1
$\begingroup$

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 $k[x,y,z]/I(Y)$?

$\endgroup$
1
  • 2
    $\begingroup$ 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. $\endgroup$
    – Mohan
    Commented May 22, 2013 at 14:16

1 Answer 1

3
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .