# Degree of an algebraic number over a field

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.

• Suppose $(x^2 + ax+b)(x^2 +cx+d)$ and compare coefficients. You should get a contradiction. – FormerMath Nov 7 '14 at 19:25

## 1 Answer

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.

• Thanks. Your suggestion was a much cleaner formulation of the solution. – Enigma Nov 7 '14 at 20:03