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?

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). 


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. 

