Does "finitely presented" mean "always finitely presented"? (Answered: Yes!)

Question

Precisely, if an R-module M has a finite presentation, and R^k → 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 R^k → 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"

Answer 1

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

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.

$\endgroup$

Answer 2

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 $R_0$.

QED

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.

Answer 3

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

0-->K-->R^p+R^q-->M-->0

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+R^q and K <--> B+R^p; these get produced using lifts R^p--> R^q and R^q--> R^p of the maps to M. Then A is finitely generated if and only if B is.

Answer 4

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.

Answer 5

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 $\kappa$-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 $A_i \rightrightarrows B$.