# Algebraic, analytic, formal modules

Consider torsion free modules over the germ of a fixed isolated algebraic hypersurface singularity {$f=0$}$\subset\mathbb{C}^n$. There are natural functors (using categories of finitely generated modules):

modules over $\mathbb{C}[x_1,..,x_n]_{(x_1,\dots,x_n)}/(f)$--> modules over $\mathbb{C}${$x_1,..,x_n$}$/(f)$--> modules over $\mathbb{C}[[x_1,..,x_n]]/(f)$.

Are they faithful, surjective? I know they are not surjective for an arbitrary local ring, but isolated hypersurface singularity is quite special.

upd:

1. The example of $x^2=y^2+y^3$ certainly counts, but can you suggest smth similar in the case of a locally irreducible analytic hypersurface?

2. Sorry I'm outsider in algebra. By surjectivity I meant smth like: every formal module over an analytic hypersurface arises from a locally analytic module. (Or maybe weaker: if a formal module has a submodule of the same rank that arises from locally analytic category, then the initial formal module arises from locally analytic category.)

up2: Thanks to everybody, sorry for delay

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I think it fails for a node in the plane, i.e. $f=y^2-x^2-x^3$. When you pass to completion it becomes reducible. $\mathbb{C}[[x,y]]/(y-x(1+x)^{1/2})$ doesn't have a preimage. – Manish Kumar Aug 6 '10 at 23:40
The ring on the left should be henselization (just think of Artin-Popescu approximation, noting that algebraic and analytic local rings are excellent), which would also remove Manish Kumar's example. Faithfulness is obvious from faithful flatness of completion. By "surjective" do you mean what is usually called "essentially surjective"? – BCnrd Aug 7 '10 at 0:31
I don't know of a counter example if you restrict to analytically irreducible singularities. I am not sure why you exactly need this for but as Brian said, looking at the Henselianization of the left ring might be a good idea. If you are interested in GAGA type result then there is a formal GAGA. Of course the setup is different but its in similar spirit. – Manish Kumar Aug 7 '10 at 2:18
Dear Manish: I don't think that formal GAGA can be useful here, since it requires completeness of the base ring. It's closer in spirit to issues around Artin approximation (but the latter doesn't seem to help much either, as far as I can tell). – BCnrd Aug 7 '10 at 12:00
oh yes. that's right! Thanks. – Manish Kumar Aug 7 '10 at 13:39

## 3 Answers

The most relevant result I am aware of is the following: suppose $R$ is a local Gorenstein ring, essentially of finite type over a field, with an isolated singularity (so certainly include your case) then the stable categories of maximal Cohen-Macaulay modules over $R$ and the completion $\hat R$ are equivalent up to direct summand via $\hat R \otimes_R?$. See Proposition 1.6 of this paper by Keller-Murfet-Van den Bergh.

In everyday English, this means that every maximal Cohen-Macaulay (MCM) module over $\hat R$ (which are torsion-free, and include high enough syzygies of any given module) is a direct summand of a completion of some MCM module over $R$. So restricting to MCM modules, the functor: (MCM modules over) $\mathbb C[x_1,\cdots,x_n]_{(x_1,\cdots,x_n)}/(f) \to \mathbb C[[x_1,\cdots,x_n]]/(f)$ is "surjective" up to direct summand.

In general, as BCnrd remarked, one can only hope to descend to the henselization of $R$. There is a big literature on this topic, see for example the references here.

ADDED: The nodal curve example by Manish is discussed in example A.5 of the Keller-Murfet-Van den Bergh paper. The completion is isomorphic to $S=\mathbb C[[u,v]]/(uv)$ and each $S/(u), S/(v)$ is not extended, but their direct sum is.

For your updated question 1 and Manish's comment: I think one can build higher dimension examples from Manish's original example using a result known as Knörrer's periodicity theorem (see Invent. Math., 88 (1987), 153–-164), namely that there is an equivalence of the stable categories of MCM modules over $\mathbb C[[x_1,\cdots,x_n]]/(f)$ and $\mathbb C[[x_1,\cdots,x_n,u,v]]/(f+uv)$. In particular, there should be an example for $f=y^2-x^2-x^3+uv$, which will be analytically irreducible.

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That's neat. Thanks. I expected a counter example in higher dimension. – Manish Kumar Aug 7 '10 at 14:21

Dear Hailong (and all others reading his), I apologize for using answer' for a comment, since my openid does not work (I am traveling) or I have forgotten it and MO does not give me a way to comment.

Yes, there are examples over any field. If my memory serves me right, the first example I saw was in Samuel's TIFR lecture notes on UFD's. There are also examples (as an aside) in my paper on rational double points which appeared in the 80s in the Inventiones.

For the question of Brian, in general class group is used for (over normal domains) rank one torsion free finitely generated modules upto isomorphism modulo free modules. If one wants to look at invertible modules, nowadays it seems more common (especially people with an alg. geom. bend) to use Picard group. The examples I mention above are algebraic' and not Henselizations.

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Dear Mohan, thanks for the reference. I will check through it for the example. – Hailong Dao Aug 9 '10 at 21:18

There are well studied hypersurfaces of dimension 2 which are UFDs whose completions are not. So, any ideal representing a non-trivial element in the class group of the completion will not come from the algebraic ring.

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Dear Mohan, by the way, do you know any example over $\mathbb C$? I looked in Fossum but could not find any. Thanks. – Hailong Dao Aug 7 '10 at 10:21
This "class group" terminology is meant in the sense of "Weil divisors modulo principal divisors" (e.g., a non-principal height-1 prime ideal) rather than the alternative meaning with invertible modules/sheaves (which is trivial for a local ring). But this notion of "class group" doesn't make sense for abstract modules, merely ideals in the ring. So can you rule out that the non-principal height-1 prime over the completion may be module-isomorphic to the completion of a finitely generated module over the henselized algebraic local ring (as in the correct form of the question)? – BCnrd Aug 7 '10 at 11:55