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.