Ideal of the twisted cubic

Solution 1:

Here is a purely algebraic proof that $I(X)=I$.
It is of course sufficient to prove that $I(X) \subset I$ and for that it suffices to prove that every homogeneous polynomial $P(a,b,c,d)$ which vanishes on $X$ is in $I$.

Lemma
Any homogeneous polynomial $P(a,b,c,d)\in k[a,b,c,d]$ can be written $$P(a,b,c,d)=R(a,d) +S(a,d)b+T(a,d)c+i(a,b,c,d) $$ for some polynomials $R,S,T\in k[a,d]$ and a polynomial $i\in I$
The easy proof is by induction on the degree of $P$ and I'll leave it to you.

Now back to our problem. If now that homogeneous $P$ is in $ I(X)$ , we write it as in the lemma and get by using that $P$ vanishes on $X$ that for all $(x:y)\in \mathbb P^1_k$ $$0=P(x^3,x^2y,xy^2,y^3)=R(x^3,y^3) +S(x^3,y^3)x^2y+T(x^3,y^3)xy^2+0 $$
By considering exponents modulo $3$ for $x$ and $y$, we see that no cancellation occurs, hence that $R=S=T=0$ and thus $P=i\in I$ as required.

Solution 2:

The ideal $I$ is prime if and only if its associated projective scheme $V_p(I)\subset \mathbb P^3_k$ is integral.
This in turn can be checked on the four standard affine open subschemes covering $\mathbb P^3_k$.
For example in the affine open subscheme (isomorphic to $\mathbb A^3_k$) $U_d\subset \mathbb P^3_k$ corresponding to $d=1$, the scheme $V_p(I)\cap U_d$ is defined by the ideal $(a-bc,b^2-ac,c^2-b)$ which is trivially prime in $k[a,b,c]$.
Three similar calculations will imply that indeed the original ideal $I$ is prime.

Solution 3:

There are several algorithms that can compute whether your ideal $I$ is prime or not, most of the algorithms use Groebner basis. I gave your ideal to a maple package, namely PrimDecomp, obtainable from http://wwwb.math.rwth-aachen.de/~markus/ and it says that your ideal is prime.