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I am struggling with the following example in Chapter 18 of Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry , in which the author uses the unmixedness theorem to show that a genus 1 degree 4 irreducible curve $C$ in $\mathbb{P}^3_k$ has a homogeneous ideal generated by two quadrics in $S = k[x_0,...,x_n]$.

He shows first that there exist linearly independent irreducible quadrics $Q_1$, $Q_2$ such that $(Q_1,Q_2) \subset I$, where $I$ is the homogeneous ideal of $C$. Next he uses irreducibility to assert that $Q_1, Q_2$ form a regular sequence in $S$. I am okay with this part.

Now he says that the curve cut out by $Q_1, Q_2$ has degree 4 by Bezout's theorem. I believe I understand this part. The degree of the scheme-theoretic intersection is equal to the product of the degrees (I believe this is what Bezout's theorem says, although I can't find it phrased in these terms), which is 2*2 since we deal with two quadric hypersurfaces. This does not depend on the fact that $Q_1, Q_2$ are a regular sequence. Please correct me if I'm mistaken anywhere here.

Now, here is the part I do not understand. "Since $(Q_1,Q_2)$ is contained in $I$, we must have $(Q_1,Q_2) = I \cap J$ where codim$J > 2$." Where does this decomposition come from?

Finally, he concludes that $J=\emptyset$ since by the Unmixedness Theorem every associated prime of $(Q_1,Q_2)$ has codimension 2. This part again I believe I understand.

Can anyone explain the middle part to me? I do not get it at all. Disclaimer: I am not interested in the actual assertion, only the technique of his proof, so while an alternate approach may be edifying, it's not what I'm looking for.

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  • $\begingroup$ What do you mean by a curve? Is it reduced and irreducible? $\endgroup$
    – Youngsu
    Dec 15, 2013 at 23:41
  • $\begingroup$ Irreducible, yes. $\endgroup$
    – Cass
    Dec 15, 2013 at 23:42

2 Answers 2

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I'm not sure exactly how Eisenbud intends to argue, because I don't know what he's assuming known at this point. But here is some kind of explanation, which perhaps you can adapt to what you know:

The inclusion $(Q_1,Q_2) \subset I$ shows that $V(Q_1,Q_2) \supset V(I) = C$. Now $V(Q_1,Q_2)$ is a curve of degree $4$ (by Bezout), as is $C$ (by assumption) and so this inclusion has to be an equality at generic points of $C$. (Essentially by definition, ``degree'' of a curve measures the behaviour of a curve (or whatever dimensional objects we are applying it to, but in this particular case it is curves) at its generic points.)

Thus the only way that $V(Q_1,Q_2)$ and $V(I)$ can differ is by the former having some "extra" scheme-structure at some finite number of closed points. These points are what are being cut out by $J$. (They are what are called embedded points of $V(Q_1,Q_2)$, and ring-theoretically correspond to associated primes that are not minimal.)

Unmixedness then shows that these embedded points don't exists.

Technically, the decomposition $(Q_1,Q_2) = I \cap J$ is probably the primary decomposition (although I've never been very sure about the algebraic terminology for these sorts of considerations), and $J$ can't have any codimension $2$ contributions because these would be extra curve components of $V(Q_1,Q_2)$ (above and beyond $C$), and we've already seen for degree reasons that these don't exist.

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  • $\begingroup$ Thanks Matt. This gives me a lot to think about. I'm going to upvote it, but not accept it as an answer, because I'd like to see something more precise and algebraic (e.g. going into those primary decomposition details). $\endgroup$
    – Cass
    Dec 16, 2013 at 6:35
  • $\begingroup$ Basically, the statement you make that I'd like to see proven is the following: If L contained in I are two ideals of S of the same height, cutting out subschemes of the same degree, then I = L. More generally, without the degree condition, how does the primary decomposition of L relate to that of I? $\endgroup$
    – Cass
    Dec 16, 2013 at 6:43
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    $\begingroup$ @Cass: Dear Cass, Probably you just misphrased what you meant, but to (try to) be clear: if $L \subset I$ both cut out curves (say) of the same degree, it needn't imply that $I = L$, since there could be emdedded components. Best wishes, $\endgroup$
    – Matt E
    Dec 16, 2013 at 12:17
  • $\begingroup$ @MattE and Cass: Do you know where the condition genus $1$ is used? Does the statement hold true without the condition of genus $1$? $\endgroup$
    – Youngsu
    Dec 17, 2013 at 6:08
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    $\begingroup$ @Youngsu: Dear Youngsu, The intersection of two quadrics is a curve of genus one (in the sense of arithmetic genus, if it is singular). So the statement is not true for higher genus curves. Regards, $\endgroup$
    – Matt E
    Dec 17, 2013 at 6:40
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Recasting Matt E's answer in terms I feel more comfortable with:

Let $L=(Q_1,Q_2)$. By the unmixedness theorem, the associated primes of $L$ are precisely the minimal primes over $L$ and all such have height 2. Since $I$ is such a prime, the minimal primary decomposition of $L$ looks like $q \cap q_1 \cap ... \cap q_l$, where $q$ is $I$-primary. Since each prime in the decomposition has codimension equal to that of $L$, we have deg $S/L$ = deg $S/q$ + deg $S/q_1$ + ...+ deg $S/q_l$ (i.e. the degree is the sum of the degrees of the components).

I believe I can prove, using an exercise in Ravi Vakil's notes (18.6F), that if $q$ is $I$ primary, deg $S/q$ $\geq$ deg$S/I$ with equality if and only if $q=I$ (in other words, putting nilpotent structure on a component of a curve always strictly increases the degree). Since we have

4 = deg $S/L$ $\geq$ deg $S/q$ $\geq$ deg $S/I$ = 4,

all the inequalities must be equalities, so that $q=I$ and $q_1$,...,$q_l$ are not really there at all. That is, $L=I$.

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  • $\begingroup$ Dear Cass, In case you want confirmation: it's certainly true that if you added nilpotents along a whole component of a variety in $\mathbb P^r$, then you would increase its degree. Regards, $\endgroup$
    – Matt E
    Dec 16, 2013 at 12:19
  • $\begingroup$ @Cass: Why is $I$ a prime ideal? $\endgroup$
    – Youngsu
    Dec 17, 2013 at 6:06
  • $\begingroup$ Well, Eisenbud is not too clear on what he means by "curve." He starts out by just saying "Let C be a curve of genus 1 and degree 4" and then in the middle of the proof he says "Since C is irreducible," so I just assumed that he uses the convention of Hartshorne IV that "curve" without qualifiers means integral. That said, the proof I used above still works if C is irreducible but nonreduced (so that I and q are both p-primary for some p, and I contains q). $\endgroup$
    – Cass
    Dec 17, 2013 at 7:44
  • $\begingroup$ Cass: I agree that you can show that there exists $q$ which is contained in $I$. Have you had any luck with Eisenbud's argument which Matt E called probably primary decomposition? This is the part I'd like understand. $\endgroup$
    – Youngsu
    Dec 18, 2013 at 2:03

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