Strategies for finding $[\mathbb{Q} (\sqrt{2} + \sqrt{3}) : \mathbb{Q} ]$ I need to find the degree of the extension $\mathbb{Q}(\sqrt{2} + \sqrt{3})$ over $\mathbb{Q}$. I don't quite know how to do it, nor can I exhibit any polynomial with root $\sqrt{2} + \sqrt{3}$, but I think it has to have at least degree $4$. I tried to work with $\mathbb{Q}(\sqrt{2} + \sqrt{3})$ as a subspace of $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ over $\mathbb{Q}$, but I also don't know if that is the case.
 A: $\frac1{\sqrt2+\sqrt3}=\frac1{\sqrt2+\sqrt3}\cdot \frac{\sqrt2-\sqrt3}{\sqrt2 -\sqrt3}=\sqrt3-\sqrt2$.
Hence we get $\sqrt2 $ and $\sqrt3$, and so $\mathbb Q(\sqrt2 +\sqrt3) =\mathbb Q(\sqrt2, \sqrt3)$.
A: Hint
\begin{align*}
x &= \sqrt{2}+\sqrt{3}\\
x - \sqrt{2}&=\sqrt{3}\\
(x - \sqrt{2})^2&=3\\
x^2-1&=2x\sqrt{2}\\
(x^2-1)^2&=8x^2\\
x^4-10x^2+1&=0.
\end{align*}
A: As you say, if $\alpha=\sqrt2+\sqrt3$ then $\alpha\in\Bbb Q(\sqrt2,\sqrt3)$.
It follows that $\Bbb Q(\alpha)\subseteq\Bbb Q(\sqrt2,\sqrt3)$. Can we show
equality? The field $\Bbb Q(\sqrt2,\sqrt3)$ is spanned over $\Bbb Q$ by
$1$, $\sqrt2$, $\sqrt3$ and $\sqrt2\sqrt3=\sqrt6$.
We have
$$\alpha^2=5+2\sqrt6$$
and
$$\alpha^3=11\sqrt2+9\sqrt3.$$
Therefore $\sqrt6=\frac12(\alpha^2-5)$, $\sqrt2=\frac12(\alpha^3-9\alpha)$
and $\sqrt3=-\frac12(\alpha^3-11\alpha)$
are all elements of $\Bbb Q(\alpha)$. Therefore $\Bbb Q(\alpha)=\Bbb Q(\sqrt2,\sqrt3)$.
To show that $|\Bbb Q(\sqrt2,\sqrt3):\Bbb Q|=4$, we need $\sqrt3\notin
\Bbb Q(\sqrt2)$. To prove this, assume $\sqrt3=a+b\sqrt2$ with $a$, $b\in
\Bbb Q$ and get a contradiction from $(a+b\sqrt2)^2=3$.
A: Let $w=\sqrt{2}+\sqrt{3}$. Note that $w^2=2\sqrt{6}+5$. Then $\frac{(w^2-5)^2}{4}=6$ is in $\mathbb{Q}$. So the polynomial $f(X)=X^4-10X^2+1$ has $w$ as one of its roots. You can factorize $f$ linearly with the substitution $Y=X^2$. So you can get the minimal polynomial of $w$ from $f$.
