Degree of a projective scheme and its defining equations 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 $d$?
 A: 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 $n$.
A: 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?
