# Math Help - Subfields.

1. ## Subfields.

I have an irreducible polynomail f(x) = x^4 +bx^2 +d over rationals.

And I have that the Galois group of this polynomail is D_4, i.e., dihedral
group of degree 8.
Then I want to show that there are only three subfields of degree 2, and I
want to show that these are the following.

Q(d^{1/2}), Q({b^2-4d}^{1/2}), and Q({d(b^2-4d)}^{1/2}).

Since there are only three subgroups of order 4 in D_4, we have by
fundamental theorem of Galois theory that there are three subfields of
degree 2. But how can I find them, i.e., how can I show that these are the
subfields.

2. Originally Posted by ZetaX
I have an irreducible polynomail f(x) = x^4 +bx^2 +d over rationals.

And I have that the Galois group of this polynomail is D_4, i.e., dihedral
group of degree 8.
Then I want to show that there are only three subfields of degree 2, and I
want to show that these are the following.

Q(d^{1/2}), Q({b^2-4d}^{1/2}), and Q({d(b^2-4d)}^{1/2}).

Since there are only three subgroups of order 4 in D_4, we have by
fundamental theorem of Galois theory that there are three subfields of
degree 2. But how can I find them, i.e., how can I show that these are the
subfields.
Stop stealing people's usernames from other forums.

If you are told the Galois group is $D_4$ you know there are three subgroups that have index 2, these correspond to three subfields over degree 2 over $\mathbb{Q}$ by Galois correspondence. Therefore, you just need to find 3 quadratic fields that are distinct. The quantity $b^2 - 4d$ cannot be a square lest the Galois group fail to be $D_4$. Thus, $\sqrt{b^2-4d}$ is quadradic over $\mathbb{Q}$. Hence, $\mathbb{Q}(\sqrt{b^2-4d})$ is quadradic number field. The same argument is used for $\mathbb{Q}(\sqrt{d})$ if you can show that $d$ is not a square in $\mathbb{Q}$. And so on. Afterwards you need to argue that that all these three fields are distinct. Thus, there is some work that needs to be done in this argument but if you can do it, it will show these are the only subfields.

The problem however is this approach is not going to work in general. Let $f(x) = x^4 - 2$, this is a counterexample.

3. How do you know that if b^2-4d is a square then the galois group is not D_4?

And how can it be shown that these subfields are different.

4. Originally Posted by ZetaX
How do you know that if b^2-4d is a square then the galois group is not D_4?
The equation $x^4 + bx^2 + d= 0$ implies $x^2 = \tfrac{1}{2}(-b\pm \sqrt{b^2 - 4d})$. If $b^2-4d$ is a square (and not zero) then we see that $x^2 = r_1,r_2$ where $r_1,r_2\in \mathbb{Q}$. Therefore, the splitting field of $x^4+bx^2+d$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt{r_1},\sqrt{r_2})$ however, $\left[ \mathbb{Q}(\sqrt{r_1},\sqrt{r_2}) : \mathbb{Q}\right] \leq \left[ \mathbb{Q}(\sqrt{r_1}):\mathbb{Q}\right]\left[\mathbb{Q}(\sqrt{r_2}):\mathbb{Q}\right] \leq 2\cdot 2 =4$. Thus, the splitting field cannot be a degree $8$ extension and so it is impossible for the Galois group to be $D_4$ because $|D_4| = 8$.

And how can it be shown that these subfields are different.
(What I wrote above about $x^4-2$ being a counterexample is wrong.)

We have shown $b^2-4d$ is not a square so $\mathbb{Q}(\sqrt{b^2 - 4d})$ is quadradic over $\mathbb{Q}$. In a similar manner it can be argued that that $d$ cannot be a square. I give you a hint. Assume that $d=e^2$ then $x^4+bx^2 + d = (x^2+e)^2 + (b^2-2e)x^2$. Thus, if $(x^2+e)^2 + (b^2-2e)x^2 = 0 \implies x^2 \pm \sqrt{b^2-2e}x + e = 0$. Thus, $x = \tfrac{1}{2}\left( \pm \sqrt{2e-b^2} \pm \sqrt{-2e-b^2}\right)$ and so the splitting field is contained in $\mathbb{Q}\left(\sqrt{2e-b^2},\sqrt{-2e-b^2} \right)$ which can never be made to equal to degree $8$ over $\mathbb{Q}$. Thus, we have shown that $\sqrt{d}$ is quadradic over $\mathbb{Q}$. Furthermore, it can be shown that $\mathbb{Q}(\sqrt{b^2-4d}) \not = \mathbb{Q}(\sqrt{d})$. Assume to the contrary, then $\sqrt{d} = x_0 + x_1\sqrt{b^2-4d}$ for some $x_0,x_1\in \mathbb{Q}$ now argue that it is impossible to find such rational numbers (bring $x_0$ to the other side and square).

This is still not the full argument you are left with $\sqrt{d(b^2-4d)}$, but I think the idea is however the same.