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I'm wondering why (and therefore also if) the notions of "a projective variety/submanifold of projective space is a complete intersection" as used in algebraic geometry and the theory of, say, Riemann surfaces agree.
The following are the precise versions of these notions I refer to:

In algebraic geometry:
An algebraic $A$ subset of $\mathbb CP^n$ of dimension $k$ is called a complete intersection if its vanishing ideal $I(A) \subseteq \mathbb C[X_0,...X_n]$ can be generated by $n-k$ polynomials.

In complex geometry:
A (complex) submanifold $A$ of $\mathbb CP^n$ of dimension $k$ is called a complete intersection if it arises as the (projective) zero-set of a homogeneous, holomorphic map $f:\mathbb C^{n+1} \rightarrow \mathbb C^{n-k}$ such that the Jacobian of $f$ has rank $n-k$ at every $0 \neq x \in \mathbb C^{n+1}$ with $f(x) = 0$.

The precise question now is:
Given a submanifold that is also an algebraic subset (e.g. smooth algebraic subset), do these notions of complete intersection coincide? And if so, do the required sets of maps agree?
Edit: Of course I am also interested in a proof of the result.

I am aware that every holomorphic, homogeneous map is actually a polynomial (just by Taylorexpansion).
As an example I can easily see that the twisted cubic curve in $\mathbb CP^3$ is not an algebraic geometry complete intersection, but how does one see that this is also not the case using the complex geometry defintion?

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Hi Francesco, why Chow's theorem? Sorry, I don't understand. –  diverietti Jan 22 '13 at 15:21
right, I misread the question. I delete the comment –  Francesco Polizzi Jan 22 '13 at 15:43
These notions will coincide if one demands, in the second of them, that rank of the Jacobian be $n-k$ everywhere on $f^{-1}(0)\setminus \{0\}$. –  Serge Lvovski Jan 22 '13 at 16:15
I mean, they coincide if in the first definition, which makes sense for singular subvarieties as well, you assume that the algebraic subset $A$ is smooth. –  Serge Lvovski Jan 22 '13 at 16:17
@Serge Lvovski: Of course, the condition only has to hold on the actual solution set, I edited that above. Thank you! But why is the statement true? –  FJH Jan 22 '13 at 17:03
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1 Answer

up vote 6 down vote accepted

$\star$ By Thm. I.5.1, page 32 of Hartshorne the Jacobian condition is equivalent to that every local ring of $A$ is a regular local ring, i.e., that $A$ is a non-singular variety.

From AG to CG:
This is (probably) the easy direction: Let the coordinate functions of $f$ be the generators of the ideal $I(A)$. By $\star$ and the assumption that $A$ is smooth/non-singular it follows that the Jacobian of $f$ has maximal rank.

From CG to AG:
Consider the ideal generated by the coordinate functions of the map $f:\mathbb C^{n+1}\to \mathbb C^{n-k}$ and call it $J$. The zero set of $J$ is $A$ and the proof of $\star$ shows that the Jacobian condition implies that localizing $\mathbb C[X_0,\dots,X_n]/J$ at any maximal ideal gives a regular local ring, so $A$ is smooth. Regular local rings are domains, so in particular this implies that $\mathbb C[X_0,\dots,X_n]/J$ has no nilpotents or in other words $J=\sqrt{J}$ and therefore $J=I(A)$. By design it is generated by $n-k$ elements.

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Is that a high-brow way of saying "a bunch of polynomials with full-rank Jacobian on their common zero-set generate a radical ideal"? If so, why is that? Or do you know a good reference? –  FJH Jan 22 '13 at 17:26
I'm sorry if my answer came across as smug, it was not intended. I was simply trying to understand your answer (which I still do not, simply because I don't know a whole lot about schemes). My question was only whether, that is the statement your answer boils down to, should one try to understand it "non-schematically". Also I do know how I would finish one direction of the proof, given that this is the case, but not how your answer comes into play. again this may well be due to ignorance on my part. –  FJH Jan 22 '13 at 19:40
OK, no problem. Sometimes comments come out differently than they were intended. You don't need to know anything about schemes actually to understand this answer. I hadn't known your background.... So, yes, that's what it means. Anyway, I will try to include a more detailed answer later, but I can't do it right now. Cheers! –  Sándor Kovács Jan 22 '13 at 21:34
OK, here is a more detailed proof. If this is still not clear enough, then try to be more specific in what is not clear and with which direction you are having trouble. –  Sándor Kovács Jan 23 '13 at 0:22
From AG to CG: Why does being a submanifold imply algebraic smoothness? From CG to AG: being a domain locally implies being a domain? On a different level: This uses "guns", where for the moment I consider the Nullstellensatz and the behaviour of local rings "guns", because they use more notion/maschinery than the question. I had been hoping for a more elementary proof, can this one be decoded into a short elementary one? –  FJH Jan 23 '13 at 18:13
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