I am looking for a topology $\mathcal{T}$ on $\mathbb{Z}$, such that
- $(\mathbb{Z},+)$ is an abelian topological group,
- $(\mathbb{Z},\mathcal{T})$ is Hausdorff,
- there is no proper ideal $I:=n\mathbb{Z}, 1<n\in\mathbb{N}$, such that $I$ is open in $\mathcal{T}$,
or, alternatively, for a proof there is none. It would be nice for this topology to be locally compact, but not necessary.
I am perfectly aware that such a topology would not be "sensible" in a number of ways, eg we'd have a non-continuous quotient map $\pi_n: \mathbb{Z}\to\mathbb{Z}_n$ for some $n$, where $\mathbb{Z}_n$ is equipped with the discrete topology, or, to phrase it differently, such that the quotient topology on $\mathbb{Z}_n$ does not agree with continuous group operations. But I am not looking for a reasonable topology, but any topology at all.
For any topology satisfying the first two conditions, I know that one can construct open sets $U$ containing an infinite arithmetic sequence $a_n=kn$ for $k\neq 1$, where all elements $x\in U$ with $|x|<l$ are within that sequence for arbitrarily large - but finite - $l$. However, I haven't managed the final step in showing there must also be an open set containing only this sequence.
I also tried something along the lines of taking some symmetrical neighbourhood $U$ of $0$ not containing $1$, and then choosing $V$ such that $V+V\subseteq U$ - if $V$ were to contain the minimal nonzero $a\in U$, then this would directly lead to an open ideal, but I can't see a way to force this.
I am starting to think that I might have overlooked something really, really basic here - or that the problem is a lot more complicated than I originally thought.