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?



Complete intersections in complex and algebraic geometry
$\star$ 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.


