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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?

Zev Chonoles
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The Substitute
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  • 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
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    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
  • @john how is $\sqrt {-7}$ a generator? For example, I don't see how to get $a$ from this element. – The Substitute Jul 18 '15 at 05:09
  • For example an invariant under $g$, is the element $a+b=\sqrt{7}$. (You can easily compute a^2+b^2+2ab). – i. m. soloveichik Jul 18 '15 at 17:14
  • @TheSubstitute: I don't mean that $\sqrt{-7}$ is the only generator. In fact both it and $\sqrt{7}$ can be used. What you can find, with a little bit of work is that $K$ can be generated by $\sqrt{\pm 7}$ (either one), and $\sqrt{-1}$. A $\mathbb{Q}$-basis for it is then ${1,\sqrt{7},\sqrt{-7},\sqrt{-1}}$. Using this basis, it is a lot clearer to determine the intermediate fields $\mathbb{Q} \subset L \subset K$. Coming up with the generators $\sqrt{\pm 7}$ and $\sqrt{-1}$ is not trivial, but probably arises out of the work you do to confirm that $[K : \mathbb{Q}] = 4$. – john Jul 18 '15 at 21:03
  • Related: https://math.stackexchange.com/questions/204709 – Watson Dec 26 '16 at 13:33

1 Answers1

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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 $<f>=Q(\sqrt{-7})$.

i. m. soloveichik
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