Set $K = \mathbb{Q}(\sqrt2, \sqrt7)$.
I can construct a basis for this extension as a vector space over $\mathbb{Q}$ like this:
$\{1, \sqrt{2}, \sqrt{7}, \sqrt{14} \}$.
So I expect the degree to be 4, but here's my issue. I cannot find a minimal polynomial of degree 4 that is irreducible. Here is what I was able to find:
$\sqrt2$ has a minimal irreducible polynomial of $x^2 - 2$ and $\sqrt{7}$ has a minimal irreducible polynomial of $x^2 - 7$. However, their product $x^4 - 9x + 14$ is not irreducible. What am I missing? How can I find the proper polynomial?