2
$\begingroup$

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, enter image description here there exists a lift like below: enter image description here 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, enter image description here there exists a lift like below: enter image description here

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?

$\endgroup$
3
  • $\begingroup$ Formally smooth morphisms only have the property above with respect to nilpotent embeddings of affine schemes. $\endgroup$
    – Angelo
    Oct 8 '19 at 17:30
  • $\begingroup$ @Angelo I believe that is equivalent; see here en.wikipedia.org/wiki/…. $\endgroup$ Oct 9 '19 at 1:21
  • $\begingroup$ No, they are not. See mathoverflow.net/questions/22015/…. And any case Wikipedia is not very authoritative. $\endgroup$
    – Angelo
    Oct 9 '19 at 4:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.