# Does “finitely presented” mean “always finitely presented”? (Answered: Yes!)

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"

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Thanks to all for removing incorrect answers :) – Andrew Critch Oct 22 '09 at 5:58

Lack of diagrams does make this difficult but let's give it a shot.

Suppose that we have a short exact sequence 0 → K → Rm → M → 0 with K finitely generated over R and that 0 → K' → Rn → M → 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:

Write the sequences like

0 → K → Rm → M → 0

0 → K' → Rn → M → 0

and imagine that I've drawn in the identity map from M to itself. Using the fact that free modules are projective we get a vertical map in the middle Rm → Rn making the right hand square commute. We can fill in the last square by restricting this to a map K → K'.

Now using Snake's lemma we find that there is an isomorphism K'/image(K) = Rn / image(Rm). We have squeezed K' between two finitely generated R modules and it follows that it is itself finitely generated.

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Excellent! This is exactly what I've been looking for! – Andrew Critch Oct 22 '09 at 6:21
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. – Anton Geraschenko Oct 22 '09 at 20:09
@Anton: I tried clicking on your link but it's telling me 404. – Yosemite Sam Feb 3 '12 at 22:46

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.

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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." – KConrad Feb 26 '10 at 0:00
Very instructive, 1+. – Martin Brandenburg 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

0-->K-->Rp+Rq-->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+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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This is Schanuel's lemma, no? and I think it answers the original question, but maybe I'm missing something – Yemon Choi Oct 22 '09 at 9:20
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! – Charles Rezk Oct 22 '09 at 14:06