# Find the degree of $K = \mathbb{Q}(\sqrt2, \sqrt7)$ over $\mathbb{Q}$

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?

• Helps to guess at a generator. Hint: try $\sqrt 2 + \sqrt 7$. – lulu Jun 30 '17 at 2:10
• Would it be $x^4 - 18x^2 +25$? – Yabbadule Jun 30 '17 at 2:22
• That looks good. – lulu Jun 30 '17 at 2:23
• Note; you still have to demonstrate that $K=\mathbb Q(\sqrt 2 +\sqrt 7)$. Not terribly difficult, but still. – lulu Jun 30 '17 at 2:26
• +lulu, this would basically mean showing that the span of my basis is equal to the span of $\{\sqrt2 + \sqrt7, -\sqrt2 + \sqrt7, -\sqrt2 - \sqrt7, \sqrt2 - \sqrt7 \}$ correct? – Yabbadule Jun 30 '17 at 16:09