# Does “finitely presented” mean “always finitely presented”, considered in general

I'm wondering about the question, "If we have a finitely presented __, is it necessarily finitely presented with respect to any finite generating set for it?" I know this is true for groups and for R-modules. Does anyone know whether this is true for A-algebras? Commutative A-algebras? Other things people might happen to know about it for?

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BCnrd's answer to the question you link to says the statement is true for commutative A-algebras. –  Peter Samuelson Jan 14 '11 at 7:32
True for groups. –  Steve D Jan 14 '11 at 8:28
Ah, thanks, I missed that. –  Harry Altman Jan 14 '11 at 8:54

Yes in general.

See Adámek and Rosicky, Locally Presentable and Accessible Categories Cambridge University Press, Cambridge, (1994).

T. 3.12 p. 143.

Of course "in general" I mean: every "algebraic theory" (many sorted) Set models.

For topological algebraic structures, (like profinite groups) this equivalence isn't true for the Yiftach Barnea example

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It is not true in "general", but under some conditions, see my answer below. So you might like to elaborate a bit. –  Yiftach Barnea Jan 15 '11 at 16:47
Well, I still need to actually work through this, but it's general enough for me. :) Thank you! –  Harry Altman Feb 14 '11 at 3:58
I'll post here the idea of the proof, as it's pretty simple. It works by proving that (in an algebraic theory) an object A is finitely presented iff Hom(A,-) commutes with directed colimits. First, A is fingen iff Hom(A,-) commutes with directed colimits of monomorphisms; => is straightforward, and for <=, look at A as a colimit of its fingen subalgebras. Then you show finitely presented => the property above. Finally, given that property and a finite generating set, look at A as a colimit of the free algebra on that set modulo what's generated by finite subsets of kernel of projection to A. –  Harry Altman Feb 19 '11 at 7:50

It is true for any algebras (sets with operations) satisfying any set of identities (laws). Indeed, if $X, Y$ are finite generating sets $R=\{u_i,v_i\mid i\in I\}$ be a finite set of defining relations over $X$, $Q$ is any set of defining relations over $Y$, then for every relation from $R$ there exists a proof of that relation using relations from $Q$. The proof involves a finite number of elements of $Q$. If $Q'$ is the (finite) collection of all elements of $Q$ involved, then the algebra has a finite presentation $\langle Y, Q'\rangle$.

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I'm not too clear on how this works. In more detail: We pick a map from free algebra on X to that on Y so that things commute; now for each relation in R, we can map it to free alg on Y, prove that using finitely many relations in Q, call the set of those used Q'. Now say we have a relation in Q, need to show it can be proved from Q. So pick a map from free alg on Y to that on X, map it, prove it using the finitely many relations in R, map back... except our two maps need not be inverses, so while we get a proof from Q', it need not be of the same relation we started with. What am I missing? –  Harry Altman Jan 15 '11 at 1:48

This is not a definite answer, but I suspect that for profinite groups this is wrong. The reason is that there are finitely generated profinite groups which have minimal generating set which is infinite. For example if you look at $\Pi_{n \geq 5}A_n$, then it is $2$-generated as a profinite group, but it also has a minimal generateing set which is inifinite. The point is that profinite words could be infinite and may contain infinite number of distinct letters.

It would be nice to see an example.

EDIT: First, a simpler example of a profinite group which is cyclic, but has a minimal set of generators is $\hat{\mathbb{Z}}$. Now, look at the category of abelian profinite groups. We can look at $\hat{\mathbb{Z}}$ as a qoutient of $\hat{\mathbb{Z}}^2$ by $\hat{\mathbb{Z}}$. Then we found a presentation of $\hat{\mathbb{Z}}$ with two generators and in one case with one relation and in the other case with infinite number of relations.

Admittedly, this is not the minimal number of generators. Maybe this can be improved.

Final Answer (I hope): Okay, so I can even give an example with one generator. Take $G=\Pi_p \mathbb{Z}/ p \mathbb{Z}$, where $p$ goes over all the primes, in the category of profinite groups or the category of abelian profinite groups. $G \cong \hat{\mathbb{Z}}/N$, where $N=\Pi_{p} p \mathbb{Z}_p$, where $\mathbb{Z}_p$ are the $p$-adic integers. Now, $p\mathbb{Z}_p \cong \mathbb{Z}_p$. So, $N \cong \Pi_p \mathbb{Z}_p \cong \hat{\mathbb{Z}}$ and again $N$ is one generated but has a minimal generating set which is infinite.

I cannot add a comment now for some reason, so Richard thanks a lot!

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For some reason in the bottome couple of lines of my answer the math is not processed. Does anyone have an idea what the problem is? –  Yiftach Barnea Jan 15 '11 at 16:55
I fixed the TeX. I couldn't see what was wrong, but retyping the offending sentence seemed to do it. You should make sure it says what you wanted. –  Richard Kent Jan 15 '11 at 17:18