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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 $f$: enter image description here 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: enter image description here 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).

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  • $\begingroup$ You want to take $A$ noetherian here. $\endgroup$ – KReiser Oct 10 at 0:54
  • $\begingroup$ @KReiser oh I see, that is what gives me the desired maximal element. $\endgroup$ – Dean Young Oct 10 at 0:55
  • $\begingroup$ @KReiser I see how it let's us lift against étale maps, but is it necessary or just sufficient to require? If $X \rightarrow Y$ lifts against $\text{Spec}(A/I_n) \rightarrow \text{Spec}(A)$ for $I_1 \subset I_2 \subset I_3 \subset \cdots \subset \text{rad}(A)$, then shouldn't it automatically lift against $\text{Spec}(A/ \cup I_n) \rightarrow \text{Spec}(A)$? Isn't $\text{Spec}(A / \cup I_n) \cong \text{limit} \text{Spec}(A/I_n)$? $\endgroup$ – Dean Young Oct 10 at 1:16
  • $\begingroup$ First, when I said you want $A$ noetherian, it was to get the requested maximal element. I don't spend much time with formally smooth/etale morphisms and lifting against non-noetherian reductions, so I'm not totally familiar with this context and don't know the answer right off the top of my head. Next, as $\operatorname{Spec}: CRing^{op} \to Sch$ is a right adjoint, it preserves all limits. If the rest of your logic is correct, then what you want is true, but I'm not confident enough to say that you've produced a full solution here. $\endgroup$ – KReiser Oct 10 at 5:07

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