Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
3  
@Martin Well, it's algebraic, so either notation is fine. –  Dylan Moreland Jun 12 '12 at 17:25
2  
@DylanMoreland Thanks! joachim: Notice that $\sqrt2\in\mathbb Q(\sqrt2+\sqrt3)$. Se e.g. here. –  Martin Sleziak Jun 12 '12 at 17:28
4  
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
add comment

4 Answers 4

up vote 6 down vote accepted

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

share|improve this answer
1  
Remark $\ $ To understand where that polynomial comes from see my answer. –  Bill Dubuque Jun 12 '12 at 20:06
add comment

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 $2$.

share|improve this answer
    
@DylanMoreland: Thanks. –  Arturo Magidin Jul 6 '12 at 0:42
add comment

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.