Sorry for the recent question spam, but I've been striving furiously over the last few days to solve a problem in Hartshorne (Algebraic Geometry) discussed in this question. The problem (II.8.4a) says:

Let $Y$ be a closed subscheme of codimension $r$ in $\mathbb{P}^n_k$ ($k$ algebraically closed). Then the homogeneous ideal of $Y$ is generated by $r$ elements in $S=k[x_0,...,x_n]$ if and only if $Y$ is equal to the scheme-theoretic intersection of $r$-many hypersurfaces (locally principal closed subschemes), i.e., $\mathcal{I}_Y=\mathcal{I}_{H_1}+\cdots+\mathcal{I}_{H_r}$ where these are ideal sheaves.

Hartshorne gives the hint to use the unmixedness theorem for $S$.

The direction $\implies$ is easy. I'm trying to prove the other (where we assume $Y$ is a scheme theoretic intersection of hypersurfaces).

In exploring this problem, I think I've twigged to how it's "supposed" to be solved, but it relies on something not obvious to me, but perhaps obvious to the sources I'm appealing to. The something is this:

Let $I_{H_1},...,I_{H_r}$ denote the homogeneous ideals of the hypersurfaces, $I_Y$ the saturated homogeneous ideal of their scheme theoretic intersection. Then the primary components of $\sum{I_{H_j}}$ corresponding to minimal primes, and the primary components of $I_Y$ corresponding to minimal primes coincide. In other words, these ideals only differ in their embedded components.

(Matt E seems to use something like the above in answering my question here).

Once this is established, the result follows easily from the unmixedness theorem, but again I'm not sure why this should be obvious. These ideals clearly have the same minimal primes, since they cut out the same irreducible subsets of $\mathbb{P}^n$, but as for why the minimal primary components should agree, I'm at a loss. Thanks in advance for any tips.

  • $\begingroup$ I don't think it's possible to translate the problem purely into commutative algebra. Shall I remove the tag? In brief, the problem is to show that the scheme theoretic intersection of hypersurfaces in P^n gives you the closed subscheme whose homogeneous ideal is generated by the polynomials cutting out the hypersurfaces. $\endgroup$
    – Cass
    Commented Dec 17, 2013 at 0:45
  • $\begingroup$ I've edited the OP to include what "hypersurfaces" means in this context, in case that was unclear, namely locally principal closed subschemes. $\endgroup$
    – Cass
    Commented Dec 17, 2013 at 0:50
  • $\begingroup$ Hi. Your previous question was also interesting. It seems that the case $r=1$ implies the general case. In other words, can you show that it is globally principal? $\endgroup$
    – Youngsu
    Commented Dec 17, 2013 at 6:05
  • $\begingroup$ The $r=1$ case is handled in another question I asked: 607479. I don't see how it implies the general case though. $\endgroup$
    – Cass
    Commented Dec 17, 2013 at 8:07
  • $\begingroup$ Cass: If $f_i$'s in $S$ correspond to the ideal $I_{H_i}$, then doesn't $I_Y$ correspond to the ideal $(f_1,\dots, f_r)$? $\endgroup$
    – Youngsu
    Commented Dec 17, 2013 at 8:13

1 Answer 1


Let me try to give slightly different reasoning from what you already have (dealing with minimal primary components, etc.):

Call $I$ the ideal of the scheme-theoretic complete intersection, and $J$ the homogeneous ideal generated by the $r$ homogeneous forms defining the hypersurfaces. Since $J \subseteq I$, there is an exact sequence

$$0 \to I/J \to S/J \to S/I \to 0$$

We want to show that $I/J = 0$. By assumption, $S/J$ and $S/I$ define the same scheme, so $I$ and $J$ have the same saturation, hence $(I/J) : \mathfrak{m}^\infty = 0$, i.e., $I/J$ is a module of finite length (where $\mathfrak{m} = (x_0,...,x_n)$ is the irrelevant ideal). But $S/J$ is Cohen-Macaulay (and has $\dim > 0$), hence cannot contain a nonzero submodule of finite length (since e.g. $H^0_\mathfrak{m}(S/J)$, the largest finite length submodule of $S/J$, is $0$, as $\operatorname{depth} S/J > 0$).

  • $\begingroup$ I have no problem once I know that I and J define the same scheme. You say this holds by assumption, but I'm sorry I do not see it. Since in general the ideals won't be the same (as in my example to Youngsu) the only way I could guess to show that they define the same subscheme would be to show that they have the same saturation, but that would render your argument circular. When you say "I and J define the same subscheme" you are saying that I and J only differ in their embedded components. Again, this is precisely what I do not understand. $\endgroup$
    – Cass
    Commented Dec 23, 2013 at 18:38
  • $\begingroup$ And nevermind, I understand. I and J define the same schemes if and only if they define the same schemes in each of the standard affine open sets U_0,...,U_n. The latter condition is clear from the hypothesis. Thanks. You didn't exactly answer my question, but your response made me see that the answer to my question was obvious! $\endgroup$
    – Cass
    Commented Dec 23, 2013 at 18:57
  • 1
    $\begingroup$ Hi, Why is it that the homogeneous primes of the hyper surfaces are principal? Is it true that any locally principal codimension 1 subscheme of $\mathbb{P}^n$ has principal homogeneous ideal? $\endgroup$ Commented Mar 14, 2022 at 22:36

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