Vrftsjtryk

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.

As you say, if $alpha=sqrt2+sqrt3$ then $alphainBbb Q(sqrt2,sqrt3)$.
It follows that $Bbb Q(alpha)subseteqBbb Q(sqrt2,sqrt3)$. Can we show
equality? The field $Bbb Q(sqrt2,sqrt3)$ is spanned over $Bbb Q$ by
$1$, $sqrt2$, $sqrt3$ and $sqrt2sqrt3=sqrt6$.
We have
$$alpha^2=5+2sqrt6$$
and
$$alpha^3=11sqrt2+9sqrt3.$$
Therefore $sqrt6=frac12(alpha^2-5)$, $sqrt2=frac12(alpha^3-9alpha)$
and $sqrt3=-frac12(alpha^3-11alpha)$
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 $sqrt3notin
Bbb Q(sqrt2)$. To prove this, assume $sqrt3=a+bsqrt2$ with $a$, $bin
Bbb Q$ and get a contradiction from $(a+bsqrt2)^2=3$.

$frac1{sqrt2+sqrt3}=frac1{sqrt2+sqrt3}cdot frac{sqrt2-sqrt3}{sqrt2 -sqrt3}=sqrt3-sqrt2$.

Hence we get $sqrt2 $ and $sqrt3$, and so $mathbb Q(sqrt2 +sqrt3) =mathbb Q(sqrt2, sqrt3)$.

Hint
begin{align*}
x &= sqrt{2}+sqrt{3}\
x - sqrt{2}&=sqrt{3}\
(x - sqrt{2})^2&=3\
x^2-1&=2xsqrt{2}\
(x^2-1)^2&=8x^2\
x^4-10x^2+1&=0.
end{align*}

Let $w=sqrt{2}+sqrt{3}$. Note that $w^2=2sqrt{6}+5$. 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 $f$.