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  1. I’d like to tentatively suggest replacing the current statement of Nakayama’s lemma on the relevant page with the following (non-local) version:

    Nakayama’s lemma (proposed version): Let R be a commutative ring and M be a finitely generated R-module. Then:

    1. supp(M)=𝒱(Ann(M)).

    2. If moreover IR is an ideal with supp(M)V(I) inhabited, then MRR/I is nonzero.

    (Recall that supp(M)={𝔭Spec(R)s.t.MRR𝔭is nonzero}, Ann(M)={fRs.t.fM=0}, and 𝒱(I)={𝔭Spec(R)s.t.𝔭I}. The first claim is not usually considered part of Nakayama’s lemma but will be useful here and can be proved with a closely related argument to that which we will use for the main result.)

    The proof uses relatively little pre-existing commutative algebraic theory (i.e., just some basic facts about 𝒱, nonzeroness in short exact sequences, localization, tensoring, and localizing/tensoring cyclic modules) as follows:

    Proof of Nakayama’s lemma: The argument proceeds in the following four steps:

    1. As M is finitely generated, it admits a filtration with cyclic cokernels. To be explicit, if the n-tuple of elements (mkM)k=0,,n1 generates M, then it begets an n-tuple of short exact sequences

      (0MkMk+1R/Ik0)k=0,,n1,

      where Mk is the submodule of M spanned by (mkM)k=0,,k1 and in particular M0=0 and Mn=M. (Here the maps MkMk+1 are the evident inclusions; as the cokernel is generated by the image of mk, it is in particular isomorphic to R/Ik for some ideal IkR.) The basic idea is that the process of extension by which M is built up from these R/Iks interacts sufficiently nicely with the operations of localization/taking fibers that the properties of the latter determine those of the former.

    2. For instance, it is clear that any fAnn(M) must descend to 0 on all the R/Iks; on the other hand, for all k=0,,n1 we have that IkMk+1Mk. We conclude that n1k=0IkAnn(M)n1k=0Ik, whence 𝒱(Ann(M))=n1k=0𝒱(Ik).

    3. Let us now show that supp(M)=n1k=0𝒱(Ik); this will complete the proof of our first claim. Recall that for any 𝔭Spec(R) the functor RR𝔭 is exact; we therefore in any case obtain an n-tuple of short exact sequences

      (0MkRR𝔭Mk+1RR𝔭R/IkRR𝔭0)k=0,,n1.

      If 𝔭k=0,,n1𝒱(Ik), then each R/IkRR𝔭 vanishes, so the above SESs collapse to isomorphisms

      (0MkRR𝔭Mk+1RR𝔭0)k=0,,n1

      and thus

      MRR𝔭MnRR𝔭M0RR𝔭0.

      Conversely, if 𝔭k=0,,n1𝒱(Ik), then there is a maximal index kmax such that 𝔭Ikmax and in particular an epimorphism

      Mkmax+1RR𝔭R/IkmaxRR𝔭0,

      with R/IkmaxRR𝔭 nonzero. But there are also, as before, isomorphisms

      (0MkRR𝔭Mk+1RR𝔭0)k=kmax+1,,n1

      whence

      MRR𝔭MnRR𝔭Mkmax+1RR𝔭

      is epic onto a nonzero module and thus itself nonzero.

    4. The proof of the second claim is in the same vein, requiring just a little bit more delicacy. Suppose that 𝔭 inhabits supp(M)V(I); by the above this means that 𝔭I and that there exists some index k for which 𝔭Ik. Again consider the maximal index kmax for which 𝔭Ikmax. As before, there is an epimorphism

      Mkmax+1RR𝔭R/IkmaxRR𝔭0

      and isomorphisms

      (0MkRR𝔭Mk+1RR𝔭0)k=kmax+1,,n1

      By the right-exactness of the functor RR/𝔭, we obtain from the former an epimorphism

      Mkmax+1RR𝔭/𝔭R/IkmaxRR𝔭/𝔭0

      (implicitly using that (RR𝔭)RR/𝔭RR𝔭/𝔭). By its functoriality, there are likewise isomorphisms

      (0MkRR𝔭/𝔭Mk+1RR𝔭/𝔭0)k=kmax+1,,n1

      Now,

      R/IkmaxRR𝔭/𝔭R𝔭/𝔭

      is nonzero (here we use that 𝔭Ikmax). It follows that

      MRR𝔭/𝔭MnRR𝔭/𝔭Mkmax+1RR𝔭/𝔭

      is epic onto a nonzero module and thus itself nonzero. But

      MRR𝔭/𝔭(MRR/I)RR𝔭/𝔭

      (here we use that 𝔭I), so the latter must itself be nonzero, as claimed.

