# Formally étale maps and lifting along reductions

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).

• You want to take $A$ noetherian here. – KReiser Oct 10 at 0:54
• @KReiser oh I see, that is what gives me the desired maximal element. – Dean Young Oct 10 at 0:55
• @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 $\text{Spec}(A / \cup I_n) \cong \text{limit} \text{Spec}(A/I_n)$? – Dean Young Oct 10 at 1:16
• 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. – KReiser Oct 10 at 5:07