# Formally smooth maps of schemes and factorization systems

I am thinking about how formally smooth maps of schemes relate to factorization systems.

Let $$C$$ be the category of schemes. Let $$E$$ be the class of morphisms of schemes consisting of closed immersions $$X \rightarrow X_{th}$$, where $$X$$ is determined by a nilpotent sheaf of ideals on $$X_{th}$$.

Let $$M$$ be the class of morphisms in $$C$$ which lift against $$E$$. i.e., the maps $$f : Y \rightarrow Z$$ such that, for each $$g : X \rightarrow X_{th}$$, and for each commutative diagram as below, there exists a lift like below: These are called formally smooth morphisms.

Let $$\overline{E}$$ be the class of morphisms of schemes which lift against $$M$$, i.e., the maps $$g : X \rightarrow X_{th}$$ such that, for each map $$f : Y \rightarrow Z$$ in $$M$$, and for each commutative diagram as below, there exists a lift like below: My questions are:

Question: What is this class of maps $$\overline{E}$$? Obviously it contains the elements of $$E$$, but what other maps does it contain?

Question: Can we factor each map of schemes $$f : X \rightarrow Y$$ as the composition $$m \circ e$$ for $$m \in M$$ and $$e \in \overline{E}$$? That is, can we factor any map as a map in $$\overline{E}$$ followed by a formally smooth map?

• Formally smooth morphisms only have the property above with respect to nilpotent embeddings of affine schemes. Oct 8 '19 at 17:30
• @Angelo I believe that is equivalent; see here en.wikipedia.org/wiki/…. Oct 9 '19 at 1:21
• No, they are not. See mathoverflow.net/questions/22015/…. And any case Wikipedia is not very authoritative. Oct 9 '19 at 4:44