Can taking the projective closure of an affine variety increase the degrees of its ideal generators?

Question

Say we have some equations $f_1(x)=0, \ldots f_k(x)=0$ defining a variety $X$ in ${\mathbb C}^n$ (not necessarily a minimal number of generators, and not necessarily of minimal degree), and suppose we want to know the ideal $\bar{I}$ of its closure $\overline{X}$ in ${\mathbb P}^n$. A naive question:

If all the original generators $f_i$ were of degree at most $d$, can we generate $\bar{I}$ using polynomials of degree at most $d$?

If not, what degree bound $d'$ can we give for a minimal generating set of $\bar{I}$?

Motivation, in short: Ideal saturation is an extremely slow operation, but solving a linear system to identify vanishing forms up to a certain degree is polynomial time, which would give a more practical way to compute $\bar{I}$ in high dimensions.

(Sorry, I'm new to computational algebra, and maybe people already know a fast way to compute projective closure...)

Motivation, in long: To compute $\bar{I}$, it's not enough to simply homogenize the generators with respect to a new variable $z$. Denote such homogenizations by $\hat{f}$, their ideal by $\hat{I}$, and their variety by $\hat{X}$. The problem is that $\hat{X}$ might contain irreducible components on the hyperplane at infinity $(z=0)$, so we need to saturate it, $\bar{I}=(\hat{I}:z^\infty)$, to eliminate these components and get the ideal of $\bar{X}$. But saturation is extremely slow! (I think it's at least doubly exponential time in the number of variables.) So I'd like to find $\bar{I}$ instead by solving a linear system on the vector space of homogeneous polynomials of degree at most $d$ to see which ones vanish, and declare that they cut out the variety $\bar{X}$, yay! But do they? If searching up to degree $d$ is not enough, how far do I have to go?

A simple example: Say $f_1 = x$ and $f_2 = y-x^2$ in ${\mathbb C}^2$, so $X$ is just the origin $(0,0)$. Suppose we first homogenize the generators $f_i$. The parabola $\hat{f}_2=yz-x^2=0$ is tangent to the line at infinity at $(x:y:z)=(0:1:0)$, which also lies on $\hat{f}_1=0$. So together they cut out $\hat{X} = \{ (0:0:1) \cup (0:1:0)\}$, and we don't get $\bar{X}=(0:0:1)$ until we saturate $\hat{I}=\langle x,yz-x^2\rangle$ by $z$ to eliminate the component at infinity.

In this example, we already know the answer is the origin, and saturation is very fast for such a small system anyway. But for many variables, we don't already know what the variety looks like, and the saturation might not finish in a human lifetime, so it would be nice to solve a linear system instead.

Thanks for any help on this question!

Answer 1

Consider $R = \Bbbk[x_1, \ldots, x_5]$ and $I = (x_1x_2^2-x_3^2, x_1x_4^2-x_5^2)$. When we homogenize w.r.t $z$, we get $(x_1x_2^2-zx_3^2, x_1x_4^2-zx_5^2)$ whose saturation w.r.t $z$ contains the binomial $x_3^2x_4^2-x_2^2x_5^2$ of degree $4$. The original generators are of degree at most $3$.

Look at Section 15.10 of Eisenbud's commutative algebra book; one should homogenize a Groebner basis of $I$. Therefore, bound for $d'$ in terms of $d$ might turn out to be quite large. (This is just my feeling!)

PS: in the `Simple Example', do you mean that $\hat{f_2} = x^2-yz$? As stated the ideal $\hat{I}$ is already saturated w.r.t. $z$ and does not have a component at $[0:1:0]$.