Minimal polynomial of $\sqrt2+1$ in $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$ I'm trying to find the minimal polynomial of $\sqrt2+1$ over $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$. The minimal polynomial of $\sqrt2+1$ over $\mathbb{Q}$ is 
$$ (X-1)^2-2.$$
So I look at $\alpha = \sqrt2 + \sqrt3$
$$ \alpha^0 = 1$$
$$ \alpha^1 = \sqrt2 + \sqrt3.$$
I cannot find $a,b$ such that $a\alpha + b\alpha = \sqrt2+1$. So the degree 2 is mimimal. Is that correct?
 A: Letting  $t = \sqrt{2} + \sqrt{3}$, $\sqrt{2} + 1 = 1 + \frac{1}{2}(t^3 - 9t)$ so the minimal polynomial is $x - \sqrt{2} - 1$.
A: The minimal polynomial of $\sqrt{2}+1$ in $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$ must divide $(x-1)^2 - 2$. So the question is whether $\sqrt{2}+1$ does not lie in $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$ (in which case the minimal polynomial is again $(x-1)^2-2$), or whether it already lies in $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$ (in which case the minimal polynomial is linear, hence equal to $x-(\sqrt{2}+1)$ ).
In fact, $\mathbb{Q}(\sqrt{2}+\sqrt{3}) = \mathbb{Q}(\sqrt{2},\sqrt{3})$, since for example
$$\alpha^3=(5+2\sqrt{6})(\sqrt{2}+\sqrt{3}) = 11\sqrt{2}+9\sqrt{3} = 2\sqrt{2}+9\alpha$$
so $2\sqrt{2}\in\mathbb{Q}(\sqrt{2}+\sqrt{3})$. Since $\sqrt{2}\in\mathbb{Q}(\sqrt{2}+\sqrt{3})$, then $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$, so the minimal polynomial of $\sqrt{2}-1$ over $\mathbb{Q}(\sqrt{2}+\sqrt{3})$ is of degree $1$, not $2$.
A: Hint $\ $ For $\rm\ w = \sqrt{3}+\!\sqrt{2},\ \  w^2= 5+2\sqrt{6}\in\mathbb Q[w]\:\Rightarrow\color{#C00}{\sqrt{6}} \in \mathbb Q[w]\:\Rightarrow\:\sqrt{2}\in \mathbb Q[w]\:$ since
$$ \begin{eqnarray} \sqrt{2}\ &=&\ \sqrt{2}\ (\ 3\,\ -\,\ 2\ ) \\
&=& \ \sqrt{2}\ (\sqrt{3}-\!\sqrt{2})&\!(\sqrt{3}+\!\sqrt{2})\\
 &=&\quad\ \ \  (\sqrt{6}\,\ -\ 2)\:&\rm\!(\sqrt{3}+\!\sqrt{2}) = (\color{#C00}{\sqrt{6}}-2)w\, \in\, \mathbb Q[w] \end{eqnarray}\quad\ \ $$
Remark $\ $ See here for a more general approach related to the Primitive Element Theorem.
Eliminating $\:\sqrt{6}\:$ yields $\rm\:\sqrt{2}\, =\, ((t^2\!-\!5)/2\!-\!2)\,t\:$ for $\rm\: t = \sqrt{3}+\!\sqrt{2},\:$ which, not surprisingly, is precisely the same polynomial in Serkan's answer.
A: Let $u=\sqrt{2}+\sqrt{3}$ then $u^3=11\sqrt{2}+9\sqrt{3}=2\sqrt{2}+9u$ so $2\sqrt{2}\in Q(u)$ and hence also $\sqrt{2}-1\in Q(u)$, so its minimal polynomia has degree 1.
