The Stacks project

Theorem 41.5.1. Let $f : X \to S$ be a morphism of schemes. Assume $S$ is a locally Noetherian scheme, and $f$ is locally of finite type. Then the following are equivalent:

  1. $f$ is unramified,

  2. the morphism $f$ is formally unramified: for any affine $S$-scheme $T$ and subscheme $T_0$ of $T$ defined by a square-zero ideal, the natural map

    \[ \mathop{\mathrm{Hom}}\nolimits _ S(T, X) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ S(T_0, X) \]

    is injective.

Proof. See More on Morphisms, Lemma 37.6.8 for a more general statement and proof. What follows is a sketch of the proof in the current case.

Firstly, one checks both properties are local on the source and the target. This we may assume that $S$ and $X$ are affine. Say $X = \mathop{\mathrm{Spec}}(B)$ and $S = \mathop{\mathrm{Spec}}(R)$. Say $T = \mathop{\mathrm{Spec}}(C)$. Let $J$ be the square-zero ideal of $C$ with $T_0 = \mathop{\mathrm{Spec}}(C/J)$. Assume that we are given the diagram

\[ \xymatrix{ & B \ar[d]^\phi \ar[rd]^{\bar{\phi }} & \\ R \ar[r] \ar[ur] & C \ar[r] & C/J } \]

Secondly, one checks that the association $\phi ' \mapsto \phi ' - \phi $ gives a bijection between the set of liftings of $\bar{\phi }$ and the module $\text{Der}_ R(B, J)$. Thus, we obtain the implication (1) $\Rightarrow $ (2) via the description of unramified morphisms having trivial module of differentials, see Theorem 41.4.1.

To obtain the reverse implication, consider the surjection $q : C = (B \otimes _ R B)/I^2 \to B = C/J$ defined by the square zero ideal $J = I/I^2$ where $I$ is the kernel of the multiplication map $B \otimes _ R B \to B$. We already have a lifting $B \to C$ defined by, say, $b \mapsto b \otimes 1$. Thus, by the same reasoning as above, we obtain a bijective correspondence between liftings of $\text{id} : B \to C/J$ and $\text{Der}_ R(B, J)$. The hypothesis therefore implies that the latter module is trivial. But we know that $J \cong \Omega _{B/R}$. Thus, $B/R$ is unramified. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 024R. Beware of the difference between the letter 'O' and the digit '0'.