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.
Ideal of the twisted cubic
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1@Mariano: I've already indicated that I've tried the monoid approach, but that it doesn't work without many case distinctions etc. ... – 2012-06-21
3 Answers
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.
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0This is an example on page 389 of Ideals, Varieties and Algorithms, by Cox, Little and O'Shea. It turns out in this case the affine basis in Georges' answer is already Groebner, and hence the given homogenizations generate the homogenization of a prime ideal, which is prime. – 2017-07-16
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.
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0Okay but irreducibility is trivial because it's the image $\phi(\mathbb{P}^1)$. – 2012-06-22
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.
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0Ah, thanks! Actually this shows that $\phi^*$ is injective. – 2012-06-22