In my previous question: M Shahryari (mathoverflow.net/users/29488), Normal Subgroups of Free Products, Normal Subgroups of Free Products (version: 20121128), I asked if a group $A$ has maxn property, is it true that the free product $A\ast \mathbb{Z}$ has also maxn? The answer was NO in that case. Now suppose $F$ is free group of finite rank and $A$ is a group having maxn (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?

$\begingroup$ 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/… . $\endgroup$ – HJRW Dec 6 '12 at 14:15

$\begingroup$ In particular, you might be interested in Denis Osin's answer to that question. $\endgroup$ – HJRW Dec 6 '12 at 14:16

$\begingroup$ @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 maxn 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. $\endgroup$ – M. Shahryari Dec 6 '12 at 20:15

$\begingroup$ I hope the answer for my question will be negative. $\endgroup$ – M. Shahryari Dec 6 '12 at 20:19

$\begingroup$ @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. $\endgroup$ – HJRW Dec 8 '12 at 5:07
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 $G_{n1}$ by adding a long smallcancellation relator in $b_1,b_2$.
Smallcancellation 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 wordhyperbolic 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).

$\begingroup$ I just noticed that this is similar to a remark that Ashot Minasyan made on your previous question. $\endgroup$ – HJRW Dec 6 '12 at 14:00

$\begingroup$ @HW: I have few knowledge of small cancellation theory ( in the level of last chapter of LyndonSchupp). So, may I ask you to write some details? $\endgroup$ – M. Shahryari Dec 6 '12 at 20:24

$\begingroup$ I'll write some details when I'm next in the office and have access to Lyndon and Schuppprobably on Monday. $\endgroup$ – HJRW Dec 8 '12 at 20:40

$\begingroup$ @HW: Thank you so much, now I understand your argument. I will send my results to your email in next days. $\endgroup$ – M. Shahryari Dec 11 '12 at 20:51
You do not need to know smallcancellation thery. Take your favorite finitely generated but nonfinitely 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.

$\begingroup$ My favourite finitelygeneratedbutnotfinitelypresented group is provided by smallcancellation theory. :) $\endgroup$ – HJRW Dec 11 '12 at 15:51

1$\begingroup$ And my is an amalgamated free product of two free groups. $\endgroup$ – Anton Klyachko Dec 11 '12 at 15:58

1$\begingroup$ And mine is the wreath product of $\mathbb Z$ with $\mathbb Z$. ;) $\endgroup$ – Ashot Minasyan Dec 12 '12 at 18:56