a root of $x^2-2$ to a root of $x^2-2$. That is $\sigma(\sqrt{2})=\pm\sqrt{2}$. Similarly $\sigma(\sqrt{3}) = \pm \sqrt{3}$

The value of both these roots can be taken independently. The following lists all the automorphisms of that fix $F$.

Thus $G=Gal(E/F)=\{\sigma_1=Id,\sigma_2,\sigma_3,\sigma_4\}$

It is interesting to note that we have intermediate fields between $F$ and $E$. Naming them as $B_1=F$ and $B_5=E$, we have other fields $B_2=Q(\sqrt{2}),B_3=Q(\sqrt{3}),B_4=Q(\sqrt{6})$.

These nested fields are shown below:

 As vector spaces $Q(\sqrt{2})$ has dimension $2$. Same is true of other intermediate field $Q(\sqrt{3})$.Field $Q(\sqrt{2},\sqrt{3})$ has dimension $4$ over $E=Q$ has dimensions $2$ over $Q(\sqrt{2})$ and $Q(\sqrt{3})$.

These examples lay foundation for understanding theorems about field extensions.