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



Ascending chain condition on ideals of free products
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 $b\_1,b\_2$.




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**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.




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**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](http://arxiv.org/abs/0809.0467)).


