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In math.stackexchange answer #239445, Makoto Kato quoted a statement from the paper

Morris Orzech, Onto Endomorphisms are Isomorphisms, Amer. Math. Monthly 78 (1971), 357--362.

The statement (Theorem 1 in said paper) claims that if $A$ is a commutative ring, if $M$ is a finitely-generated $A$-module, if $N$ is an $A$-submodule of $M$, and if $f : N \to M$ is a surjective $A$-linear map, then $f$ is also injective.

This claim generalizes the well-known fact that a surjective endomorphism of a finitely-generated $A$-module is injective (which appears, e.g., as Lemma 10.15.4 in the Stacks project, Version 5b422bc, compiled on Nov 05, 2014 -- the numbering will probably change but it is Lemma {lemma-fun} in algebra.tex -- and I am citing this because the Stacks project has an interesting alternative proof of this fact). Orzech's proof first deals with the case of $A$ noetherian and then reduces the general case to it. Unfortunately, the reduction step is wrong: the map $f$ is applied to elements of $M$ which are not known to belong to $N$. (The proof in Orzech's paper is the same as the one in Kato's post.)

Is the claim itself true for non-Noetherian $A$ ? I cannot even prove the $M = A$ case, nor can I obtain a counterexample from the usual suspects (polynomials in infinitely many variables, or idempotent variables, or nilpotent variables).

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  • $\begingroup$ note that the person who first gave the reference was math.stackexchange.com/users/619/kcd $\endgroup$ Commented Dec 13, 2014 at 4:48
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    $\begingroup$ @Darij: The stacks project provides tags which are version-independent; they can even be used in papers. See stacks.math.columbia.edu/tags for further information. Here it's Tag 05G8. Great question by the way; I really wondered about that statement when I heard of it recently. $\endgroup$ Commented Dec 13, 2014 at 9:15
  • $\begingroup$ I think the proof of the Noetherian case is also flawed. $f^n$ doesn't make sense there. $\endgroup$ Commented Dec 13, 2014 at 11:25
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    $\begingroup$ Proof when $M=A$: Choose $n \in N$ with $f(n)=1$. If $f(x)=0$, then $0=f(x)n=f(xn)=f(n)x=x$. $\endgroup$ Commented Dec 13, 2014 at 11:34

5 Answers 5

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Let $A$ be a commutative ring. Let $M$ be a finitely generated $A$-module and $N$ be an $A$-submodule of $M$. Let $f\colon N \rightarrow M$ be a surjective homomorphism of $A$-modules. Then $f$ is injective.

Proof. Let $0 \neq x'_0 \in N$. It suffices to prove $f(x'_0) \neq 0$. Set $f(x'_0) = x_0$.

Let $x_1, \dots, x_n$ be generators for $M$. Then $x'_0=\sum_{i=1}^na'_ix_i$ and $x_0=\sum_{i=1}^na_ix_i$.

Let $x'_i\in N$ such that $f(x'_i) = x_i$ and write $x'_i = \sum_{j = 1}^{n} a_{ij} x_j$.

Let $A' = \mathbb{Z}[a_{ij}, a_{i}, a'_i]$. $A'$ is a Noetherian subring of $A$.

Let $N' = A'x'_0 + A'x'_1 + \cdots + A'x'_n$, $M' = A'x_1 + \cdots + A'x_n$, and $f':N'\to M'$ defined in an obvious way.

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    $\begingroup$ Looks good! Not the kind of proof I was looking for, but it fixes the gap in Orzech's. $\endgroup$ Commented Dec 13, 2014 at 15:33
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user26857's answer shows how to repair the reduction to the Noetherian case. Here is how to repair the proof of the Noetherian case:

Let $M$ be a noetherian $A$-module and let $N \subseteq M$ be a submodule. Let $f : N \to M$ be a surjective linear map. Then $f$ is injective.

Proof: Let $n \geq 0$. Although $f^n$ is not a well-defined homomorphism this is a partial homomorphism, defined on the submodule $D_n = N \cap f^{-1}(N) \cap f^{-1}(f^{-1}(N)) \cap \dotsc \cap f^{-1}(f^{-1}(\dotsc))$ (intersection of $n$ submodules). Let $K_n \subseteq D_n$ be the kernel of $f^n$. Then $K_n \subseteq K_{n+1}$ for all $n$. Since $M$ is noetherian, there is some $n$ such that $K_n = K_{n+1}$. Let $x \in N$ with $f(x)=0$. Since $f$ is surjective, the same is true for $f^n : D_n \to M$. Choose some $y \in D_n$ with $x = f^n(y)$. Then $y \in D_{n+1}$ and actually $f^{n+1}(y)=0$, so that $x=f^n(y)=0$.

