MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

The following is an Exercise 1.1.11 of Hartshorne's Algebraic Geometry.

Let $Y\subset \mathbb{A}^3$ be the curve given parametrically by $x=t^3, y=t^4, z=t^5$. Show that $I(Y)$ is a prime ideal of height $2$ in $k[x,y,z]$ which can not be generated by $2$ elements. We say $Y$ is "not a local complete intersection".

My question is why is it called "local"? Should not we look some localization of $k[x,y,z]/I(Y)$?

share|cite|improve this question
You can easily see that $I$ localized at any prime ideal of the polynomial ring is in fact a complete intersection except for the prime (maximal) ideal $(x,y,z)$. So, I is not a local complete intersection. – Mohan May 22 '13 at 14:16

A ideal defining a complete intersection has a regular sequence as a generating set. Then a local complete intersection would be a quotient ring which has a regular sequence as a generating set after some localization.

share|cite|improve this answer

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.