Subfields of splitting field of $x^4+25$ over $ℚ$.

Question

Let $F$ be the splitting field of the polynomial $x^4+25$ over $ℚ$. List all subfields in $F$ and the corresponding subgroups in the Galois group.

is problem $1$ on this pdf. The solution is:

As we proved in class $(F / ℚ)=4$. The Galois group $G$ is the Klein subgroup of $S_4$, isomorphic to $ℤ_2 × ℤ_2$. Note that $F$ contains $i$ and $\sqrt{5}$, each subgroup of $G$ of index 2 corresponds to a subfield of degree 2. There are 3 such subfields $ℚ(i)$, $ℚ(\sqrt{5})$ and $ℚ(\sqrt{-5})$. The trivial subgroup of $G$ corresponds to $F$ and $G$ corresponds to $ℚ$.

I agree that the Galois group $G$ is the Klein four group.

It said “(the splitting field) $F$ contains $\sqrt5$”.

I think this is wrong, $F$ doesn't contain $\sqrt5$ or $\sqrt{-5}$.

The roots of $x^4+25$ over $ℚ$ are $r_1=\frac{1+i}{\sqrt2}\sqrt{5},r_2=\frac{-1+i}{\sqrt2}\sqrt{5},r_3=\frac{-1-i}{\sqrt2}\sqrt{5},r_4=\frac{1-i}{\sqrt2}\sqrt{5}$.

$r_2/r_1=i$

$r_1+r_4=\sqrt{10}$

so I think the three proper subfields of $F$ should be $ℚ(i)$, $ℚ(\sqrt{10})$ and $ℚ(\sqrt{-10})$. Is this correct?

Answer 1

I think, you are right. Of course, $r_1+r_2=\sqrt{10}i\in L$ and $r_1+r_4=\sqrt{10}\in L$, so that the splitting field is given by $L=\Bbb Q(\sqrt{10},i)$. This field doesn't contain $\sqrt{5}$, or $\sqrt{-5}$. Otherwise there exist rational numbers $a,b,c,d$ with $a+b\sqrt{10}+ci+di\sqrt{10}=\sqrt{5}$, which leads to $10d^2=-5$, a contradiction.

The splitting field is given by $KL$ with $K=\Bbb Q(i)$ and $L=\Bbb Q(\sqrt[4]{-25})$, see this post. However, we have $\sqrt[4]{-25}=\frac{\sqrt{10}}{2}(1+i)$, and not $\sqrt{-5}$, which was probably the error in the above pdf-file.