The projective closure of the twisted cubic curve

Question

I'm now reading Hartshorne's Algebraic Geometry and trying to solve Exercise 2.9(b).

Let $Y$ be an affine variety in $\mathbb{A}^n$. Identifying $\mathbb{A}^{n}$ with the open subset $U_0$ of $\mathbb{P}^n$ by the homeomorphism $\varphi_{0}: (x_0,x_1,..,x_n)\mapsto (\dfrac{x_1}{x_0},...,\dfrac{x_n}{x_0})$. Then we can speak about $\bar{Y}$, the projective closure of $Y$ in $\mathbb{P}^{n}$.

(a) Show that $I(\bar{Y})$ is an ideal generated by $\beta(I(Y))$

(b) Prove that if $f_1,...,f_r$ generate $I(Y)$, then $\beta(f_1),...,\beta(f_r)$ do not necessarily generate $I(\bar{Y})$, using the example of the twisted cubic curve.

I have proved (a). My question is on the projective closure of the twisted cubic curve.

In exercise 1.2 of Hartshorne, the affine twisted cubic curve $Y$ is defined as the image of the map $$ v: k \rightarrow \mathbb{A}^3; \quad t \mapsto (t, t^2, t^3). $$

Many questions and answers related to this question on MSE seems to take it for granted or claim that it is trivial that the projective closure of $Y$ is the image of the map $$ \bar{v}: \mathbb{P}^1 \rightarrow \mathbb{P}^3; [X_0, X_1] \mapsto [X_0^3, X_0^2 X_1, X_0 X_1^2, X_1^3]. $$

My question is: Why the projective closure $\bar{Y}$ is $\mathrm{image}(\bar{v})$ ? I have tried to prove but have no ideas on this.

I have looked up book like Joe Harris' Algebraic Geometry: A First Course and the definition of $v$ and $\bar{v}$ appears in Example 1.10. But Prof. Joe Harris did not show that as well.

Thank you for your helps!

P.S. The LaTeX code for the quoted exercise is adapted from MSE question 275034, i.e. Projective closure: How to determine?

Answer 1

First we can verify that the image of $\overline{v}$ is closed: it's the vanishing locus of $xw=yz$, $xz=y^2$, and $z^2=yw$. Thus the image of $\overline{v}$ is a closed set containing $Y$, and it's obtained by adding the point $[0:0:0:1]$ to $Y$. So either the closure of $Y$ is $Y$, or the closure of $Y$ is the image of $\overline{v}$. If the first option was the case, there would be a homogeneous polynomial vanishing on $Y$ which doesn't vanish on $[0:0:0:1]$. We show that can't happen.

Consider an arbitrary homogeneous polynomial of degree $d>0$ in $x,y,z,w$. If it evaluates to zero on $Y$, then it can't have a term of the form $\lambda w^d$ with $\lambda\neq 0$: plugging in $x=1$, $y=t$, $z=t^2$, $w=t^3$, we would have a nonzero polynomial in $t$ of degree $3d>0$ which evaluates to $0$ for all values of $t$. On the other hand, the only way for a homogeneous polynomial of degree $d>0$ to fail to vanish at $[0:0:0:1]$ is if it has a term of the form $\lambda w^d$. So every homogeneous polynomial vanishing on $Y$ must also vanish on $[0:0:0:1]$, and we're done.

Answer 2

What one has to show is that any homogeneous polynomial $F$ vanishing on $$Y=\{(1,t,t^2,t^3):t\in k\}$$ also vanishes at $(0,0,0,1)$. This means that $f(t)=F(1,t,t^2,t^3)$ vanishes at all $t\in k$, and as $k$ is algebraically closed, then $f$ is the zero polynomial.

Let the degree of $F$ be $N$ and consider $g(u)=F(u^3,u^2,u,1)$. Then $$g(u)=u^{3N}f(u^{-1})$$ so $g$ is also the zero polynomial. Setting $u=0$ gives $$F(0,0,0,1)=g(0)=0.$$