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1. • CommentRowNumber1.
   • CommentAuthorthesnakefromthelemma
   • CommentTimeDec 6th 2023
   • (edited Dec 6th 2023)
   • PermaLink
   Author: thesnakefromthelemma
   Format: MarkdownItexI'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. $\text{supp}(M) = \mathcal{V}(Ann(M))$. > > 2. If moreover $I\subseteq R$ is an ideal with $supp(M)\cap V(I)$ inhabited, then $M\otimes_{R} R/I$ is nonzero. > > (Recall that $supp(M) = \{\mathfrak{p}\in Spec(R)\ \text{ s.t. }\ M\otimes_{R} R_{\mathfrak{p}}\ \text{ is nonzero}\}$, $Ann(M) = \{f \in R\ \text{ s.t. }\ f M = 0\}$, and $\mathcal{V}(I) = \{\mathfrak{p} \in Spec(R)\ \text{ s.t. }\ \mathfrak{p} \supseteq I\}$. 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 $\mathcal{V}$, 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 $(m_{k}\in M)_{k=0,\dots,n-1}$ generates $M$, then it begets an $n$-tuple of short exact sequences \[\left(\ 0 \to M_{k} \to M_{k+1} \to R/I_{k} \to 0\ \right)_{k=0,\dots,n-1},\] where $M_{k}$ is the submodule of $M$ spanned by $(m_{k}\in M)_{k=0,\dots,k-1}$ and in particular $M_{0}=0$ and $M_{n}=M$. (Here the maps $M_{k} \to M_{k+1}$ are the evident inclusions; as the cokernel is generated by the image of $m_{k}$, it is in particular isomorphic to $R/I_{k}$ for some ideal $I_{k}\subseteq R$.) The basic idea is that the process of extension by which $M$ is built up from these $R/I_{k}$s 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 $f \in Ann(M)$ must descend to $0$ on all the $R/I_{k}$s; on the other hand, for all $k=0,\dots,n-1$ we have that $I_{k}M_{k+1} \subseteq M_{k}$. We conclude that $\prod_{k=0}^{n-1} I_{k} \subseteq Ann(M) \subseteq \bigcap_{k=0}^{n-1} I_{k}$, whence $\mathcal{V}(Ann(M)) = \bigcup_{k=0}^{n-1} \mathcal{V}(I_{k})$. > > 3. Let us now show that $supp(M) = \bigcup_{k=0}^{n-1} \mathcal{V}(I_{k})$; this will complete the proof of our first claim. Recall that for any $\mathfrak{p}\in Spec(R)$ the functor $-\otimes_{R} R_{\mathfrak{p}}$ is exact; we therefore in any case obtain an $n$-tuple of short exact sequences \[\left(\ 0 \to M_{k} \otimes_{R} R_{\mathfrak{p}} \to M_{k+1} \otimes_{R} R_{\mathfrak{p}} \to R/I_{k} \otimes_{R} R_{\mathfrak{p}} \to 0\ \right)_{k=0,\dots,n-1}.\] If $\mathfrak{p} \notin \bigcup_{k=0,\dots,n-1} \mathcal{V}(I_{k})$, then each $R/I_{k} \otimes_{R} R_{\mathfrak{p}}$ vanishes, so the above SESs collapse to isomorphisms \[\left(\ 0 \to M_{k} \otimes_{R} R_{\mathfrak{p}} \to M_{k+1} \otimes_{R} R_{\mathfrak{p}} \to 0\ \right)_{k=0,\dots,n-1}\] and thus \[M \otimes_{R} R_{\mathfrak{p}} \simeq M_{n} \otimes_{R} R_{\mathfrak{p}} \simeq M_{0} \otimes_{R} R_{\mathfrak{p}} \simeq 0.\] Conversely, if $\mathfrak{p} \notin \bigcup_{k=0,\dots,n-1} \mathcal{V}(I_{k})$, then there is a maximal index $k_{max}$ such that $\mathfrak{p} \supseteq I_{k_{max}}$ and in particular an epimorphism \[M_{k_{max}+1} \otimes_{R} R_{\mathfrak{p}} \to R/I_{k_{max}} \otimes_{R} R_{\mathfrak{p}} \to 0,\] with $R/I_{k_{max}} \otimes_{R} R_{\mathfrak{p}}$ nonzero. But there are also, as before, isomorphisms \[\left(\ 0 \to M_{k} \otimes_{R} R_{\mathfrak{p}} \to M_{k+1} \otimes_{R} R_{\mathfrak{p}} \to 0\ \right)_{k=k_{max}+1,\dots,n-1}\] whence \[M \otimes_{R} R_{\mathfrak{p}} \simeq M_{n} \otimes_{R} R_{\mathfrak{p}} \simeq M_{k_{max}+1} \otimes_{R} R_{\mathfrak{p}}\] 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 $\mathfrak{p}$ inhabits $\text{supp}(M)\cap V(I)$; by the above this means that $\mathfrak{p}\supseteq I$ and that there exists some index $k$ for which $\mathfrak{p}\supseteq I_{k}$. Again consider the maximal index $k_{max}$ for which $\mathfrak{p} \supseteq I_{k_{max}}$. As before, there is an epimorphism \[M_{k_{max}+1} \otimes_{R} R_{\mathfrak{p}} \to R/I_{k_{max}} \otimes_{R} R_{\mathfrak{p}} \to 0\] and isomorphisms \[\left(\ 0 \to M_{k} \otimes_{R} R_{\mathfrak{p}} \to M_{k+1} \otimes_{R} R_{\mathfrak{p}} \to 0\ \right)_{k=k_{max}+1,\dots,n-1}\] By the right-exactness of the functor $-\otimes_{R} R/\mathfrak{p}$, we obtain from the former an epimorphism \[M_{k_{max}+1} \otimes_{R} R_{\mathfrak{p}} /\mathfrak{p} \to R/I_{k_{max}} \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p} \to 0\] (implicitly using that $(- \otimes_{R} R_{\mathfrak{p}})\otimes_{R} R/\mathfrak{p}\ \simeq\ - \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p}$). By its functoriality, there are likewise isomorphisms \[\left(\ 0 \to M_{k} \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p} \to M_{k+1} \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p} \to 0\ \right)_{k=k_{max}+1,\dots,n-1}\] Now, \[R/I_{k_{max}} \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p} \simeq R_{\mathfrak{p}}/\mathfrak{p}\] is nonzero (here we use that $\mathfrak{p} \supseteq I_{k_{max}}$). It follows that \[M \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p} \simeq M_{n} \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p} \simeq M_{k_{max}+1} \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p}\] is epic onto a nonzero module and thus itself nonzero. But \[M \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p} \simeq (M \otimes_{R} R/I) \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p}\] (here we use that $\mathfrak{p} \supseteq I$), so the latter must itself be nonzero, as claimed. $\blacksquare$ 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 $\mathfrak{m}\subseteq R$ and $M$ be a finitely generated $R$-module. Then if $M$ is nonzero, so too is $M \otimes_R R/\mathfrak{m}$. Indeed, $\mathfrak{m}$ is practically tautologically in (and so $\mathcal{V}(\mathfrak{m})$ practically tautologically intersects) the support of any nonzero $R$-module; the conditions of our version of the lemma apply, and we conclude that $M \otimes_R R/\mathfrak{m}$ is nonzero as advertised. $\blacksquare$ Sometimes (for instance in the [Stacks Project](https://stacks.math.columbia.edu/tag/07RC)), 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 $I\subseteq R$ be an ideal with $I M = M$. Then there exists $f\in R$ such that $f = 1\ (mod\ I)$ and $f M=0$. Of course, $I M = M\ \iff\ M \otimes_{R} R/I \simeq 0$; it follows from our version of the lemma that $I$ is comaximal with $Ann(M)$. The [Chinese Remainder theorem](https://stacks.math.columbia.edu/tag/00DT) then asserts the existence of $f\in R$ such that $f = 1\ (mod\ I)$ and $f = 0\ (mod\ Ann(M))$; the desired $f$. $\blacksquare$ (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!)

