# A problem about the twisted cubic

I have some difficulty with the following problem:

Let $f : k → k^3$ be the map which associates $(t, t^2, t^3)$ to $t$ and let $C$ be the image of $f$ (the twisted cubic). Show that $C$ is an affine algebraic set and calculate $I(C)$. Show that the affine algebra $Γ(C)=k[X,Y,Z]/I(C)$ is isomorphic to the ring of polynomials $k[T]$.

I found a solution in the case when $k$ is an infinite field:

Clearly $C$ is the affine algebraic set $C=\big<X^3-Z,\;X^2-Y\big>$ and moreover $\big<X^3-Z,\;X^2-Y\big>\subseteq I(C)$. To prove the inclusion "$\supseteq$", let $F$ be a polynomial with $F\in I(C)$. Then we can write (thanks to successive divisions): $$F(X,Y,Z)=(X^3-Z)\cdot Q(X,Y,Z)+(X^2-Y)\cdot P(X,Y)+R(X).$$ For all $t\in k$ we have $$0=(t,t^2,t^3)=R(t)$$ and since $k$ is infinite then $R$ is the zero polynomial and $I(C)=\big<X^3-Z,\;X^2-Y\big>$.

Now since the function $f$ is an isomorfism between $C$ and $k$ follows that $\Gamma(C)\cong \Gamma(k)=k[T]$.

I can not find the ideal $I(C)$ if the field $k$ is finite.

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If $k$ is finite then $C$ is finite and so $\Gamma(C)$ is a finite-dimensional vector space over $k$, so it is certainly not isomorphic to $k[T]$. – user8268 May 3 '12 at 12:05
Thanks! your comment helped me a lot. So the statement of the problem is wrong. – Dubious May 3 '12 at 13:03
Algebraic closed field is always infinite. And when we are talking about (most part of) algebraic geometry, $k$ is implicitly assumed to be algebraic closed. – asatzhh Oct 28 '13 at 6:25
Aha, I always forget that the affine twisted cubic is actually a complete intersection (as opposed to the projective case) – Theo Mar 30 '15 at 16:00