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If $p,q:\mathbb R^3\rightarrow\mathbb R$ are two polynomials, such that $\{p=0\}\cap\{q=0\}$ is two-dimensional, does it follow that $p$ and $q$ have a common factor? (I believe it does.) How to prove that?

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We consider the rimg $A = R[x,y,z]$, $dim A = 3$. We have $V((p)) \cap V((q)) = V((p,q))$. If $gcd(p,q) = 1$, then $ht(p,q) = 2$. So $V((p)) \cap V((q))$ is not 2-dimension –  Pham Hung Quy Oct 5 '11 at 10:47
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@Pham Hung Quy: Why don't you make this an answer? Otherwise this question may linger around until someone else comes in and gives your comment as an answer just to stop the Mathoverflow-bot from bumping the question –  David White Oct 5 '11 at 12:19
    
I am sorry David White. I am not good at algebraic geometry. So I posed it as a comment, and I hope to see a better anwser. Now I pose it as an anwser. –  Pham Hung Quy Oct 5 '11 at 15:53
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1 Answer 1

Here is my answer. We consider the rimg $A=R[x,y,z]$ , $dimA=3$. We have $V((p))∩V((q))=V((p,q))$. If $gcd(p,q)=1$, then $ht(p,q)=2$. So $V((p))∩V((q))))=V((p,q))$ is not 2-dimension.

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unfortunately, i don't understand the proof. what is ht(p,q)? –  filipm Oct 6 '11 at 8:26
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$ht(I)$ is the height of $I$, (see, en.wikipedia.org/wiki/Krull_dimension). Notice that $\mathbb{R}[x,y,z]$ is a Gassian ring, hence a prime ideal of height 1 is a principal ideal $(f)$ with $f$ is a irreducible polynomial. If $ht(p,q) = 1$, then $(p,q) \subset (f)$ for some $f$, so $f$ is a common factor of $p$ and $q$, a contradiction. Thus $ht((p,q))>1$ (= 2), so $\dim A/(p,q) < 2$. So $V((p,q))$ is not dimension two (see, Atiyah-Macdonald: Introduction to Commutative algebra, chapter 11 as example). –  Pham Hung Quy Oct 6 '11 at 17:01
    
thx Pham Hung Quy, now it's much clearer, I'll try to digest it. –  filipm Oct 7 '11 at 15:39
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