    It is now straightforward to recover more familiar forms of Nakayama’s lemma as a corollary of the above. For instance, the nLab’s current version states:

    Nakayama’s lemma (nLab’s version): Let R be a local ring with maximal ideal 𝔪R and M be a finitely generated R-module. Then if M is nonzero, so too is MRR/𝔪.

    Indeed, 𝔪 is practically tautologically in (and so 𝒱(𝔪) practically tautologically intersects) the support of any nonzero R-module; the conditions of our version of the lemma apply, and we conclude that MRR/𝔪 is nonzero as advertised.

    Sometimes (for instance in the Stacks Project), Nakayama’s lemma is stated instead as follows:

    Nakayama’s lemma (Stacks Project’s version): Let R be a commutative ring, M be a finitely generated R-module, and IR be an ideal with IM=M. Then there exists fR such that f=1(modI) and fM=0.

    Of course, IM=MMRR/I0; it follows from our version of the lemma that I is comaximal with Ann(M). The Chinese Remainder theorem then asserts the existence of fR such that f=1(modI) and f=0(modAnn(M)); the desired f.

    (I could not find a discussion page linked to the Nakayama’s Lemma entry; I’m not sure if this is the proper way to discuss pages without linked nForum threads. Please advise if not!)

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeDec 6th 2023

    Thanks for this, this looks like great material. I think you should feel invited to add this to the entry.

    (The dedicated discussion page for the entry will appear once you make the first edit there, together with a log-message. The page has probably been created so long ago (and left essentially untouched since) that it was before this announcement mechanism was in place.)

    • CommentRowNumber3.
    • CommentAuthorJ-B Vienney
    • CommentTimeDec 6th 2023
    • (edited Dec 6th 2023)

    I would be glad that you add this to the entry, because it looks very helpful to me, especially the clear and detailled proof.

    • CommentRowNumber4.
    • CommentAuthorIngoBlechschmidt
    • CommentTimeDec 6th 2023
    • (edited Dec 6th 2023)

    I agree, very nice, and I wonder what a constructive version of statement 2 looks like. If I should venture a guess, it would be:

    If MRR/I is zero, then the geometric theory of a prime filter 𝔭 with x𝔭 for xI proves, for every xM, that sAsx=0s𝔭.

    I arrived at this formulation as follows. First let’s take the original formulation:

    (1) If supp(M)V(I) is inhabited, then MRR/I is nonzero.

    This looks like the contrapositive of a more direct statement:

    (2) If MRR/I is zero, then supp(M)V(I)=.

    The set V(I) is the set of models of a geometric theory, and also Spec(A)supp(M) is; but supp(M) is not. So let’s rewrite as follows:

    (3) If MRR/I is zero, then V(I)Spec(A)supp(M).

    The inclusion states that all models of the first geometric theory are also models of the second. So let’s do the leap to the theories themselves instead of their models (equivalently, referring to models in all toposes, not just the standard topos). In this way we arrive at the proposed constructivization.

    To check that the proposed constructivization is indeed constructively provable, we can either appeal to Barr’s theorem or use the direct description of provability modulo the theory of prime filters. I’m guessing that I’m talking mostly to myself right now, thank you for the nice nudge, but if anyone wants details I’m happy to fill them in. :-)

  2. Thanks for the feedback everybody! I'm a little behind on some work for the next few days, but will try to add the suggested edits by the weekend.

    Re @above, I admittedly don't entirely follow, but it seems interesting! :)
    • CommentRowNumber6.
    • CommentAuthorzskoda
    • CommentTimeDec 7th 2023
    • (edited Dec 7th 2023)

    In modern algebra, Nakayama’s lemma is often presented in generality of noncommutative rings. See for example http://alpha.math.uga.edu/~pete/noncommutativealgebra.pdf.

    • CommentRowNumber7.
    • CommentAuthorUrs
    • CommentTimeDec 12th 2023

    I have now merged the contribution from #1 into the entry Nakayama’s lemma (as announced there).