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}.

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