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May 19 at 14:19 comment added LittleBear @JasonStarr: Sorry for a silly question: why the graded ideal generated by degree 3 polynomials is not $(x^2y,xy^2,x^2z,xz^2,y^2z,yz^2,xyz,xyw,xzw,yzw)$? If so, then it seems that its associated closed subscheme is equal to $X$.
Nov 28, 2014 at 15:56 comment added Jason Starr Just to make that last point a bit more clear, if one starts with the homogeneous coordinate ring $k[x,y,z,w]$ of $\mathbb{P}^3$, for the degree 3 closed subscheme $X$ with graded ideal $\langle xy, xz, yz \rangle$, the graded ideal generated by the degree $3$ homogeneous polynomials above is $\langle x^2y, xy^2, x^2z, xz^2, y^2z, yz^2 \rangle$, whose associated closed subscheme does not equal $X$ (although they are set-theoretically equal).
Nov 28, 2014 at 15:03 comment added Jason Starr Just to make one comment: this result is "set-theoretic", not "scheme-theoretic". If $X$ is smooth, the result is scheme-theoretic. (Mumford discusses this in an old paper -- forgot the reference, but Lazarsfeld's "Positivity in Algebraic Geometry" gives the Mumford reference in the discussion of C-M regularity for smooth schemes.) For a reducible counterexample, consider the union of the three coordinate axes in $\mathbb{P}^3$: correct set-theoretic intersection, but wrong scheme structure at the singular point.
Nov 28, 2014 at 11:47 comment added Turbo Is there a textbook reference for this?
Nov 27, 2014 at 19:46 vote accept aglearner
Nov 27, 2014 at 19:45 history answered abx CC BY-SA 3.0