   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. $\text{supp}(M) = \mathcal{V}(Ann(M))$.

   2. If moreover $I\subseteq R$ is an ideal with $supp(M)\cap V(I)$ inhabited, then $M\otimes_{R} R/I$ is nonzero.

   (Recall that $supp(M) = \{\mathfrak{p}\in Spec(R)\ \text{ s.t. }\ M\otimes_{R} R_{\mathfrak{p}}\ \text{ is nonzero}\}$, $Ann(M) = \{f \in R\ \text{ s.t. }\ f M = 0\}$, and $\mathcal{V}(I) = \{\mathfrak{p} \in Spec(R)\ \text{ s.t. }\ \mathfrak{p} \supseteq I\}$. 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 $\mathcal{V}$, 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 $(m_{k}\in M)_{k=0,\dots,n-1}$ generates $M$, then it begets an $n$-tuple of short exact sequences

      $$\left(\ 0 \to M_{k} \to M_{k+1} \to R/I_{k} \to 0\ \right)_{k=0,\dots,n-1},$$

      where $M_{k}$ is the submodule of $M$ spanned by $(m_{k}\in M)_{k=0,\dots,k-1}$ and in particular $M_{0}=0$ and $M_{n}=M$. (Here the maps $M_{k} \to M_{k+1}$ are the evident inclusions; as the cokernel is generated by the image of $m_{k}$, it is in particular isomorphic to $R/I_{k}$ for some ideal $I_{k}\subseteq R$.) The basic idea is that the process of extension by which $M$ is built up from these $R/I_{k}$s 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 $f \in Ann(M)$ must descend to $0$ on all the $R/I_{k}$s; on the other hand, for all $k=0,\dots,n-1$ we have that $I_{k}M_{k+1} \subseteq M_{k}$. We conclude that $\prod_{k=0}^{n-1} I_{k} \subseteq Ann(M) \subseteq \bigcap_{k=0}^{n-1} I_{k}$, whence $\mathcal{V}(Ann(M)) = \bigcup_{k=0}^{n-1} \mathcal{V}(I_{k})$.

   3. Let us now show that $supp(M) = \bigcup_{k=0}^{n-1} \mathcal{V}(I_{k})$; this will complete the proof of our first claim. Recall that for any $\mathfrak{p}\in Spec(R)$ the functor $-\otimes_{R} R_{\mathfrak{p}}$ is exact; we therefore in any case obtain an $n$-tuple of short exact sequences

      $$\left(\ 0 \to M_{k} \otimes_{R} R_{\mathfrak{p}} \to M_{k+1} \otimes_{R} R_{\mathfrak{p}} \to R/I_{k} \otimes_{R} R_{\mathfrak{p}} \to 0\ \right)_{k=0,\dots,n-1}.$$

      If $\mathfrak{p} \notin \bigcup_{k=0,\dots,n-1} \mathcal{V}(I_{k})$, then each $R/I_{k} \otimes_{R} R_{\mathfrak{p}}$ vanishes, so the above SESs collapse to isomorphisms

      $$\left(\ 0 \to M_{k} \otimes_{R} R_{\mathfrak{p}} \to M_{k+1} \otimes_{R} R_{\mathfrak{p}} \to 0\ \right)_{k=0,\dots,n-1}$$

      and thus

      $$M \otimes_{R} R_{\mathfrak{p}} \simeq M_{n} \otimes_{R} R_{\mathfrak{p}} \simeq M_{0} \otimes_{R} R_{\mathfrak{p}} \simeq 0.$$

