Recall that a ring map $R \to A$ is called formally unramified (see Algebra, Definition 10.148.1) if for every commutative solid diagram
where $I \subset B$ is an ideal of square zero, at most one dotted arrow exists which makes the diagram commute. This motivates the following analogue for morphisms of schemes.
Definition 37.6.1. Let $f : X \to S$ be a morphism of schemes. We say $f$ is formally unramified if given any solid commutative diagram where $T \subset T'$ is a first order thickening of affine schemes over $S$ there exists at most one dotted arrow making the diagram commute.
\[ \xymatrix{ X \ar[d]_ f & T \ar[d]^ i \ar[l] \\ S & T' \ar[l] \ar@{-->}[lu] } \]
We first prove some formal lemmas, i.e., lemmas which can be proved by drawing the corresponding diagrams.
Lemma 37.6.2. If $f : X \to S$ is a formally unramified morphism, then given any solid commutative diagram where $T \subset T'$ is a first order thickening of schemes over $S$ there exists at most one dotted arrow making the diagram commute. In other words, in Definition 37.6.1 the condition that $T$ be affine may be dropped.
\[ \xymatrix{ X \ar[d]_ f & T \ar[d]^ i \ar[l] \\ S & T' \ar[l] \ar@{-->}[lu] } \]
Proof. This is true because a morphism is determined by its restrictions to affine opens. $\square$
Lemma 37.6.3. A composition of formally unramified morphisms is formally unramified.
Proof. This is formal. $\square$
Lemma 37.6.4. A base change of a formally unramified morphism is formally unramified.
Proof. This is formal. $\square$
Lemma 37.6.5. Let $f : X \to S$ be a morphism of schemes. Let $U \subset X$ and $V \subset S$ be open such that $f(U) \subset V$. If $f$ is formally unramified, so is $f|_ U : U \to V$.
Proof. Consider a solid diagram
as in Definition 37.6.1. If $f$ is formally ramified, then there exists at most one $S$-morphism $a' : T' \to X$ such that $a'|_ T = a$. Hence clearly there exists at most one such morphism into $U$. $\square$
Lemma 37.6.6. Let $f : X \to S$ be a morphism of schemes. Assume $X$ and $S$ are affine. Then $f$ is formally unramified if and only if $\mathcal{O}_ S(S) \to \mathcal{O}_ X(X)$ is a formally unramified ring map.
Proof. This is immediate from the definitions (Definition 37.6.1 and Algebra, Definition 10.148.1) by the equivalence of categories of rings and affine schemes, see Schemes, Lemma 26.6.5. $\square$
Here is a characterization in terms of the sheaf of differentials.
Lemma 37.6.7. Let $f : X \to S$ be a morphism of schemes. Then $f$ is formally unramified if and only if $\Omega _{X/S} = 0$.
Proof. We recall some of the arguments of the proof of Morphisms, Lemma 29.32.5. Let $W \subset X \times _ S X$ be an open such that $\Delta : X \to X \times _ S X$ induces a closed immersion into $W$. Let $\mathcal{J} \subset \mathcal{O}_ W$ be the ideal sheaf of this closed immersion. Let $X' \subset W$ be the closed subscheme defined by the quasi-coherent sheaf of ideals $\mathcal{J}^2$. Consider the two morphisms $p_1, p_2 : X' \to X$ induced by the two projections $X \times _ S X \to X$. Note that $p_1$ and $p_2$ agree when composed with $\Delta : X \to X'$ and that $X \to X'$ is a closed immersion defined by a an ideal whose square is zero. Moreover there is a short exact sequence
and $\Omega _{X/S} = \mathcal{J}/\mathcal{J}^2$. Moreover, $\mathcal{J}/\mathcal{J}^2$ is generated by the local sections $p_1^\sharp (f) - p_2^\sharp (f)$ for $f$ a local section of $\mathcal{O}_ X$.
Suppose that $f : X \to S$ is formally unramified. By assumption this means that $p_1 = p_2$ when restricted to any affine open $T' \subset X'$. Hence $p_1 = p_2$. By what was said above we conclude that $\Omega _{X/S} = \mathcal{J}/\mathcal{J}^2 = 0$.
Conversely, suppose that $\Omega _{X/S} = 0$. Then $X' = X$. Take any pair of morphisms $f'_1, f'_2 : T' \to X$ fitting as dotted arrows in the diagram of Definition 37.6.1. This gives a morphism $(f'_1, f'_2) : T' \to X \times _ S X$. Since $f'_1|_ T = f'_2|_ T$ and $|T| =|T'|$ we see that the image of $T'$ under $(f'_1, f'_2)$ is contained in the open $W$ chosen above. Since $(f'_1, f'_2)(T) \subset \Delta (X)$ and since $T$ is defined by an ideal of square zero in $T'$ we see that $(f'_1, f'_2)$ factors through $X'$. As $X' = X$ we conclude $f_1' = f'_2$ as desired. $\square$
Lemma 37.6.8. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:
The morphism $f$ is unramified (resp. G-unramified), and
the morphism $f$ is locally of finite type (resp. locally of finite presentation) and formally unramified.
Proof. Use Lemma 37.6.7 and Morphisms, Lemma 29.35.2. $\square$
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