Ideal of the twisted cubic The twisted cubic is the image of the morphism $\phi : \mathbb{P}^1 \to \mathbb{P}^3 , (x:y) \mapsto (x^3:x^2 y:x y^2:y^3)$, it is given by $X = V(ad-bc,b^2-ac,c^2-bd)$. Now I would like to compute $I(X)$, which equals by the Nullstellensatz the radical of the ideal $I := (ad-bc,b^2-ac,c^2-bd) \subseteq k[a,b,c,d]$. I think that that $I$ is already a radical ideal, even a prime ideal. Namely, I suspect that 
$$\phi^* : Q:=k[a,b,c,d]/I \to k[s,t] , a \mapsto s^3, b \mapsto s^2 t , c \mapsto s t^2 , d \mapsto t^3$$
is an injection. If this is true: How can we prove that? I've already tried to find a $k$-basis of the quotient, but this turned out to be a big mess. Even the representation of the quotient as a monoid algebra doesn't seem to help. Another idea is the following: A formal manipulation of generators and relations implies $Q_a \cong k[a,b]_a$. Thus it suffices to prove that $Q \to Q_a$ is injective, i.e. that $a$ is not a zero divisor.
 A: 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.
A: 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.
A: 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.
