# What is the intuition behind the proof of the algebraic version of Cartan's theorem A?

I am trying to understand the idea behind the proof of GAGA. A crucial step is the following:

Theorem: Let $X=\mathbb{P}^r_{\mathbb{C}}$ (either as a variety or as an analytic space), and let $\mathcal{M}$ be a coherent sheaf on $X$. Then for $n>>0$, the twisted sheaf $\mathcal{M}(n)$ is generated by finitely many global sections.

In the algebraic case, this is Theorem 5.17 in Hartshorne chapter II. If one tries to read the proof of Theorem 5.17, one sees that it depends on Lemma 5.14, which in turn is a generalization of Lemma 5.3. Lemma 5.3 seems to me to be a completely algebraic lemma with no geometric intuition. It will disrupt the flow of the question to state it here, so I will put it at the end. My point is that I don't see any intuition in this statement

In the analytic case this is equivalent to Cartan's Theorem A. To quote the wikipedia page: "Naively, they imply that a holomorphic function on a closed complex submanifold $Z$ of a Stein manifold $X$ can be extended to a holomorphic function on all of $X$". I must confess that I have not read a proof of Cartan's Theorem A itself. But I would like to get some intuition about why it is true, and how it translates to the nilpotent proof found in Hartshorne...

### Appendix

Lemma 5.3 in Hartshorne chapter II: Let $X=Spec(A)$ be an affine scheme, let $f\in A$, let $D(f)\subset X$ be the corresponding open set, and let $\mathcal{F}$ be a quasi-coherent sheaf on $X$.

a. If $s\in \Gamma(X,\mathcal{F})$ is such that its restriction to $D(f)$ is $0$, then for some $n>0$, $f^ns=0$.

b. Given a section $t\in \mathcal{F}(D(f))$ of $\mathcal{F}$ over the open set $D(f)$, then for some $n>0$, $f^nt$ extends to a global section of $\mathcal{F}$ over $X$.

The reason I call this a "nilpotent method" is because in complex algerbaic geometry $f^ns=0$ would imply that either $f=0$ or $s=0$.

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(b) does have geometric meaning; you should think of $t$ as having a pole of some high order over the subscheme of $X$ cut out by $f$; then multiplying by a large power of $f$ cuts down the order of the pole so that $f^nt$ is regular over all of $X$. Think of e.g. the function $1/x^n$ on D(x) in $\mathbb{A}^1_\mathbb{C}$. –  Daniel Litt Nov 13 '11 at 19:10

Since we're looking for intuition, let's assume that $\mathscr F=\mathscr O_X$, the sheaf of regular functions.

The geometric meaning of this Lemma is that not-everywhere-defined regular functions are more like meromoprhic functions than holomorphic ones. In other words it says that regular functions can't have essential singularities.

Part (b) says regular functions (or sections of $\mathscr F$ in general) on $D(f)$ can be written as $g/f^n$ for some $n>0$ and some global regular function $g$.

Part (a) says that the expression $g/f^n$ is unique as much as possible: If there were $g$ and $h$ such that $g/f^n=h/f^m$ on $D(f)$ then $s=g/f^n-h/f^m$ is $0$ on $D(f)$ and hence by (a) there exists a $q>0$ such that $$0=f^qs=f^q(g/f^n-h/f^m)$$ i.e., $$f^{m+q}g=f^{n+q}h$$ as global regular functions, so the original function (or section) can be written as a fraction in essentially a unique way. (The ambiguity of multiplying both the numerator and the denominator by a power of $f$ is always there and unless we are working with a UFD we can't say that $g$ and $f$ should have no common divisors).

In other words, this Lemma says that the ring of regular functions on the complement of the zero set of a global regular function is simply the localization of the ring of global regular functions. Part (b) says that everything is a fraction and part (a) says that two different representation as a fraction gives the same element in the localization. If this sentence does not feel like a geometric description, then consider this: this is actually a geometric description of the process of localization by a single element.

Finally let me point out that "nilpotent" method is not a good name, because $s$ does not have to be nilpotent for this. The condition only means that the support of $s$ is contained in the zero locus of $f$. This can happen even on smooth (complex) manifolds: Take the disjoint union of two manifolds and take a holomorphic function on each. Of course it can't happen on connected manifolds if we are talking about regular functions, but if we're talking about sheaves, then this only needs that the sheaf be not torsion free. For instance the structure sheaf of a proper closed submanifold would do. So, instead of "nilpotent" you should think "torsion" (zerodivisor).

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