So I'm working on a question on splitting fields and I'm a little confused by the result I keep getting. Any help would be very much appreciated.
I have to construct the splitting field of $(x^4-10)(x^2-20)$ over $\Bbb Q$. We can factorise the given polynomial as
$$(x^4-10)(x^2-20)=(x-\sqrt[4]{10})(x+\sqrt[4]{10})(x-i\sqrt[4]{10})(x+i\sqrt[4]{10})(x-2\sqrt{5})(x+2\sqrt{5}).$$
So now we know the roots. Testing if $\sqrt[4]{10}$ is a an element of $\Bbb Q$ goes as follows:
Suppose it is. Then $\sqrt[4]{10}=\frac{a}{b}$ for some $a,b, \in \Bbb Q \Rightarrow a= \sqrt[4]{10}b$, a contradiction. So it is not an element of $\Bbb Q$.
The polynomial $x^4-10$ is irreducible by Eisenstein's Criterion with $p=5$. So we conclude that
$$\Bbb Q(\sqrt[4]{10}) \cong \Bbb Q[x]/\langle x^4-10\rangle. $$
and now we know that $\Bbb Q(\sqrt[4]{10}) := \{a+b\sqrt[4]{10}+c(\sqrt[4]{10})^2+d(\sqrt[4]{10})^3 \mid a,b,c,d \in \Bbb Q \}$.
This next part is where I feel I must be messing something up because it doesn't make sense to me. To see where the next extension will come from I tested to see if $i^4\sqrt{10}$ is an element of $\Bbb Q(\sqrt[4]{10})$.
I tried to aim for a contradiction. That is, suppose $i\sqrt[4]{10}=a+b(\sqrt[4]{10})+c(\sqrt[4]{10})^2+d(\sqrt[4]{10})^3$. Then
$$10=a^2+(2ab+20cd)(\sqrt[4]{10})+(2ac+b^2+10d^2)(\sqrt[4]{10})^2+(2ad+2bc)(\sqrt[4]{10})^3+20bd+10c^2 $$
Obviously we don't want any of the quartic roots involved so we want either $a=0, b=0, d=0,$ or $c=0, b=0, d=0$.
If we take the former option, however, we are left with $10=10c^2$. This implies that $c=\pm 1$, but this is not a contradiction. So does this mean that $i\sqrt[4]{10} \in \Bbb Q(\sqrt[4]{10})$? Or does it just mean that I can't use a proof by contradiction argument here?
The key is that $\mathbb{Q}(\sqrt[4]{10})$ is not the splitting field of your polynomial, but it is a subfield of it. The reason is, as said in the comments, that $i\sqrt[4]{10}\notin \mathbb{Q}(\sqrt[4]{10})$, since $\mathbb{Q}(\sqrt[4]{10})\subset\mathbb{R}$. Adding $i\sqrt[4]{10}$ as a generator gives you all the roots, so the splitting field is $\mathbb{Q}(\sqrt[4]{10}, i\sqrt[4]{10})$.