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I am trying to show that the ideal $I = (x^2 -2, y^2 +1, z)$ is a proper ideal of $\mathbb{Q}[x,y,z]$. I have been trying to show that 1 is not in the ideal with a degree argument and evaluating at a particular $(x,y,z)$, but I keep getting stuck.

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3 Answers 3

up vote 2 down vote accepted

An idea: suppose

$$I=(1)=R:=\Bbb Q[x,y,z]\Longrightarrow 1\in I\Longrightarrow \,\exists\,\,a,b,c\in R\,\,s.t.\,\, $$

$$x=a(x^2-2)+b(y^2+1)+cz=1$$

Now use the evaluation homomorphism

$$\Phi_{(r,s,t)}\,:T\to\Bbb C\;\;,\;\;\Phi_{(r,s,t)}(f(x,y,z)):=f(r,s,t)$$

But with the embedding $\,R\subset \Bbb R[x,y,z]\subset\Bbb C[x,y,z]=:T\,$ , and we then get

$$1=\Phi_{(\sqrt 2, i,0)}=\Phi_{(\sqrt 2,i,0)}(a(x^2-2)+b(y^2+1)+cz)=0$$

which gives a straightforward contradiction

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This is the idea I had in mind, but I was a bit uneasy about the embedding. Thanks! –  Joe Jan 9 '13 at 4:04
    
Dear DonAntonio, Your map $\Phi$ has target $\mathbb C$, not $\mathbb Q$. Regards, –  Matt E Jan 9 '13 at 4:26
    
Indeed so, @Matt...Thanks –  DonAntonio Jan 9 '13 at 4:28

I have another proof which perhaps you like it. Pay attention that $I\neq \mathbb{Q}[x,y,z]$ if and only if $\frac{\mathbb{Q}[x,y,z]}{I}\neq\{0_{\frac{\mathbb{Q}[x,y,z]}{I}}\}$. But look that every element in $\frac{\mathbb{Q}[x,y,z]}{I}$ is class of following relation "$\forall f,g\in\mathbb{Q}[x,y,z]\; :\; f\sim g\Longleftrightarrow\; f-g\in I$. And we denote these classes (elements of $\frac{\mathbb{Q}[x,y,z]}{I}$) with $f+I$. It is easy to check $f+I=g+I$ if and only if the reminder of their division to polynomials which generate $I$ by "Division Algorithm in K[x,y,z]" (that you can find it with simple description in book "Ideals, Varieties, Algorithm" written by "David Cox") will be same. So it is enough to know is the set of reminders to dividing in $\{x^{2}-2,y^{2}+1,z\}$ larger than $\{0\}$ or not. But it's brightly from that Algorithm which by paying attention to degrees of those polynomials with respect to $x$, $y$ and $z$ that the set of reminders is polynomials which have no term including $z$ and powers of $x$ and $y$ larger than two. So there are a lot of polynomials out of $I$ like $x,y,x+y,xy,x+1,...$.

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As I said one can, in fact without using any division algorithms in multivariate polynomial rings, write down a basis of the quotient: $1,x,y,xy$ –  Martin Brandenburg Jan 9 '13 at 10:20
2  
@MartinBrandenburg, Yes. And as you mentioned in your note, your lemma and actually a lot of computational methods in Algebraic Geometry or Commutative Algebra are using Algorithm of Division. And about what I wrote here, I only explained a sight, view that I was watching by reading the question. –  AmirHosein SadeghiManesh Jan 9 '13 at 10:30

Lemma: If $R$ is a commutative ring, $f \in R[x]$ is a monic polynomial of degree $n$, then $R[x]/(f)$ is free of rank $n$ over $R$. Proof: Polynomial division.

Applying this twice, we see that $\mathbb{Q}[x,y,z]/(x^2-2,y^2+1,z) = (\mathbb{Q}[x]/(x^2-2)) [y]/(y^2+1)$ is free of rank $4$ over $\mathbb{Q}$.

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Maybe you should say for the OP too that you used $\Bbb{Q}[x,y,z]/(z) \cong \Bbb{Q}[x,y]$. –  user38268 Jan 9 '13 at 4:17
    
Yes, $\{(a,b,c):c=0\}=\{(a,b)\}$ and Yoneda. –  Martin Brandenburg Jan 9 '13 at 4:36

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