## Compute the degree of a given extension of the rationals

Prove that $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$. Conclude that $\mathbb{Q}(\sqrt{2}+\sqrt{3})$ has degree 4 over $\mathbb{Q}$. Find an irreducible polynomial satisfied by $\sqrt{2} + \sqrt{3}$.

The $(\supseteq)$ inclusion is clear. To see the $(\subseteq)$ inclusion, let $\zeta = \sqrt{2} + \sqrt{3}$. Then evidently $\frac{1}{2}(\zeta^3 - 9\zeta) = \sqrt{2}$ and $-\frac{1}{2}(\zeta^3 - 11\zeta) = \sqrt{3}$. (WolframAlpha agrees; see here and here.) Thus the extensions are equal.

Recall that $\mathbb{Q}(\sqrt{2})$ has degree 2 over $\mathbb{Q}$. Now we claim that $\mathbb{Q}(\sqrt{2},\sqrt{3})$ has degree 2 over $\mathbb{Q}(\sqrt{2})$. To see this, suppose to the contrary that $\sqrt{3} \in \mathbb{Q}(\sqrt{2})$. Say $(a+b\sqrt{2})^2 = 3$ for some rationals $a$ and $b$. Coparing coefficients, we have $a^2 + 2b^2 = 3$ and $2ab = 0$; if $a = 0$, then $b^2 = \frac{3}{2}$, and if $b = 0$, then $a^2 = 3$. Either case yields a contradiction.

So $\mathbb{Q}(\sqrt{2} + \sqrt{3})$ has degree 4 over $\mathbb{Q}$.

The minimal polynomial of $\zeta$ has degree 4 over $\mathbb{Q}$. We can solve the system $\zeta^4 + a\zeta^3 + b\zeta^2 + c\zeta + d = 0$ over $\mathbb{Q}$, and in so doing we see that $p(x) = x^4-10x^2+1$ is the minimal polynomial of $\sqrt{2} + \sqrt{3}$.