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I've recently been reading about rational normal curves and how they may be represented and have come to the following question: (for simplicity's sake, the problem is stated in terms of the RNC in $\mathbb{P}^3$, i.e., the twisted cubic)
The twisted cubic is the image of the map $$ v : \mathbb{P}^1 \to \mathbb{P}^3 $$ which sends
$$[X_0 : X_1] \mapsto [X_0^3 : X_0^2 X_1 : X_0 X_1 ^2 : X_1 ^3]$$ In Harris' book on Algebraic Geometry it is stated that the twisted cubic is also the set of common zeros of the three homogeneous quadratic polynomials $$ F_0 (Z) = Z_0 Z_2 - Z_1 ^2$$ $$F_1 (Z) = Z_0 Z_3 - Z_1 Z_2$$ $$ F_2 (Z) = Z_1 Z_3 - Z_2 ^2 $$
I have filled in the blanks as Harris (implicitly) suggests: that is, I have verified that any point on the twisted cubic is a point on each of the quadrics $F_i$ and that any common zero of the quadrics lies on the twisted cubic. What continues to mystify me is the question: where do these quadratic polynomials come from? When they are given to me, I can certainly verify that they have the prescribed properties, but given the first description of the twisted cubic, how does one arrive at the quadrics defining it as a variety?
Any guidance will be greatly appreciated!

EDIT: I have found the quadrics simply by "playing around:"
Denote by $C$ the twisted cubic in $\mathbb{P}^3$. Let $P = [Z_0 : Z_1 : Z_2 : Z_3] \in C$ be any point. Then we know that $$Z_0 = X_0^3, \ Z_1 = X_0^2 X_1, \ Z_2 = X_0 X_1^2, \ Z_3 = X_1 ^3 $$ Observing, for example, that $$(X_0 ^3)(X_1^3) - (X_0 ^2 X_1)(X_0 X_1^2) = 0 $$ Implies that $$ Z_0 Z_3 - Z_1 Z_2 = 0$$ and yields the quadratic polynomial $F_1 (Z)$. The same sort of trick yields the other two polynomials $F_0 (Z)$ and $F_2 (Z)$.
Now I've come to (perhaps) a more sophisticated set of questions:
(1) How did we know to search for quadratic polynomials?
(2) How did we know to search for three of them?
I suppose we could proceed in a very ad hoc way: Well, it wasn't that bad to come up with the polynomial equations, and they happened to be quadratic (how fortuitous!) Supposing I stopped after only finding two of the quadratics I could have manually checked "are these enough to get all the points of the twisted cubic?" If yes, great. If no, search for more polynomials and repeat.
Sure. The above procedure would have worked just fine. But what about when we try to generalize to the rational normal curve in $\mathbb{P}^n$? Certainly we can (and should) be smarter about things...

aherring
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  • It's really pretty hard to naively figure out generators for the ideal of given variety. I guess one thing to say about this case is that these things are determinental and there are a lot of techniques in that case — I think Harris talks about some of them later. – Hoot May 29 '15 at 23:23
  • Some other ideas: on the affine patch $Z_0 \neq 0$ it really is pretty easy to get generators for the ideal. You could try homogenizing those and see where that goes, although if you choose poorly then you'll run into an interesting issue. Also, this thing is the image of a projective morphism, and the algebraic keyword for the resulting equations here is "resultant". I think there's some discussion of this in Ch 5 of Eisenbud-Harris (the book is written in the language of schemes and there are new features in that case; maybe it's unsuitable here). – Hoot May 29 '15 at 23:30

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I will answer your "more sophisticated" questions. I am on a mobile phone though, so I will be a bit brief.

For Question 1, this is because we can describe the ideal I as the kernel of the morphism $k[w,x,y,z] \to k[s,t]$ mapping each variable to monomials $s^3,s^2t$ etc.. These types of kernels are called Toric ideals and are always generated by binomials—see Sturmfels, Gröbner Bases and Convex Polytopes, Lemma 4.1. Why we can expect the binomials to be quadratic is part of Question 2.

For Question 2, I don't know what exactly you want but you can easily show you need three quadratic generators (plus possibly others) as follows. The short exact sequence $$0 \to I \to k[w,x,y,z] \to k[s^3,s^2t,st^2,t^3] \to 0$$ restricted to degree 2 shows that the k-dimension of these pieces satisfies $\operatorname{dim} I_2 + 7 =10 = \binom{3+2}{3}$ since dimension is additive on short exact sequences of vector spaces. This shows the dimension of $I_2$ is 3, and since $I$ had no degree 0 or 1 elements by a similar analysis on the degree 0 and 1 parts, we see I must have three quadratic generators.

Takumi Murayama
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