Determine the splitting field (polynomial/degree)

Question

How to determine the splitting field $L$ of the polynomial $x^4-7x^2+10$ over $\mathbb{Q}$ and $[L:\mathbb{Q}]$?

My way was:

$x^4-7x^2+10=(x^2-2)(x^2-5)$

So the roots are $x_1=\sqrt{2}, x_2=-\sqrt{2}, x_3=\sqrt{5}$ and $x_4=-\sqrt{5}$

Then the splitting field is $L=\mathbb{Q}(\sqrt{2},-\sqrt{2},\sqrt{5},-\sqrt{5})$

I'm not sure if this is enough? Or is it false?

Also, I don't know how to compute $[L:\mathbb{Q}]$ now.

Answer 1

This is correct, though note that this field is simply equal to $\mathbb{Q}(\sqrt{2},\sqrt{5})$. (because a field is closed to additive inverses anyway). Now to find the degree we can use the tower rule:

$[L,\mathbb{Q}]=[L:\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}]$

We have $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$ because $x^2-2$ is the minimal polynomial of $\sqrt{2}$ over $\mathbb{Q}$. So now we just have to find $[L:\mathbb{Q}(\sqrt{2})]$. Note that $L=\mathbb{Q}(\sqrt{2})(\sqrt{5})$, so the extension degree is the degree of the minimal polynomial of $\sqrt{5}$ over $\mathbb{Q}(\sqrt{2})$. Obviously, $x^2-5$ is a polynomial in $\mathbb{Q}(\sqrt{2})[x]$ such that $\sqrt{5}$ is its root, so the degree is at most $2$. Now we just have to show the degree can't be $1$. Suppose $[\mathbb{Q}(\sqrt{2})(\sqrt{5}):\mathbb{Q}(\sqrt{2})]=1$. Then $\sqrt{5}\in\mathbb{Q}(\sqrt{2})$ and hence we can write $\sqrt{5}=a+b\sqrt{2}$ for some $a,b\in\mathbb{Q}$. Using the fact that $\sqrt{2},\sqrt{5}$ are irrational this easily leads to a contradiction. So $[L:\mathbb{Q}(\sqrt{2})]=2$, and hence by the tower rule $[L:\mathbb{Q}]=4$.

Answer 2

Yes, it is correct. But that is also equal to $\mathbb Q[\sqrt2,\sqrt5]$ and still equal to $\mathbb Q[\sqrt 2+\sqrt 5]$. So\begin{align}[L:\mathbb Q]&=\text{degree of minimal polynomial of }\sqrt2+\sqrt5\\&=4.\end{align}