Formally unramified morphism

Let $f\text{\hspace{0.17em}⁣}:X\to S$ be a morphism of schemes (Morphisms of schemes, Schemes and morphisms over a base).

Square-zero thickenings. A square-zero thickening of a scheme $T_0$ is a closed immersion $i\text{\hspace{0.17em}⁣}:T_0\hookrightarrow T$ (Closed immersions of schemes) whose ideal sheaf $\mathcal{I}=\ker ({\mathcal{O}}_T\to i_{\ast }{\mathcal{O}}_{T_0})$ (Ideal sheaves) satisfies ${\mathcal{I}}^2=0$, meaning that the product of any two local sections of $\mathcal{I}$ over a common open set is zero. Such a thickening is an $S$-thickening when $T$ is an $S$-scheme and $i$ is an $S$-morphism.

Formally unramified. The morphism $f$ is formally unramified if for every commutative diagram of schemes

$\begin{array}{ccc}{T}_0& \stackrel{a}{\to }& X\\ \downarrow i& & \downarrow f\\ T& \stackrel{b}{\to }& S\end{array}$

in which $i\text{\hspace{0.17em}⁣}:T_0\hookrightarrow T$ is a square-zero thickening and the square is over $S$ — that is, $a$ and $b$ are compatible with $f$ — there is at most one $S$-morphism $T\to X$ whose restriction to $T_0$ is $a$. In other words, two $S$-morphisms $T\to X$ agreeing on a square-zero closed subscheme agree everywhere.

The condition is a uniqueness condition only: no existence is required, no finite-type, finite-presentation, flatness or separatedness hypothesis is imposed on $f$, and the test thickenings are required to be square-zero but are otherwise arbitrary, in particular not assumed to be affine or of finite type over $S$. For the affine case $X=\mathrm{Spec}B$, $S=\mathrm{Spec}A$ with $f$ induced by $A\to B$, the condition is the algebraic one: for every $A$-algebra $C$ with an ideal $I\subseteq C$ satisfying $I^2=0$, two $A$-algebra maps $B\to C$ that agree modulo $I$ are equal.