Ascending chain condition on ideals of free products - MathOverflow

Ascending chain condition on ideals of free products

2012-12-06T05:36:59Z

In my previous question: M Shahryari (mathoverflow.net/users/29488), Normal Subgroups of Free Products, http://mathoverflow.net/questions/114801 (version: 2012-11-28), I asked if a group $A$ has max-n property, is it true that the free product $A\ast \mathbb{Z}$ has also max-n? The answer was NO in that case. Now suppose $F$ is free group of finite rank and $A$ is a group having max-n (maximal condition on normal subgroups). A normal subgroup $N$ of $A\ast F$ is called an ideal if $N\cap A=1$. Is it true that $A\ast F$ has maximal property of ideals?

Answer by HW for Ascending chain condition on ideals of free products

2012-12-06T13:53:55Z

The answer to your question is 'no'. Consider any homomorphism $f:A*F\to G$ which is injective on $A$. Then $\ker f$ is an ideal in your sense (and this is necessary and sufficient). An infinite increasing chain of ideals is therefore equivalent to an infinite sequence of surjections

$A*F=G_0\to G_1\to G_2\to\ldots $

so that $A$ embeds into $G_n$ for all $n$. For a specific example, take $F=\langle b_1,b_2\rangle$ and construct $G_n$ from $G_{n-1}$ by adding a long small-cancellation relator in $b_1,b_2$.

Small-cancellation theory

At the OP's request, here are some references for the last sentence; they are all from Lyndon and Schupp's book Combinatorial group theory.

Take $r_1,r_2,\ldots$ to be an infinite sequence of elements of $F(b_1,b_2)$ such that, for every $n$, $R_n={r_1,\ldots,r_n}$ satisfies condition $C'(1/6)$ (as defined on p. 240 of Lyndon and Schupp). It's a nice exercise to confirm that such sequences exist.

For each $n$, take $G_n=A*\langle b_1,b_2\mid R_n\rangle$. It's easy to check that there are infinitely many elements $g_i$ so that, for all distinct $i,j$, $g_ig_j^{-1}$ is $R$-reduced in the sense of p. 251 of Lyndon ad Schupp. Therefore, by Dehn's algorithm, $G_n$ is infinite for all $n$, as claimed.

Equational Noetherian groups

In response to a remark of the OP's in the comments, I want to point out that it does not follow that $G_0$, or any $G_n$, is not equationally Noetherian. Indeed, the $G_n$ constructed above are all word-hyperbolic and hence equationally Noetherian by a theorem of Sela. (Alternatively, for $C'(\lambda)$ for small enough $\lambda$, they are all linear by the work of Wise and friends, and hence are equationally Noetherian by Hilbert's Basis Theorem.)

To prove that $G_0$ (say) is not equationally Noetherian, you need an infinite sequence of proper epimorphisms

$L_0\to L_1\to L_2\to\cdots$

where each $L_n$ is residually $G_0$.

For more details, I suggest you look at the beginning of Bestvina and Feighn's paper Notes on Sela's work (here).

Answer by Anton Klyachko for Ascending chain condition on ideals of free products

2012-12-11T15:14:39Z

You do not need to know small-cancellation thery. Take your favorite finitely generated but non-finitely presented group $$ \langle x_1,\dots,x_n\;|\;w_1=1,w_2=1,\dots\rangle. $$ Then $A*F(x_1,\dots,x_n)$ has an ascending chain of `ideals' $$ \langle\langle w_1\rangle\rangle \subset \langle\langle w_1,w_2\rangle\rangle \subset \dots. $$ Here, $\langle\langle\dots\rangle\rangle$ means the normal closure in $A*F(x_1,\dots,x_n)$.

The group $A$ plays no role here.