TITLE: How to show that $\sqrt{2}+\sqrt{3}$ is algebraic?
QUESTION [30 upvotes]: In Abbot's Understanding Analysis I am asked to show that $\sqrt{2}+\sqrt{3}$ is an algebraic number. I have shown that those two are algebraic separately (that was simple), but I can't figure out what to do to show that their sum is algebraic, too. For example, I tried $(\sqrt{2}+\sqrt{3})^{2}=5+\sqrt{24}$ and then I tried to think of a polynomial of form $ax^2-bx^{0}=0$ e.g. $c(\sqrt{2}+\sqrt{3})^{2}-c(5+\sqrt{24})=0$ that would work, but couldn't find the $c$ value that would make $b=c(5+\sqrt{24})$ integer, but couldn't find one. Maybe I can somehow conjecture that the sum of two algebraic numbers must be algebraic, too, but I was wondering if there's a way to find a polynomial to show this. Thanks!

REPLY [10 votes]: Note $\rm\ \alpha = \sqrt{2}\:+\sqrt{3}\in F = \mathbb Q(\sqrt{2},\sqrt{3})\:.\:$ But $\rm\:F = \mathbb Q\langle1,\sqrt{2},\sqrt{3},\sqrt{6}\rangle\:$ is a $\rm\:\mathbb Q$-vector space of dimension $4\:,\:$ so the $\:5\:$ elts $\:1,\alpha,\alpha^2,\alpha^3,\alpha^4$ are $\rm\:\mathbb Q$-linearly dependent. Thus $\rm\:f(\alpha) = 0\:$ for some $\rm\:0 \ne f(x)\in\mathbb Q[x]\:$ of degree $\le 4\:.\:$ This proof by dimension works generally - see Zhen's answer.