9
$\begingroup$

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 $C=\big$ and moreover $\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 $I(C)=\big$.

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.

  • 0
    Aha, I always forget that the affine twisted cubic _is_ actually a complete intersection (as opposed to the projective case)2015-03-30

1 Answers 1

0

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 $C$.

  • 0
    By definition, $k[x_0,\ldots,x_n]$ is a finitely-generated $k$-algebra (also known as a $k$-algebra of finite type). However, it is not a finite-dimensional vector space over $k$. Indeed, a basis for $k[x_0,\ldots,x_n]$ is $\{1,x_0,x_0^2,\ldots\}\cup\{x_1,x_1^2,\ldots\}\cup\ldots\cup\{x_n,x_n^2,\ldots\}$, which is clearly not a finite set.2018-09-12