Subfields of a splitting field using Galois correspondence

Question

My question concerns the techniques for finding the subfields of a given splitting field using Galois correspondence. I still struggle sometimes with these kind of problems and thus shall ask this question applied to a concrete example.

Given the irreducible polynomial $f = X^4 + 4X^2 + 9$ over $\mathbb{Q}$, we can denote its roots by

$a = \sqrt{-2 + \sqrt{-5}}$

$b = -\sqrt{-2 + \sqrt{-5}}$

$c = \sqrt{-2 - \sqrt{-5}}$

$d = -\sqrt{-2 - \sqrt{-5}}$.

The splitting field of $f$ over $\mathbb{Q}$ is given by $L = \mathbb{Q}(a,b,c,d) = \mathbb{Q}(a)$.

The elements of the Galois-group are exactly the $\mathbb{Q}$-automorphisms of L, and thus are determined by the image of $a$. If we define

$\phi_1 = Id$

$\phi_2: a \mapsto b$

$\phi_3: a \mapsto c$

$\phi_4 = \phi_2 \circ \phi_3: a \mapsto d$

Results in Galois theory yield that the Galois-group of $f$ over $\mathbb{Q}$ is the Klein four-group $V_4$, which is isomorphic to the direct product of 2$\mathbb{Z}$ with itself.

Using Galois-correspondence, one can determine the subgroups of $L$. Obviously, the (non-trivial) subgroups of $G = Gal(f/\mathbb{Q})$ are $H_2 = \left\{Id, \phi_2\right\}$, $H_3 = \left\{Id, \phi_3\right\}$ and $H_4 = \left\{Id, \phi_4\right\}$.

What follows is my attempt to find the corresponding subfields of L.

The non-trivial subfield of $L$ corresponding to $H_2$ is given by $\mathbb{Q}(a^2)$. This can be seen by the fact that $\phi_2(a^2) = a^2$. Now, in the same manner one could argue that it can be easily seen that $\mathbb{Q}(a + c)$ and $\mathbb{Q}(b + c)$ are the other subfields of $L$, corresponding with $H_3$ and $H_4$ respectively.

Is there a more general way to determine these subfields? One strategy that sometimes seems to work (but not in this particular example), is to determine the $\mathbb{Q}$-basis of L. This basis is equal to $\left\{1, a, a^2, a^3\right\}$. Then, one can see that $\phi_2(v + wa + xa^2 + ya^3) = v - wa + xa^2 - ya^3$ for $v, w, x, y \in \mathbb{Q}$ and thus $\phi_2(a^2) = a^2$, making $\mathbb{Q}(a^2)$ the subfield of $L$ corresponding with $H_2$. If I try the same approach for $\phi_3$ for example, I don't get anything useable. So really my question is, when does this approach work, when doesn't it, is there any more general approach to these kinds of problems?

Answer 1

The problem of finding the subfields of a finite Galois extension is settled elegantly by the normal basis theorem. The normal basis theorem says that every finite Galois extension $ L/K $ admits a normal basis, that is, a basis of the $ K $-vector space $ L $ consisting of some $ \beta \in L $ and its $ K $-conjugates. If you can find such a basis, one can use a technique analogous to the Gaussian periods in cyclotomic field theory, which is actually a special case of the normal basis theorem.

To find a normal basis for this extension, let $ x = \sqrt{-2 + \sqrt{-5}} $, then $ \{1, x, x^2, x^3\} $ is a basis. Let $ \gamma = c_1 + c_2 x + c_3 x^2 + c_4 x^3 $ be an element. Recalling that $ 1/x = -(x^3 + 4x)/9 $, we write down the conjugates of $ \gamma $:

$$ \gamma_1 = \gamma = (c_1, c_2, c_3, c_4) $$ $$ \gamma_2 = (c_1, -c_2, c_3, -c_4) $$ $$ \gamma_3 = \left(c_1 + \frac{4c_3}{9}, \frac{4c_2}{3} + \frac{25c_4}{81}, \frac{c_3}{9}, \frac{c_2}{3} + \frac{4c_4}{81} \right) $$ $$ \gamma_4 = \left(c_1 + \frac{4c_3}{9}, -\frac{4c_2}{3} - \frac{25c_4}{81}, \frac{c_3}{9}, -\frac{c_2}{3} - \frac{4c_4}{81} \right) $$

We wish to find $ (c_1, c_2, c_3, c_4) $ such that the $ 4 \times 4 $ matrix with rows $ \gamma_1, \gamma_2, \gamma_3, \gamma_4 $ is invertible. The normal basis theorem guarantees their existence, and simple row reduction gives that this matrix is row equivalent to

$$ \begin{bmatrix} 2c_1 + c_3 & 0 & 0 & 0 \\ 0 & c_2 & 0 & c_4 \\ c_1 + 4c_3/9 & 0 & c_3/9 & 0 \\ 0 & 4c_2/3 + 25c_4/81 & 0 & c_2/3 + 4c_4/81 \end{bmatrix} $$

We want this matrix to be invertible. We may obviously take $ c_1 = c_4 = 0 $ and $ c_3 = c_2 = 1 $, and thus, it follows that we may take

$$ \gamma = x + x^2 $$

as a generator for our normal basis. (Many different elements will work, we only need one.) Now, the primitive elements for the three subfields are found by taking the sum

$$ \sum_{\sigma \in H} \sigma(\gamma) $$

where $ H $ is the stabilizer of the subfield in the Galois group. Concretely, we have the following subfields:

$$ \mathbf Q, \mathbf Q(x^2), \mathbf Q(x^3 + x), \mathbf Q(x^3 + 7x), \mathbf Q(x + x^2) = \mathbf Q(x) = L $$

Note: In this particular case, we could avoid all of the work by noting that

$$ (\sqrt{-10} + \sqrt{2})^2 = -8 + 2\sqrt{-20} = 4(-2 + \sqrt{-5}) $$

and thus our mystery field $ \mathbf Q(x) $ is actually $ \mathbf Q(\sqrt{2}, \sqrt{-10}) $. However, since the question asked for a general method, I performed the somewhat tedious computation above in order to illustrate the power of this method.

Answer 2

"If I try the same approach for $\phi_3$ for example, I don't get anything useable."

Sure you do – you just have to see how to use it.

$\phi_3(a)=c$, but $ac=3$, so $\phi_3(a)=3a^{-1}$.

Now $a^4+4a^2+9=0$, so $a(a^3+4a)=-9$, so $3a^{-1}=-(1/3)(a^3+4a)$.

$\phi_3(a^2)=c^2=-2-\sqrt{-5}=-a^2-4$.

I'll let you work out $\phi_3(a^3)$ in terms of $1,a,a^2,a^3$. Then you can work out $\phi_3(v+wa+xa^2+ya^3)$, and take it from there.

Another method that often works is this: given any subgroup $H$, and any field element $r$, the element $s=\sum_{\sigma\in H}\sigma(r)$ will be in the fixed field of $H$. If you are unlucky in your choice of $r$, then $s$ might turn out to be in $\bf Q$, but then you just try to find a more clever choice of $r$.