Chapter 1, Exercise 1.2: The Twisted Cubic Curve

Let $Y\subseteq\mathbb A^3$ be the set $Y=\{(t,t^2,t^3):t\in k\}$. Show that $Y$ is an affine variety of dimension $1$. Find generators for the ideal $I(Y)$. Show that $A(Y)$ is isomorphic to a polynomial ring in one variable over $k$. We say that $Y$ is given by the parametric representation $x=t,y=t^2,z=t^3$.

At first, this problem looks like it should be extremely trivial: we are given a certain set, and we are asked to show that it is an affine variety of dimension $1$. The trick, however, appears when we realize this set is given in a parametric form, and it is in fact up to us to find the polynomials that specify or cut out this twisted cubic. Intuitively, we might expect that there are two, since the twisted cubic is a curve in $\mathbb A^3$ and thus we will have to "cut out" two dimensions. To find these polynomials, we can exploit the fact that $x=t$ along the curve and thus that $y=t^2=x^2$ and so on: in particular, we can see that $Y=Z(y-x^2,z-x^3)$. Applying $I$ to both sides, we find as expected that $I(Y)=(y-x^2,z-x^3)$. This solves one part of the problem.

Next we will show $A(Y)\cong k[x]$. As in the previous exercise, define a surjective map $k[x,y,z]\to k[x]$ taking $f(x,y,z)\mapsto f(x,x^2,x^3)$. Clearly by the Nullstellensatz this has kernel $I(Y)$, so we induce an isomorphism $A(Y)=A/I(Y)\cong k[x]$.

Finally, the remaining parts of the problem follow immediately: $Y$ is an affine variety since $k[x]$ is an integral domain, and thus $I(Y)$ is prime. Further, we have $$\dim Y = \dim A(Y) = \dim k[x]=1$$ and we are done.