Question

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?

Answer 1

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

Answer 2

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 $\langle Y, Q'\rangle$.

Answer 3

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!