How to Compute Projective Closure in General?

There have been a few questions about this on this site, but I think my question is different because a) my question isn't about Hartshorne 2.9, it's just inspired by that question, and b) the other questions don't ever seem to actually describe how to go about finding the ideal, just verifying some work.

The question in Hartshorne is to find the ideal of the projective closure of the twisted cubic parameterized by $(t,t^2,t^3)$ over some field $k$, and show it's not the same as projectivizing the generators of the twisted cubic's ideal in affine space.

All of this is well and good and I've done this with only minimal struggling. The problem is that at the end, I had to make a sort of leap of faith. By this I mean, I wrote down the projectivizations, and I could visualize that the equations I had written down were not going to cut out the twisted cubic as I wanted it by looking at the appropriate affine piece. I needed one more equation, which I was able to deduce, and then include, and convince myself that this was the projective closure.

This is highly unsatisfying, because at the end of the day I had to consult my ability to visualize a variety, rather than just doing algebra, and I would like to be able to do this in general. In Hartshorne, we prove that $I(\bar{Y}) = \beta(I(Y))$ where $\beta$ is the projectivization map. This description is not helpful really since in general this ideal will have infinitely many elements and it's really not useful to describe an ideal by listing its elements.

So, suppose that we are working in a more general setting, considering maybe $k[x_1, ..., x_n]/I$ where $I = (p_1, ..., p_m)$ for some polynomials in these variables. How can I write down the ideal for the projective closure?

• You should know that modern algebraic geometry is "highly decorated algebra", meaning everything you do is algebra, but you get a lot of motivation/intuition from geometry. The sooner you get used to this, the better. – 54321user May 6 '17 at 4:48
• @MoarCake559 the things I'm interested in pertaining to algebraic geometry are over R and C, and so they have quite a bit of geometric content in themselves. I would find it disappointing if these ideas were not able to be fleshed out nicely. – Alfred Yerger May 6 '17 at 4:59

Cox, Little, and O'Shea give a nice method of computing projective closures using Gröbner bases in $\S4$ of Chapter $8$ of Ideals, Varieties, and Algorithms.
Theorem 4. Let $I$ be an ideal in $k[x_1, \ldots, x_n]$ and let $G = \{g_1, \ldots, g_s\}$ be a Gröbner basis for $I$ with respect to a graded monomial order in $k[x_1, \ldots, x_n]$. Then $G^h = \{g_1^h, \ldots, g_s^h\}$ is a basis for $I^h \subseteq k[x_0, x_1, \ldots, x_n]$.
Here $f^h$ denotes the homogenization of the polynomial $f$ and $I^h = \langle f^h : f \in I \rangle$. Note that the monomial ordering must be graded, i.e., one that respects total degree: if $\alpha, \beta$ are multi-indices with $|\alpha| > |\beta|$, then $x^\alpha > x^\beta$.
Let's see how this works in the example of the twisted cubic. Let $C \subseteq \mathbb{A}^3$ be the affine twisted cubic and $I = \mathbb{I}(C)$ be its vanishing ideal. People often choose the generators $I = \langle y - x^2, z - x^3 \rangle$. Now $y-x^2, z - x^3$ do form a Gröbner basis for the lexicographical ordering $y > z > x$, but this is not a graded ordering. If we instead choose the graded lexicographical ordering with $y > z > x$, then Buchberger's algorithm yields the Gröbner basis $y^2 - zx, yx - z, x^2 - y$. By the above theorem, then $y^2 - zx, yx - zw, x^2 - yw$ is a Gröbner basis for $I^h$, so $I^h = \langle y^2 - zx, yx - zw, x^2 - yw \rangle$.