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In my previous question: M Shahryari (mathoverflow.net/users/29488), Normal Subgroups of Free Products, Normal Subgroups of Free Products (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?

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Looking at the comments on your previous question, I see that you are interested in groups which are not equationally Noetherian. You might be interested in this question of mine: mathoverflow.net/questions/75784/… . –  HJRW Dec 6 '12 at 14:15
    
In particular, you might be interested in Denis Osin's answer to that question. –  HJRW Dec 6 '12 at 14:16
    
@HW:Thank you for comment. You are right, I'm interested in versions of Hibert's basis theorem for groups. So, I'm trying to recognize groups $A$ having max-n such that A\ast F$ has maximal property on ideals. I need time to read your question and and D. Osin's answer. Thank you again. –  M. Shahryari Dec 6 '12 at 20:15
    
I hope the answer for my question will be negative. –  M. Shahryari Dec 6 '12 at 20:19
    
@M Shahryari, my answer worked for any group $A$, so no group $A$ has the property you want. However, are you just asking which groups are equationally Noetherian? I could write you a list of the current state of knowledge, if it would help. –  HJRW Dec 8 '12 at 5:07
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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).

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I just noticed that this is similar to a remark that Ashot Minasyan made on your previous question. –  HJRW Dec 6 '12 at 14:00
    
@HW: I have few knowledge of small cancellation theory ( in the level of last chapter of Lyndon-Schupp). So, may I ask you to write some details? –  M. Shahryari Dec 6 '12 at 20:24
    
I'll write some details when I'm next in the office and have access to Lyndon and Schupp---probably on Monday. –  HJRW Dec 8 '12 at 20:40
    
@HW: Thank you so much, now I understand your argument. I will send my results to your email in next days. –  M. Shahryari Dec 11 '12 at 20:51
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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.

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My favourite finitely-generated-but-not-finitely-presented group is provided by small-cancellation theory. :) –  HJRW Dec 11 '12 at 15:51
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And my is an amalgamated free product of two free groups. –  Anton Klyachko Dec 11 '12 at 15:58
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And mine is the wreath product of $\mathbb Z$ with $\mathbb Z$. ;-) –  Ashot Minasyan Dec 12 '12 at 18:56
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