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
$$0=F(t,t^2,t^3)=R(t)$$
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.

Just to remove this question from the list of unanswered questions. It was found in comments that the statement of the problem is wrong if $k$ is finite. In this case the ideal $I(C)$ is the product of maximal ideals corresponding to the points of $C$.
