I am trying to find the Galois Group of $f(x)=x^4 + x^2 - 12$ over $\mathbb{Q}$. I was able to show that the factors $f(x)=(x^2-3)(x^2+4)$ were irreducible over $\mathbb{Q}$ and that the splitting field $E$ of $f(x)$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt{3},i)$. Since the basis of the extension field is $\beta$={$1,\sqrt{3},i,\sqrt{3}i$} which has 4 elements, I know that $\mathbb{Q}(\sqrt{3},i)$ will have 4 automorphisms. My claim is that the automorphisms will be defined as:

(i) $\mu$: $i \rightarrow i$, $\sqrt{3} \rightarrow \sqrt{3}$.

(ii) $\tau$: $i \rightarrow i$, $\sqrt{3} \rightarrow -\sqrt{3}$

(iii) $\sigma$: $i \rightarrow -i$, $\sqrt{3} \rightarrow \sqrt{3}$

(iv) $\tau\sigma$: $i \rightarrow -i$, $\sqrt{3} \rightarrow -\sqrt{3}$

My question is if $\mathbb{Q}(\sqrt{3},i)$={$\mu, \tau, \sigma, \tau\sigma$} is the correct Galois Group. Also, in the basis of the extension I have $\sqrt{3}i$ as an element but I have not defined an automorphism for that element, that is:

$\rho$: $\sqrt{3}i \rightarrow \sqrt{3}i$, $\sqrt{3}i \rightarrow -\sqrt{3}i$.

Is that mapping already taken care of in the 4 that I already have?

Thank you for your help.


1 Answer 1


The Galois group is of order 4 so it's either the Klein 4 group or C_4 and it's obviously not C_4 cause the operation on the roots isn't transitive (since the polynomial is not irreducible over Q), so it's one of the non transitive versions of the Klein 4 group in S_4. So yeah, you got it. Those maps are taken care of since you have a map sending i to -i and stabilizing root3 thus sending iroot3 to -iroot3 and so on.


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