STATEMENT: Show that $\sqrt{2}+\sqrt{3}$ is algebraic over $\mathbb{Q}$, of degree 4.
QUESTION: I have produced the polynomial $X^4-10X^2+1$, that has the indicated number as a zero, show it is algebraic.But I am not quite sure how to show that it is of degree 4. I am guessing it is sufficient to show that the polynomial that I generated is irreducible. I know that it can't be factored linearly in $\mathbb{Q}$ by using rational root theorem. I am just not sure how to show that it can't be factored into quadratic terms except by possibly applying discriminants. Some hints or suggestions would be appreciated.
Consider the set of numbers of the form $a+b\sqrt2 + c\sqrt 3 + d\sqrt6$, where $a,b,c,d$ are rational, as a vector space of dimension 4 over the rationals; call this space $V$. Let $r=\sqrt2+\sqrt 3$. If $r$ were a root of a nonzero polynomial of degree 3 or less, say $P(x) = a_3x^3 + a_2x^2 + a_1x +a_0$, this would show that $1, r, r^2, r^3$ were not linearly independent in $V$, because we would have nonzero $a_0, a_1, a_2, a_3$ for which $a_3r^3 + a_2r^2 + a_1r +a_0=0$.
So it suffices to show that $1, r, r^2, r^3$ are linearly independent in $V$, which is easily done.