Let $f: X \rightarrow Y$ be a formally étale map of schemes. By definition, $f$ lifts against maps $\text{Spec}(A/I) \rightarrow \text{Spec}(A)$, where $I$ is nilpotent.
When does $f$ lift against maps $\text{Spec}(A_{red}) \rightarrow \text{Spec}(A)$, where $A_{red}$ is the reduction of a commutative ring $A$? That is, let $A$ be a commutative ring, $I$ the nilradical of $A$, and $A_{red} = A/I$. When is it true that we have the following lifting property for $f$:
for each such commutative ring $A$?
Is the following argument correct?
Let $S = \{ I \subset A : I \text{ an ideal }, I \subset \text{rad}(A) \}$. Let $T \subset S$ be the set of elements in $S$ such that there exists a lift as above; order $T$ by inclusion, and take a maximal element in it, $I$. If $I \neq \text{rad}(A)$, take $a \in \text{rad}(A) - I$. Then $a^n \in I$, so $J/I = (\langle a \rangle + I)/J$ is a nilpotent ideal in $A/I$, so we have a lifting:
Therefore $J \in T$ and $I \subsetneq J$, so that $I$ was not maximal, a contradiction.
But why can I choose a maximal element in $T$? I need to be able to apply Zorn's lemma.
Edit: If $\text{Spec}(A / \cup I_n) \cong \text{limit} \text{Spec}(A/I_n)$ then we are done because then Zorn's lemma applies. I think this follows from lemma 2.1 here. It comes down to $\text{Spec}(\text{colim} A_i ) \cong \text{lim} \text{Spec}( A_i)$ when the colimit is filtered (so that the limit is cofiltered).