Precisely, if an R-module M has a finite presentation, and Rk → M is some unrelated surjection (k finite), is the kernel necessarily also finitely generated?

Basically I want to believe I can choose generators for M however I please, and still get a finite presentation. I have reasons from algebraic geometry to believe this, but it seems like a very basic result, so I would like to understand it directly in terms of the commutative algebra, which I just can't seem to figure out...

(Here R is an arbitrary commutative ring, with no other hypotheses.)

Edit: All maps here are maps of R-modules. Also, the reason this is not the same as "does finite presentation imply coherent?" is that I am only asking for finite type kernels of surjections Rk → M. That the hypotheses assume surjectivity is a common misreading of the general definition of "coherent".

If the answer to the above is "yes", then coherent will mean "finite type, and all finite type submodules are finite presentation"


$\require{begingroup} \begingroup$ $\def\coker{\operatorname{Coker}}$ $\def\im{\operatorname{Im}}$

Suppose that we have a short exact sequence $0 \to K \to R^m \to M \to 0$ with $K$ finitely generated over $R$ and that $0 \to K' \to R^n \to M \to 0$ is another short exact sequence. Your question is: is $K'$ necessarily finitely generated?

The answer is yes and we can see this as follows:

First, we argue for the existence of a commutative diagram

$$ \require{AMScd} \begin{CD} 0 @>>> K @>>> R^m @>>> M @>>> 0 \\ @. @VV{\tilde{f}}V @VV{f}V @| \\ 0 @>>> K' @>>> R^n @>>> M @>>> 0 \\ \end{CD} $$

Using the fact that free modules are projective we can lift the identity map $M = M$ to an $f\colon R^m\to R^n$ which makes the right hand square commute. Restricting $f$ to a map $\tilde{f}\colon K → K'$ fills in the last square and so we have the diagram as claimed.

Now using Snake's lemma we find that there is an isomorphism $\coker{\tilde{f}} \cong \coker{f}$. Thus We have a short exact sequence; $$ 0\to \im{\tilde{f}}\to K'\to \coker{f}\to 0. $$ Since $K'$ is squeezed between two finitely generated $R$ modules, it follows (by a well-known-fact) that $K'$ is itself finitely generated.


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    $\begingroup$ Excellent! This is exactly what I've been looking for! $\endgroup$ Oct 22 '09 at 6:21
  • $\begingroup$ Nice answer! Regarding commutative diagrams: I've added a couple of examples of how to do commutative diagrams on MO (mathoverflow.net/questions/16/…). It's not a perfect solution, but you can get your point across. I think your "imagine the vertical arrows" works fine for this answer. $\endgroup$ Oct 22 '09 at 20:09
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    $\begingroup$ Part of me wants to correct "Snake's lemma" to "the Snake lemma". But another part of me wants there to have been someone called Snake, who had a lemma. $\endgroup$ Oct 10 '18 at 17:00
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    $\begingroup$ @ZachTeitler A mathematical Lieutenant Kijé. $\endgroup$ Oct 10 '18 at 20:31
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    $\begingroup$ @ZachTeitler it is hard for me to express how strenuously I would object to this "correction" $\endgroup$
    – Joel Dodge
    Dec 18 '18 at 20:45

There's a famous quote, I think due to Szego, that a technique which can be used once is a trick, but if you can use it twice then it is a method. In that spirit, here is the EGA method which is very very useful to kill all such problems by reducing to the noetherian case. One can easily extrapolate from this example how one might prove related results (such as for finitely presented algebras, in which case the module-theoretic arguments in the other answers may not apply so easily).

General Principle: If everything used can be "described" with only finitely many elements of the base ring, the objects all descend to a noetherian subring (such as $\mathbf{Z}$-subalgebra generated by the finitely many elements used in the "description") and if we increase it a bit the morphisms also descend (need to kill off some relations), and if we increase it some more we can also descend all "reasonable" properties (flatness, smoothness, radiciel, surjectivity, etc.). This last step is by far the most subtle (for things like flatness). In the end, it always comes down to the fact that if something vanishes in a direct limit then it vanishes somewhere always the way, and if a direct limit of rings is 0 then the rings eventually vanish (track whether or not $1 = 0$).

Worked example for finite presentation of modules:

Step 0: Choose some right exact sequence $F' \stackrel{f}{\rightarrow} F \rightarrow M \rightarrow 0$ with $F$ and $F'$ finite free over $R$, and another surjection $\pi:P \twoheadrightarrow M$ from another finite free module to $M$. We wish to prove $\ker \pi$ is finitely generated.

