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?

everywhere on$f^{-1}(0)\setminus \{0\}$. – Serge Lvovski Jan 22 '13 at 16:15