Zorn's Lemma states that any poset with the property that every chain has an upper has at least one maximal element. Are there necessary or sufficient conditions on the poset for the maximal element to be provably unique?
I'm thinking in the context of a poset within a larger structure. For example, in the case of ideals ordered by set containment, are there any conditions on the ideal $I_0$ such that $I_0 \subset I_1, I_2 \implies I_1 = I_2$, where $I_1$ and $I_2$ are maximal ideals? Are there concrete examples of posets where this is known to be true?
It's hard to know what exactly counts as an answer to this question - of course there's there trivial sufficient suggested in the comments, namely "the poset has a greatest element". But you want a useful condition, along the same lines as the usual statement of Zorn's Lemma, that you could apply to a poset whose structure you understand "locally" but not "globally".
Here's a suggestion of such a condition:
The proof is easy. If $P$ has a greatest element $m$, then $m$ is an upper bound for every chain and every pair of elements. Conversely, if the two conditions above are satisfied, then by (1) and Zorn's Lemma, $P$ has at least one maximal element. Now suppose $x\in P$ is any element. By (2), there exists $n\in P$ such that $m\leq n$ and $x\leq n$. But since $m$ is maximal, $m = n$, so $x\leq m$. Thus $m$ is the greatest element of $P$.
So for example: Looking at the poset of proper ideals in a nonzero commutative ring $R$, you can check (1) and apply Zorn's Lemma to get a maximal ideal in $R$. If additionally you can prove (2), then you would get that $R$ has a unique maximal ideal ($R$ is a local ring).
In this case, (2) amounts to showing that if $I$ and $J$ are proper ideals, then the ideal $(I+J)$ generated by $I$ and $J$ is proper. Equivalently, if $1_R = i+j$ for any two elements $i,j\in R$ then either $i$ or $j$ is a unit - this is a well-known characterization of local rings.