Step 1: Observe that the map $f$ involves only finitely many elements of $R$ (think of a matrix), and likewise each basis vector in $P$ goes to an element of $M$ which lifts to something in $F$, so again that only involves finitely many elements of $R$ (to describe these lifts in $F$). Let $R_0 \subset R$ be the $\mathbf{Z}$-subalgebra generated by the elements of $R$ we just mentioned.

Step 2: Consider the $R_0$-linear map $$f _0 : F' _0 \rightarrow F _0$$

between finite free $R_0$-modules given by "the same matrix" as for $f$, so $R \otimes_{R_0} f_0 = f$. Let $M_0 = {\rm{coker}}(f_0)$, so by right-exactness (!) of tensor product, $M_0$ is an $R_0$-descent of $M$. Now $f$ has done its work and we forget about it.

Step 3: We can likewise define a map $\pi_0: P_0 \rightarrow M_0$ from a finite free $R_0$-module so that scalar extension to $R$ is $\pi$. If $\pi_0$ were surjective, then right-exactness of tensor product would imply that ${\rm{ker}}(\pi)$ is a quotient of $R \otimes_{R_0} {\rm{ker}}(\pi_0)$, the latter being finitely generated since $R_0$ is noetherian. Is $\pi_0$ surjective? Maybe not. But no big deal: if it becomes surjective after scalar extension to a bigger noetherian subring of $R$ then we can rename that as $R_0$ and proceed as above.

The issue is whether ${\rm{coker}}(\pi_0)$ vanishes. By right exactness (!) of tensor product, formation of this cokernel commutes with scalar extension to intermediate rings between $R_0$ and $R$. Scalar extension all the way to $R$ makes it vanish (since $\pi$ is surjective), so by expressing $R$ as a direct limit of finitely generated $R_0$-subalgebras $R_i$ we conclude that each of the finitely many generators of ${\rm{coker}}(\pi_0)$ have vanishing image after scalar extension to some common such $R_i$. Rename that as $R_0$.


See EGA IV$_1$, 1.4.4 for the variant for finitely presented algebras. Well, better to first work it out for yourself. And then prove a module-finite algebra over a ring is finitely presented as a module if and only if it is finitely presented as an algebra. (This is not a tautology.) This may be a bit trickier to figure out, but good exercise. Solution in EGA IV$_1$, 1.4.7.

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    $\begingroup$ The quote referred to at the start is due to Polya and Szego. In volume 1 page VIII of their Problems and Propositions in Analysis, they write "An idea which can be used only once is a trick. If one can use it more than once it becomes a method." $\endgroup$
    – KConrad
    Feb 26 '10 at 0:00
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    $\begingroup$ Very instructive, 1+. $\endgroup$ Jan 14 '11 at 7:56

This is more or less the same as Andy's answer, but I'll say it differently. Suppose 0-->A-->Rp-->M-->0 and 0-->B-->Rq-->M-->0 are two exact sequences. Then we can form an exact sequence


where the thing in the middle is a direct sum, and the map to M is the sum of the two surjections. Then you can show there are isomorphisms K <--> A+Rq and K <--> B+Rp; these get produced using lifts Rp--> Rq and Rq--> Rp of the maps to M. Then A is finitely generated if and only if B is.

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    $\begingroup$ This is Schanuel's lemma, no? and I think it answers the original question, but maybe I'm missing something $\endgroup$
    – Yemon Choi
    Oct 22 '09 at 9:20
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    $\begingroup$ Yes, it is Schanuel's lemma (though I didn't know it was called that). And it still looks like a good answer to me! $\endgroup$ Oct 22 '09 at 14:06
  • $\begingroup$ Yet another way to express this can be found in Chapter IV, comment after Lemma 1.0 page 180 of arxiv.org/pdf/1605.04832.pdf for algebraic structures defined by generators and relations. $\endgroup$ Jan 31 at 10:50

Since it has not been mentioned yet here (in this question- certainly it appears somewhere on MathOF), let me mentioned that such a phenomenon holds in a considerably larger setting.

An object $K$ of a category $\mathcal{K}$ is said to be finitely presentable if its hom-functor $\mathrm{Hom}_{\mathcal{K}}(K,-):\mathcal{K}\to\mathbf{Set}$ preserves directed colimits.

In a variety of finitary algebras (in the sense of universal algebra), so it can be "groups", "monoids", "$R$-modules", etc., an algebra $A$ is finitely presentable iff it can be presented by finitely many generators and relations; in addition the definition implies that for such finitely presentable $A$,

for every quotient map $p:B\to A$ with $B$ finitely generated, the set of relations $\{(x,y)\in B^2:p(x)=p(y)\}$ is a finitely generated compatible equivalence relation on $B$.

