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

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    $\begingroup$ 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
    May 22 '13 at 14:16
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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.

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