      Conversely, if $\mathfrak{p} \notin \bigcup_{k=0,\dots,n-1} \mathcal{V}(I_{k})$, then there is a maximal index $k_{max}$ such that $\mathfrak{p} \supseteq I_{k_{max}}$ and in particular an epimorphism

      $$M_{k_{max}+1} \otimes_{R} R_{\mathfrak{p}} \to R/I_{k_{max}} \otimes_{R} R_{\mathfrak{p}} \to 0,$$

      with $R/I_{k_{max}} \otimes_{R} R_{\mathfrak{p}}$ nonzero. But there are also, as before, isomorphisms

      $$\left(\ 0 \to M_{k} \otimes_{R} R_{\mathfrak{p}} \to M_{k+1} \otimes_{R} R_{\mathfrak{p}} \to 0\ \right)_{k=k_{max}+1,\dots,n-1}$$

      whence

      $$M \otimes_{R} R_{\mathfrak{p}} \simeq M_{n} \otimes_{R} R_{\mathfrak{p}} \simeq M_{k_{max}+1} \otimes_{R} R_{\mathfrak{p}}$$

      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 $\mathfrak{p}$ inhabits $\text{supp}(M)\cap V(I)$; by the above this means that $\mathfrak{p}\supseteq I$ and that there exists some index $k$ for which $\mathfrak{p}\supseteq I_{k}$. Again consider the maximal index $k_{max}$ for which $\mathfrak{p} \supseteq I_{k_{max}}$. As before, there is an epimorphism

      $$M_{k_{max}+1} \otimes_{R} R_{\mathfrak{p}} \to R/I_{k_{max}} \otimes_{R} R_{\mathfrak{p}} \to 0$$

      and isomorphisms

      $$\left(\ 0 \to M_{k} \otimes_{R} R_{\mathfrak{p}} \to M_{k+1} \otimes_{R} R_{\mathfrak{p}} \to 0\ \right)_{k=k_{max}+1,\dots,n-1}$$

      By the right-exactness of the functor $-\otimes_{R} R/\mathfrak{p}$, we obtain from the former an epimorphism

      $$M_{k_{max}+1} \otimes_{R} R_{\mathfrak{p}} /\mathfrak{p} \to R/I_{k_{max}} \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p} \to 0$$

      (implicitly using that $(- \otimes_{R} R_{\mathfrak{p}})\otimes_{R} R/\mathfrak{p}\ \simeq\ - \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p}$). By its functoriality, there are likewise isomorphisms

      $$\left(\ 0 \to M_{k} \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p} \to M_{k+1} \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p} \to 0\ \right)_{k=k_{max}+1,\dots,n-1}$$

      Now,

      $$R/I_{k_{max}} \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p} \simeq R_{\mathfrak{p}}/\mathfrak{p}$$

      is nonzero (here we use that $\mathfrak{p} \supseteq I_{k_{max}}$). It follows that

      $$M \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p} \simeq M_{n} \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p} \simeq M_{k_{max}+1} \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p}$$

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

      $$M \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p} \simeq (M \otimes_{R} R/I) \otimes_{R} R_{\mathfrak{p}}/\mathfrak{p}$$

      (here we use that $\mathfrak{p} \supseteq I$), so the latter must itself be nonzero, as claimed. $\blacksquare$

   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 $\mathfrak{m}\subseteq R$ and $M$ be a finitely generated $R$-module. Then if $M$ is nonzero, so too is $M \otimes_R R/\mathfrak{m}$.

   Indeed, $\mathfrak{m}$ is practically tautologically in (and so $\mathcal{V}(\mathfrak{m})$ practically tautologically intersects) the support of any nonzero $R$-module; the conditions of our version of the lemma apply, and we conclude that $M \otimes_R R/\mathfrak{m}$ is nonzero as advertised. $\blacksquare$

   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 $I\subseteq R$ be an ideal with $I M = M$. Then there exists $f\in R$ such that $f = 1\ (mod\ I)$ and $f M=0$.

