Is it true that every projective sub-variety of degree $d$ in $\mathbb CP^n$ is an intersection of some number of hypersurfaces of degree $d$? Is there some simple proof of this fact? (I believe this is so)
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It is true. The idea is as follows : suppose your variety, say $X$, has dimension $m$. For any $(n-m-2)$-plane $\pi \subset\mathbb{P}^n$, take the join $\langle \pi ,X\rangle$, that is, the union of the lines $\langle p ,x\rangle$ for $p\in\pi $, $x\in X$. It is easy to see that it is a hypersurface of degree $d$ when $\pi $ is general, and that the intersection of all these hypersurfaces when $\pi $ varies is $X$.
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$\begingroup$ Is there a textbook reference for this? $\endgroup$– TurboCommented Nov 28, 2014 at 11:47
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1$\begingroup$ 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. $\endgroup$ Commented Nov 28, 2014 at 15:03
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1$\begingroup$ 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). $\endgroup$ Commented Nov 28, 2014 at 15:56
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$\begingroup$ @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 $X$. $\endgroup$ Commented May 19 at 14:19