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I have now found A constructive proof of Orzech's theorem, which relies on the Cayley-Hamilton theorem.

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    $\begingroup$ At the time Orzech proved his result Corollary 0.2 wasn't a "well-known" fact. (The references of the paper can help understanding why I'm saying this.) Moreover, Orzech's result has not been "mostly forgotten". It has some important consequences for free modules which were used freely (sic!), and in some sense the result became folklore. $\endgroup$ Commented Jan 31, 2015 at 3:34
  • $\begingroup$ @user26857: thanks; I have cut down on the historical remarks in the pdf, as I never seem to get those right. What I was trying to say is that the general $N \subseteq M$ case has rarely appeared in literature (when I tried to search for references in order to answer this very question, I found nothing that used the result in full generality). Usually, it is the $N = M$ case (and its consequences) that is being used, and that is by now well-known. $\endgroup$ Commented Jan 31, 2015 at 4:01
  • $\begingroup$ If this matters, I've used Orzech's result at least once on M.SE; see this answer. (That deleted account was mine.) $\endgroup$ Commented Jan 31, 2015 at 10:43
  • $\begingroup$ FYI: part of your proof was used (by me) to settle a similar question right here on MSE. I tried to give due credit, even though you released most of your writings (including this one) to the public domain. :-) In any case, I learned something new, so thanks for that! $\endgroup$ Commented Oct 2, 2017 at 8:29
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I don't have enough reputation points to add a comment. The paper “Épimorphismes de modules qui sont nécessairement des isomorphismes”, by Paulo Ribenboim (Séminaire Dubreil. Algèbre et théorie des nombres (1970-1971), Volume: 24, Issue: 2, page 1-5) studies this question and contains the proofs of the following results. Let's say a ring $R$ is a orzechian (Ribenboim says a $\Pi$-ring) if, for any finitely generated $R$-module $M$, any surjective homomorphism $u\colon N\to M$ from a submodule $N$ of $M$ is injective.

  • Left-noetherian rings are orzechian (there is a proof here)
  • Ribenboim proves a theorem of Djokovic according to which direct limits of orzechian rings are orzechian
  • Ribenboim then deduces that commutative rings are orzechian (as a direct limit of finitely generated subrings, which are noetherian by Hilbert's theorem).

Ribenboim notes the interesting fact: if $R$ is orzechian, $M$ is a free $R$-module generated by $m$ elements and $N$ is a free submodule of $M$ which is generated by $n$ elements, then $n\leq m$.

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Here's a direct proof using Cayley-Hamilton.

Let $M=\langle m_1,\ldots,m_k\rangle$ with $m_i=f(n_i)$ and $n_i=\sum_ja_{ij}m_j$. Set $A=(a_{ij})$, so that $\vec{n}=A\vec{m}$.

By the Cayley-Hamilton theorem, $A$ satisfies a polynomial equation $$A^k+c_{k-1}A^{k-1}+\cdots+c_2A^2+c_1A+c_0I=0.$$ Plugging in $\vec{m}$ gives $$A^k\vec{m}+c_{k-1}A^{k-1}\vec{m}+\cdots+c_2A^2\vec{m}+c_1A\vec{m}+c_0\vec{m}=0.$$ Substituting $A\vec{m}=\vec{n}$ gives $$A^{k-1}\vec{n}+c_{k-1}A^{k-2}\vec{n}+\cdots+c_2A\vec{n}+c_1\vec{n}+c_0\vec{m}=0.$$ Applying $f$ component-wise gives $$A^{k-1}\vec{m}+c_{k-1}A^{k-2}\vec{m}+\cdots+c_2A\vec{m}+c_1\vec{m}+f(c_0\vec{m})=0.$$ Substituting $A\vec{m}=\vec{n}$ gives $$A^{k-2}\vec{n}+c_{k-1}A^{k-3}\vec{n}+\cdots+c_2\vec{n}+c_1\vec{m}+f(c_0\vec{m})=0.$$ Applying $f$ component-wise gives $$A^{k-2}\vec{m}+c_{k-1}A^{k-3}\vec{m}+\cdots+c_2\vec{m}+f(c_1\vec{m}+f(c_0\vec{m}))=0.$$ Iterating this will eventually give the identity $$\vec{m}+f(c_{k-1}\vec{m}+f(c_{k-2}\vec{m}+f(\cdots))=0.$$ By taking linear combinations, this same identity holds for all $x\in M$, so $$x+f(c_{k-1}x+f(c_{k-2}x+f(\cdots))=0.$$ Plugging in $x\in\ker f$ forces $x=0$. Thus, $f$ is injective.

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