   Of course, $I M = M\ \iff\ M \otimes_{R} R/I \simeq 0$; it follows from our version of the lemma that $I$ is comaximal with $Ann(M)$. The Chinese Remainder theorem then asserts the existence of $f\in R$ such that $f = 1\ (mod\ I)$ and $f = 0\ (mod\ Ann(M))$; the desired $f$. $\blacksquare$

   (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!)

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeDec 6th 2023
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   Author: Urs
   Format: MarkdownItexThanks 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.)

   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.)

3. • CommentRowNumber3.
   • CommentAuthorJ-B Vienney
   • CommentTimeDec 6th 2023
   • (edited Dec 6th 2023)
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   Author: J-B Vienney
   Format: MarkdownItexI would be glad that you add this to the entry, because it looks very helpful to me, especially the clear and detailled proof.

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

4. • CommentRowNumber4.
   • CommentAuthorIngoBlechschmidt
   • CommentTimeDec 6th 2023
   • (edited Dec 6th 2023)
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   Author: IngoBlechschmidt
   Format: MarkdownItexI 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 $M \otimes_R R/I$ is zero, then the geometric theory of a prime filter $\mathfrak{p}$ with $x \in \mathfrak{p} \vdash \bot$ for $x \in I$ proves, for every $x \in M$, that $\top \vdash \bigvee_{\substack{s \in A\\sx=0}} s \in \mathfrak{p}$. I arrived at this formulation as follows. First let's take the original formulation: (1) If $\mathrm{supp}(M) \cap V(I)$ is inhabited, then $M \otimes_R R/I$ is nonzero. This looks like the contrapositive of a more direct statement: (2) If $M \otimes_R R/I$ is zero, then $\mathrm{supp}(M) \cap V(I) = \emptyset$. The set $V(I)$ is the set of models of a geometric theory, and also $\mathrm{Spec}(A) \setminus \mathrm{supp}(M)$ is; but $\mathrm{supp}(M)$ is not. So let's rewrite as follows: (3) If $M \otimes_R R/I$ is zero, then $V(I) \subseteq \mathrm{Spec}(A) \setminus \mathrm{supp}(M)$. 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. :-)

   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 $M \otimes_R R/I$ is zero, then the geometric theory of a prime filter $\mathfrak{p}$ with $x \in \mathfrak{p} \vdash \bot$ for $x \in I$ proves, for every $x \in M$, that $\top \vdash \bigvee_{\substack{s \in A\\sx=0}} s \in \mathfrak{p}$.

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

   (1) If $\mathrm{supp}(M) \cap V(I)$ is inhabited, then $M \otimes_R R/I$ is nonzero.

   This looks like the contrapositive of a more direct statement:

   (2) If $M \otimes_R R/I$ is zero, then $\mathrm{supp}(M) \cap V(I) = \emptyset$.

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

   (3) If $M \otimes_R R/I$ is zero, then $V(I) \subseteq \mathrm{Spec}(A) \setminus \mathrm{supp}(M)$.

   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. :-)

5. • CommentRowNumber5.
   • CommentAuthorthesnakefromthelemma
   • CommentTimeDec 7th 2023
   • (edited Dec 7th 2023)
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   Author: thesnakefromthelemma
   Format: TextThanks 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! :)

   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! :)
6. • CommentRowNumber6.
   • CommentAuthorzskoda
   • CommentTimeDec 7th 2023
   • (edited Dec 7th 2023)
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   Author: zskoda
   Format: MarkdownItexIn modern algebra, Nakayama's lemma is often presented in generality of noncommutative rings. See for example <http://alpha.math.uga.edu/~pete/noncommutativealgebra.pdf>.

   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.

7. • CommentRowNumber7.
   • CommentAuthorUrs
   • CommentTimeDec 12th 2023
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   Author: Urs
   Format: MarkdownItexI have now merged the contribution from #1 into the entry *[[Nakayama's lemma]]* (as announced [there](https://nforum.ncatlab.org/discussion/17537/nakayamas-lemma/?Focus=114599#Comment_114599)).

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