Let $X \subset \mathbb{P}^n$ be any projective scheme. Denote by $I_X$ the (saturated) ideal of $X$. Suppose the degree of $X$ is $d$. Under what assumptions there exists a polynomial in $I_X$ of degree less than or equal to $d$?
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$\begingroup$ Yes, think about the twisted cubic in $P^3$. Its ideal is generated by 3 quadrics. $\endgroup$– IMeasyCommented Nov 5, 2013 at 15:06
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$\begingroup$ Do you really want to ask only if there exists a polynomial in $I_X$ of degree $\leq d$? Of course I assume that you are asking about a non-zero polynomial. However, you could ask for much more, e.g., a collection of polynomials that set-theoretically / scheme-theoretically define $X$. I recommend that you take a look at Def. 1.8.37, p. 111, of Lazarsfeld's "Positivity in Algebraic Geometry, I", as well as the follow-up discussion. $\endgroup$– Jason StarrCommented Nov 5, 2013 at 15:18
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$\begingroup$ @IMeasy: I meant for ANY projective scheme with the above mentioned properties. $\endgroup$– JanaCommented Nov 5, 2013 at 15:26
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$\begingroup$ @Starr: Thank you for your comment. I am asking if there exist at least one non-zero polynomial of degree less than or equal to the degree of the scheme. I looked into the reference. Ex. 1.8.38 gives an answer to my question if $X$ is smooth. I am interested in the case when $X$ is not smooth (especially when $X$ is non-reduced). I do not totally understand how 1.8.37 helps me compare with the degree of the scheme. It would be very helpful if you could elaborate a bit more. $\endgroup$– JanaCommented Nov 5, 2013 at 15:47
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2 Answers
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If $X$ is not reduced the answer is negative (and doesn't make much sense) : take $X$ defined by $x^n=xy^n=0$ in $\Bbb{P}^2$ (a line with an embedded point). Then $X$ has degree 1 but the minimal degree of an element of $I_X$ is $n$.
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In general the answer is no because X could have components of different dimensions and then the degree of X is usually defined as the degree of the union of the irreducible component of highest possible dimension. But X could also have embedded components which cause trouble. To get a criterion we should assume that X is equidimensional and has no embedded components.
In this case there is a chance that it holds. The standard way to prove this kind of statement is to project X to a hypersurface via a general projection and compute the degree of the hypersurface. So you'd want to look at some references talking about the degree of images of projections... It will fortify your soul to find the argument yourself, rather than me telling you exactly how to do it here. Don't you agree?
