Here is a problem that I have only solved partially. I am (edit: "was") stuck at the second part.
Problem: Given that two posets $(\mathcal{P}_1,\leq_1)$ and $(\mathcal{P}_2,\leq_2)$ are isomorphic if there is a bijection $\psi : \mathcal{P}_1\to \mathcal{P}_2$ such that $x\leq_1 y$ $\Leftrightarrow$ $\psi(x)\leq_2 \psi(y)$.
(i) Are the posets of ideals, under inclusion, in $\mathbb{Z}$ isomorphic to the posets of ideals, under inclusion, in $\mathbb{Q}[X]$??
(ii) Are the posets of ideals, under inclusion, in $\mathbb{Q}[X]$ isomorphic to the posets of ideals, under inclusion, in $\mathbb{Q}[X,Y]$??
My partial solution: I have only been able to solve (i) which was pretty natural. I think (ii) is the more tricky part. For (i) it is sufficient to get a bijection between maximal (prime) ideals. Explicitly, a map $\psi : \mathcal{I}(\mathbb{Z})\to \mathcal{I}(\mathbb{Q}[X])$ ($\mathcal{I}(R)$ denotes the poset, under inclusion, of ideals of a unital ring $R$) that bijectively maps prime ideals $(p)$ to $(f_p(X))$ where $f_p(X)$ is an irreducible polynomial in $\mathbb{Z}[X]$ and an ideal $(p_1^{a_1}\cdots p_k^{a_k})$ to $(f_{p_1}(X)^{a_1}\cdots f_{p_k}(X)^{a_k})$, should obviously work. For (ii) I don't see how to modify this argument (for PIDs) in case of UFDs.
(Edit : The second one was not that difficult; it turns out that $(x)\subset (x,y^j)$ for all $j\geq 1$, but any ideal in $Q[X]$ say of the form $(f(X))$ has only finitely many ideals containing it, indeed if $f=f_1^{a_1}\cdots f_k^{a_k}$ (decomposing into primes/irreducibles) the ideal $(f(X))$ is contained in at most (rather exactly) $(a_1+1)\cdots (a_k+1)$ many ideals. Hence there can't be a bijection at all.)
Thanks in advance for any hints or solutions. Now that it turns out that it was kind of trivial I might vote to close.