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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.

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Checking your map is injective is more or less the same as finding the relations that present the subsemigroup of $\mathbb Z^2$ generated by vectors $v_1=(3,0)$, $v_2=(2,1)$, $v_3=(1,2)$ and $v_4=(0,3)$. Can you do that? –  Mariano Suárez-Alvarez Jun 20 '12 at 20:01
    
@Mariano: I've already indicated that I've tried the monoid approach, but that it doesn't work without many case distinctions etc. ... –  Martin Brandenburg Jun 21 '12 at 6:45

3 Answers 3

up vote 4 down vote accepted

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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Ah, thanks! Actually this shows that $\phi^*$ is injective. –  Martin Brandenburg Jun 22 '12 at 14:23

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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This may seem to be a trivial question, but why does your claim in the first sentence holds? Don't you "forget" the origin at which some zero divisors may be "supported"? In terms of commutative algebra, I think that your argument only shows that $\phi^*$ is an isomorphism when localized at $a$, $b$, $c$ or $d$; but why does this suffice? Also, is there any elementary proof which doesn't use scheme language (I'm just asking because I would like to present this to students which just have started to learn the concept of classical varieties, which are always reduced). –  Martin Brandenburg Jun 21 '12 at 6:38
    
Being reduced is a local property, but being integral is not. So I doubt that it suffices to show that each affine chart is integral. But even if I just want to prove that $I$ is reduced: Why does it suffice to prove that $V_p(I)$ is reduced? Doesn't this only say that all the localizations $Q_a,Q_b,Q_c,Q_d$ are reduced? I agree that they are reduced (in fact, localizations of polynomial rings), but why does this imply that $Q$ is reduced? –  Martin Brandenburg Jun 22 '12 at 9:01
    
Okay but irreducibility is trivial because it's the image $\phi(\mathbb{P}^1)$. –  Martin Brandenburg Jun 22 '12 at 13:59

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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If you have questions about this, feel free to contact me or the author. It's not my peace of software, but from another guy in my department. Also you can send messages in german. –  sebigu Jun 20 '12 at 20:15
    
Dear sebigu, the ideal $I(X)$ is certainly prime, since $X$ is irreducible. The question is whether $I$ is prime (and the answer is "yes"). –  Georges Elencwajg Jun 20 '12 at 20:58
    
You are right, but I also typed in the generators for $I$, since only they were given. I will correct my answer. –  sebigu Jun 20 '12 at 21:00
    
I had already checked with Sage that it is a prime ideal if the base field is $\mathbb{Q}$. Do the algorithms also work over arbitrary $k$? –  Martin Brandenburg Jun 21 '12 at 6:36
    
They wont, but since all coefficients are $1$ or $-1$, it suffices to check $\mathbb{Q}$ or, here, $\mathbb{Z}$. –  sebigu Jun 21 '12 at 6:58

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