# infinitesimal lifting criterion for non-noetherian schemes

We have the "standard criterion" which says that a morphism $f:X\rightarrow Y$ is smooth if:

1/ $Y$ is locally noetherian.

2/ $f$ is locally of finite type and satisfies lifting criterion for artinian rings.

Can we have the same satement if I remove 1/ and replace locally of finite type by locally of finite presentation?

• Yes. See for example stacks.math.columbia.edu/tag/049R – Mattia Talpo Jan 23 '16 at 2:52
• No. Let $Y = {\rm{Spec}}(R)$ for a local ring $R$ whose maximal ideal $m$ is nonzero and satisfies $m^2 = m$ (e.g., the valuation ring of the completed algebraic closure of the fraction field of a dvr), and let $X = {\rm{Spec}}(R/m)$ be the (reduced) closed point of $Y$. For any artinian local ring $A$, a map ${\rm{Spec}}(A) \rightarrow Y$ hitting the closed point factors (uniquely) through $X$ scheme-theoretically since any local map of rings $R \rightarrow A$ carries $m = m^n$ into $\mathfrak{m}_A^n = 0$ for $n \gg 0$. So the infinitesimal criterion is satisfied but $f$ is not even flat. – nfdc23 Jan 23 '16 at 3:17
• it is not finitely presented. – prochet Jan 23 '16 at 7:16
• Take $X$ to be ${\rm{Spec}}(R/(r))$ for a nonzero $r \in m$ to get a finitely presented instance of the same phenomenon. – nfdc23 Jan 23 '16 at 12:37
• @nfdc23: Would you please post this as an answer? – Jason Starr Jan 23 '16 at 14:48

Let $Y = {\rm{Spec}}(R)$ for a local ring $R$ with nonzero maximal ideal $m$ satisfying $m^2 = m$ (e.g., a valuation ring with algebraically closed fraction field). Note that the noetherian $R$ can satisfy this condition, due to the Krull Intersection Theorem.
Let $X = {\rm{Spec}}(R/(r))$ for a nonzero $r \in m$. Then the natural map $f:X \rightarrow Y$ is finitely presented but not flat (since it corresponds to a non-injective local map between local rings), so it is not smooth.
However, the "infinitesimal criterion" restricted to artin local rings is satisfied. Indeed, every functorial point of $Y$ supported at the closed point and valued in an artinian local ring $A$ uniquely factors through $X$ (as any local homomorphism $R \rightarrow A$ carries $m = m^2 = \dots = m^n$ into $\mathfrak{m}_A^n = 0$ for $n$ large, so such a homomorphism kills $r$).