# 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?

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You mean $\mathbb Q(\sqrt2+\sqrt3)$, i.e. the smallest field containing $\mathbb Q\cup\{\sqrt2+\sqrt3\}$, right? (I was little confused by the notation $\mathbb Q[\sqrt2+\sqrt3]$.) – Martin Sleziak Jun 12 '12 at 17:24
@Martin Well, it's algebraic, so either notation is fine. – Dylan Moreland Jun 12 '12 at 17:25
@DylanMoreland Thanks! joachim: Notice that $\sqrt2\in\mathbb Q(\sqrt2+\sqrt3)$. Se e.g. here. – Martin Sleziak Jun 12 '12 at 17:28
I know the question has already been answered, but the problem with your earlier approach was that you should have been looking for $a,b,c,d$ such that $a+b\alpha+c\alpha^2+d\alpha^3=\sqrt{2}+1$, since the extension is of degree 4. – alex.jordan Jun 12 '12 at 19:05

Letting $t = \sqrt{2} + \sqrt{3}$, $\sqrt{2} + 1 = 1 + \frac{1}{2}(t^3 - 9t)$ so the minimal polynomial is $x - \sqrt{2} - 1$.

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Remark $\$ To understand where that polynomial comes from see my answer. – Bill Dubuque Jun 12 '12 at 20:06

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 $2$.

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@DylanMoreland: Thanks. – Arturo Magidin Jul 6 '12 at 0:42

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.

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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.

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