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

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?

• The general case $x^4+a$, or even $x^n+a$ is well known, see for example this duplicate. Commented May 30 at 11:00
• I agree that the Galois group G is the Klein four group. My question is about the subfields. Commented May 30 at 11:03
• I think the Galois group is generated by $i\mapsto-i$ and $\sqrt{10}\mapsto-\sqrt{10}$ Commented May 30 at 11:05

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.
• Could the error in the pdf be that it meant $x^4\color{blue}-25$? Commented Jun 2 at 4:32