Let $R$ be a local ring (commutative, Noetherian, over an algebraically closed field; if needed Henselian). Suppose one wants to prove some statement.

Suppose $R$ happens to be the ring of "functions", each of which being defined in a small enough neighborhood of the origin, then one can consider some representative of $Spec(R)$, with genuine geometric points off the origin. And to prove the statement pointwise, by going over all the points. Sometimes, this is simpler than to do the proof in the general case of an "abstract" local ring.

e.g. Suppose we are given two ideals $I,J\subset R$ (defined in a complicated way). And we want to check that the ideal they generate together contains a power of the maximal ideal. Geometrically (if $R$ is an analytic ring) this means: the two subschemes $V(I)$, $V(J)\subset Spec(R)$ intersect at the origin only. Then one can just go over the points in the punctured neighborhood of the origin and to check pointwise. This might greatly simplify the proof. (At least the idea of the proof. At least for some people.)

But, when working with complete local rings, one cannot speak of the "points near the origin", etc.

${\bf Question:}$ Is there some analogue of Lefschetz principle, when working with local rings? Something like: if a statement is formulated over an arbitrary local ring, and can be proven for analytic rings (i.e. $k \{ x_1,..,x_n \} /I$), then it is true for an arbitrary local ring (at least henselian, over $k=\bar{k}$)?

${\bf upd:}$ In view of the comments, an additional example might be helpful. Consider a matrix $A$ over a local ring. Suppose one wants to re-derive the standard upper bound on the height of the fitting ideal $I_j(A)$. The bound is well known in alg.com. But if this matrix happens to be a matrix of "functions", that are computable in some open neighborhood of the origin, then one can derive this bound also geometrically, as the codimension of the corresponding degeneracy locus. The question is: after I prove geometrically some bound of this type, which invocations should be pronounced to ensure the validity over an arbitrary local ring?

$\bf upd2:$ While the general approach to treat such questions uses the model theory (as is explained below), the following standard tricks often help.

  • Suppose a statement is known over some ring R. Often it trivially follows for any quotient of R and any subring of R.

  • Suppose the statement involves a finite amount of matrices, finitely generated ideals, modules. Then instead of the whole ring R take the subring generated by all the entries. This subring often embeds into some "good" ring.

  • (If nothing else works) Suppose there exists a counterexample to the statement, over some "bad" ring. Starting from this counterexample try to construct a counterexample over the "good" ring. (e.g. cut all the tails in the Taylor expansions, to get polynomials)

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    $\begingroup$ This may not be exactly what you're looking for, but if the statement you're trying to prove can be reduced to the completion of your local ring, you can use the Cohen structure theorem (en.wikipedia.org/wiki/Cohen_structure_theorem). $\endgroup$ Feb 17, 2013 at 16:52
  • $\begingroup$ @Eric Wofsey: I speak about a statement formulated over an arbitrary local ring. Maybe complete, maybe not. Can't see how Cohen's structure theorem can be helpful here. $\endgroup$ Feb 17, 2013 at 17:29
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    $\begingroup$ @Dmitry: Many properties hold for a local ring if and only if they hold for its completion. $\endgroup$ Feb 17, 2013 at 17:58
  • $\begingroup$ @Mahdi: precisely. That's what I'm asking. For which statements about the local rings it is enough to check the statement just for e.g. localization/henselization of an affine ring? $\endgroup$ Feb 17, 2013 at 19:15
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    $\begingroup$ There is Artin's Approximation Theorem, which I think is similar to the question you are asking. But then I see that you already asked a question about Artin's Approximation Theorem before, so that tells me this is not what you are looking for? $\endgroup$ Feb 17, 2013 at 21:18

1 Answer 1


In model theory the notion of categoricity captures the concept I think you are looking for. In particular Lefschetz principle is equivalent to the fact that the theory of algebraically closed fields is categorical. By this we mean that any two algebraically closed fields of the same cardinality (and characteristic, provided it is at least the continuum) are automatically isomorphic and in fact, any first order statements about such a field only depends on this cardinality. (The countable algebraically closed fields are isomorphic to algebraic closures of QQ(t_1,...,t_n) and transcendence degree distinguishes these)


There are model theoretic statements about complete discrete valuation rings which only depend the residue field. I'm not sure if this is what you are interested it.

The statement as you stated it is false: If k =FFbar_p. Every local rings takes the form W(k)[[X_1, \ldots, X_n]]/I for some ideal I by the Cohen-Structure theorem. Here W(k) is the ring of p-typical witt vectors of k. Then the first order sentence 1 + ... + 1 =0 (p-times) is true for analytic rings while it is not true for W(k).

Suppose you want to refine the statement, by just when k is the complex numbers.

If k=CC, and you allow me to make statements about convergence (using absolute values on CC in my language) I don't think it would be hard to come up with a first order statement which would be true for elements of analytic rings but not true for CC[[T]].

You can refine further and this leads to restricting what kind of statements are allowed (essentially restricting this symbols we are allowed to use). This can lead to thinking about things like the model theory of valued fields which is quite a large body of mathematics. I'm not an expert in this so I can just point you in this direction.

  • $\begingroup$ "any two algebraically closed fields of the same cardinality are automatically isomorphic"? $\endgroup$ Jul 15, 2014 at 22:31
  • $\begingroup$ Yeah... I was omitting the business about the countable case and the characteristic $\endgroup$
    – MathChump
    Jul 15, 2014 at 22:45
  • $\begingroup$ Why not just say "same transcendence degree and characteristic"? $\endgroup$
    – Todd Trimble
    Jul 15, 2014 at 23:28
  • $\begingroup$ to MathChump: thanks! (This is the first time in my life that I need smth from logic/model theory!) $\endgroup$ Aug 7, 2014 at 7:50

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