# Finding the Fixed Fields in the Galois Correspondence for the Splitting Field of $x^4-3x^2+4$ over $\mathbb{Q}$

I have found the Galois group of the polynomial $x^4 - 3x^2 + 4$ (see below), but I am not sure how to find the fixed fields in the Galois correspondence. The roots of the polynomial are $$\pm \sqrt{\frac{3\pm i\sqrt7}{2}}$$ so the splitting field is $$K=\mathbb{Q}\Bigg(\sqrt{\frac{3+ i\sqrt7}{2}},\sqrt{\frac{3- i\sqrt7}{2}}\Bigg)$$ The degree of the extension $K/\mathbb{Q}$ is 4. Denote $$a=\sqrt{\frac{3+ i\sqrt7}{2}} \\b=\sqrt{\frac{3- i\sqrt7}{2}}\\c=-\sqrt{\frac{3+ i\sqrt7}{2}}\\d=-\sqrt{\frac{3- i\sqrt7}{2}}$$ and define $$f:a\mapsto c, b\mapsto d, c\mapsto a, d\mapsto b$$ and $$g:a\mapsto b,b\mapsto a, c\mapsto d,d\mapsto c$$ so that $f^2=g^2=1$ and that $$fg=gf:a\mapsto d,b\mapsto c,c\mapsto b, d\mapsto a$$ So $\text{Gal}(K/\mathbb{Q})=\lbrace 1,f,g,fg\rbrace\cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.

The nontrivial subgroups of the Galois group are $\langle f\rangle,\langle g\rangle,\langle fg\rangle$. How can I determine the subfields of $K$ that are fixed by each of these subgroups? I tried finding a basis for $K$ and seeing how each of these automorphisms act on the basis, but I'm having trouble finding a basis: Here's my attempt $$a_0+a_1\sqrt{\frac{3+i\sqrt7}{2}}+a_2\sqrt{\frac{3-i\sqrt7}{2}}+a_3\frac{3+ i\sqrt7}{2}+a_4\frac{3- i\sqrt7}{2}$$ which I know is incorrect since the basis should have only 4 elements. Can this be adjusted or is there a better way to proceed?

• Finding a useful basis may be easier if you choose other generators for the extension $K/\mathbb{Q}$. For example, one of them could be $\sqrt{7}$. Also, the elements you have denoted by $a,b,c$ and $d$ do not form a basis of $K$ as a $\mathbb{Q}$-vector space since $a = -c$ and $b = -d$, which means $f$ and $g$ are not well-defined. – john Jul 17 '15 at 23:26
• This is assuming that $[K: \mathbb{Q}] = 4$ which I believe is true in this case, but is not automatic. – john Jul 17 '15 at 23:28
• Note that $f(ac) = ac, f(bd) = bd$, and that $ac$ and $bd$ are both elements of degree $2$ over $\Bbb Q$, since they are the solutions to $x^2 + 3x + 4 = 0$. As for $g$, this approach doesn't work so well, since $ab = cd = 2$. – Arthur Jul 17 '15 at 23:37
• In my comment above, I should say $\sqrt{-7}$, not $\sqrt{7}$. – john Jul 17 '15 at 23:41
• Some MathJax advice: < and > mean "less than" and "greater than", and produce spacing correct for that meaning only; to make angle brackets, use \langle and \rangle. – Zev Chonoles Jul 17 '15 at 23:58

You can easily compute $a^2+b^2+2ab=7$. Thus an invariant under $g$, is the element $a+b=\sqrt{7}$. So the fixed field for $<g>=Q(\sqrt{7})$.
You can easily compute $a^2+b^2-2ab=-1$. Thus an invariant under $h=fg$, is the element $a-b=\sqrt{-1}$. So the fixed field for $<h>=Q(\sqrt{-1})$.
Then it follows easily that $<f>=Q(\sqrt{-7})$.
• what's the intuition for considering $a^2+b^2\pm 2ab$? – The Substitute Jul 19 '15 at 0:23
• @TheSubstitute We want $a+b$ or $a-b$ so by squaring we can evaluate the algebra. – i. m. soloveichik Jul 19 '15 at 1:35