(compatible meaning that the laws passes to the quotient; in groups or modules, this says that the kernel is finitely generated as normal subgroup, resp., as submodule; in monoids, this can't be described in terms of kernel).

All this can be found in Adamek-Rosicky's book, Section 1.A.

  • $\begingroup$ This is intriguing! Could I trouble you for a more precise reference? I'm looking through Adamek-Rosicky for this result and having trouble finding it in either in Ch. 1.A (which doesn't have much on algebraic categories per se) or in Ch. 3. As your answer suggests, it's a bit delicate to get a correct statement in this generality. E.g. I'd believe there exist f.p. groups $G$ where you can choose a "bad" finite generating set, whose relations are not f.g. as a group (though according to your answer, they must be f.g. as a normal subgroup of the free group on the generators). $\endgroup$
    – Tim Campion
    Jan 30 at 23:33
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    $\begingroup$ Er -- maybe I've got it. Let $\mathcal K$ be a variety and let $K \in \mathcal K$ be finitely-presentable. Let $F \twoheadrightarrow K$ be a regular epi from a finitely-generated free object. Let $C$ be the the kernel pair, so that we have a coequalizer $C \rightrightarrows F \to K$, and let $F' \twoheadrightarrow C$ be a regular epi from a free object, which we don't know is f.g. Then $F'$ is the filtered colimit of its f.g. free subobjects $F'_i$, so that $K$ is the filtered colimit of the coequalizers of $F'_i \rightrightarrows F$. $\endgroup$
    – Tim Campion
    Jan 30 at 23:50
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    $\begingroup$ By finite-presentability, $K$ is a retract of one of these coequalizers. So $K$ is a retract of the quotient of $F$ by a f.g. congruence. Just an $\epsilon$ to go to eliminate the retract. $\endgroup$
    – Tim Campion
    Jan 30 at 23:59
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    $\begingroup$ Okay -- I believe we can eliminate the retract as follows. First, convince yourself that because $K \in \mathcal K$ is f.p., it follows that $F \twoheadrightarrow K \in \mathcal K_{F/}$ is f.p., and that $\mathcal K_{F/} \to \mathcal K$ preserves filtered colimits. Denote the above coequalizers by $F_i' \rightrightarrows F \twoheadrightarrow K_i$. We have $K = \varinjlim_i K_i$ in the coslice category $\mathcal K_{F/}$, so we conclude that $K$ is a retract of $K_i$ in the coslice category. But because $F \twoheadrightarrow K_i$ is epi, the split mono we get is also epi and hence iso $\endgroup$
    – Tim Campion
    Jan 31 at 0:24

Here's a further generalization of the generalization from YCor's answer.

Claim: Let $\mathcal K$ be a locally $\kappa$-presentable category containing a coequalizer diagram $A \rightrightarrows B \twoheadrightarrow C$ where $B,C$ are $\kappa$-presentable. Then there exists a coequalizer diagram $A' \rightrightarrows B \twoheadrightarrow C$ with $A'$ $\kappa$-presentable and a map $A' \to A$ making the obvious diagram commute.

It follows, for example, that if $B$ is any finitely-presentable module and $C$ is any finitely-presentable quotient, then the kernel is finitely-generated (we need not assume that $B$ is free). The result will still be most useful in "algebraic" categories $\mathcal K$ (the context YCor described already discussed) where there is a good supply of coequalizer diagrams. Nevertheless, it's true in this more general context.

Proof: Write $A = \varinjlim A_i$ as the $\kappa$-filtered colimit of $\kappa$-presentable objects. We have resulting coequalizer diagrams $A_i \rightrightarrows B \twoheadrightarrow C_i$, and $C = \varinjlim_i C_i$. In fact, the isomorphism $C = \varinjlim_i C_i$ lifts to the coslice category $\mathcal K_{B/}$. Because $C$ is finitely-presentable (even regarded as an object of $\mathcal K_{B/}$), the identity map $C = C$ factors through some stage of the colimit, $C \rightarrowtail C_i \twoheadrightarrow C$ This is true in the coslice category, so that $(B \twoheadrightarrow C_i) = (B \twoheadrightarrow C \rightarrowtail C_i)$. Because this map is epi, it follows that $C \rightarrowtail C_i$ is epi. But it is also split mono, and hence iso. Thus $C = C_i$ is the coequalizer of $A_i \rightrightarrows B$.


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