MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
@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

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.

share|cite|improve this answer
unfortunately, i don't understand the proof. what is ht(p,q)? – filipm Oct 6 '11 at 8:26
$ht(I)$ is the height of $I$, (see, 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

Your Answer


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.