Recall that a ring map $R \to A$ is called *formally étale* (see Algebra, Definition 10.150.1) if for every commutative solid diagram

\[ \xymatrix{ A \ar[r] \ar@{-->}[rd] & B/I \\ R \ar[r] \ar[u] & B \ar[u] } \]

where $I \subset B$ is an ideal of square zero, there exists exactly one dotted arrow which makes the diagram commute. This motivates the following analogue for morphisms of schemes.

Definition 37.8.1. Let $f : X \to S$ be a morphism of schemes. We say $f$ is *formally étale* if given any solid commutative diagram

\[ \xymatrix{ X \ar[d]_ f & T \ar[d]^ i \ar[l] \\ S & T' \ar[l] \ar@{-->}[lu] } \]

where $T \subset T'$ is a first order thickening of affine schemes over $S$ there exists exactly one dotted arrow making the diagram commute.

It is clear that a formally étale morphism is formally unramified. Hence if $f : X \to S$ is formally étale, then $\Omega _{X/S}$ is zero, see Lemma 37.6.7.

Lemma 37.8.2. If $f : X \to S$ is a formally étale morphism, then given any solid commutative diagram

\[ \xymatrix{ X \ar[d]_ f & T \ar[d]^ i \ar[l] \\ S & T' \ar[l] \ar@{-->}[lu] } \]

where $T \subset T'$ is a first order thickening of schemes over $S$ there exists exactly one dotted arrow making the diagram commute. In other words, in Definition 37.8.1 the condition that $T$ be affine may be dropped.

**Proof.**
Let $T' = \bigcup T'_ i$ be an affine open covering, and let $T_ i = T \cap T'_ i$. Then we get morphisms $a'_ i : T'_ i \to X$ fitting into the diagram. By uniqueness we see that $a'_ i$ and $a'_ j$ agree on any affine open subscheme of $T'_ i \cap T'_ j$. Hence $a'_ i$ and $a'_ j$ agree on $T'_ i \cap T'_ j$. Thus we see that the morphisms $a'_ i$ glue to a global morphism $a' : T' \to X$. The uniqueness of $a'$ we have seen in Lemma 37.6.2.
$\square$

Lemma 37.8.3. A composition of formally étale morphisms is formally étale.

**Proof.**
This is formal.
$\square$

Lemma 37.8.4. A base change of a formally étale morphism is formally étale.

**Proof.**
This is formal.
$\square$

Lemma 37.8.5. Let $f : X \to S$ be a morphism of schemes. Let $U \subset X$ and $V \subset S$ be open subschemes such that $f(U) \subset V$. If $f$ is formally étale, so is $f|_ U : U \to V$.

**Proof.**
Consider a solid diagram

\[ \xymatrix{ U \ar[d]_{f|_ U} & T \ar[d]^ i \ar[l]^ a \\ V & T' \ar[l] \ar@{-->}[lu] } \]

as in Definition 37.8.1. If $f$ is formally ramified, then there exists exactly one $S$-morphism $a' : T' \to X$ such that $a'|_ T = a$. Since $|T'| = |T|$ we conclude that $a'(T') \subset U$ which gives our unique morphism from $T'$ into $U$.
$\square$

Lemma 37.8.6. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

$f$ is formally étale,

$f$ is formally unramified and the universal first order thickening of $X$ over $S$ is equal to $X$,

$f$ is formally unramified and $\mathcal{C}_{X/S} = 0$, and

$\Omega _{X/S} = 0$ and $\mathcal{C}_{X/S} = 0$.

**Proof.**
Actually, the last assertion only make sense because $\Omega _{X/S} = 0$ implies that $\mathcal{C}_{X/S}$ is defined via Lemma 37.6.7 and Definition 37.7.2. This also makes it clear that (3) and (4) are equivalent.

Either of the assumptions (1), (2), and (3) imply that $f$ is formally unramified. Hence we may assume $f$ is formally unramified. The equivalence of (1), (2), and (3) follow from the universal property of the universal first order thickening $X'$ of $X$ over $S$ and the fact that $X = X' \Leftrightarrow \mathcal{C}_{X/S} = 0$ since after all by definition $\mathcal{C}_{X/S} = \mathcal{C}_{X/X'}$ is the ideal sheaf of $X$ in $X'$.
$\square$

Lemma 37.8.7. An unramified flat morphism is formally étale.

**Proof.**
Say $X \to S$ is unramified and flat. Then $\Delta : X \to X \times _ S X$ is an open immersion, see Morphisms, Lemma 29.35.13. We have to show that $\mathcal{C}_{X/S}$ is zero. Consider the two projections $p, q : X \times _ S X \to X$. As $f$ is formally unramified (see Lemma 37.6.8), $q$ is formally unramified (see Lemma 37.6.4). As $f$ is flat, $p$ is flat, see Morphisms, Lemma 29.25.8. Hence $p^*\mathcal{C}_{X/S} = \mathcal{C}_ q$ by Lemma 37.7.7 where $\mathcal{C}_ q$ denotes the conormal sheaf of the formally unramified morphism $q : X \times _ S X \to X$. But $\Delta (X) \subset X \times _ S X$ is an open subscheme which maps isomorphically to $X$ via $q$. Hence by Lemma 37.7.8 we see that $\mathcal{C}_ q|_{\Delta (X)} = \mathcal{C}_{X/X} = 0$. In other words, the pullback of $\mathcal{C}_{X/S}$ to $X$ via the identity morphism is zero, i.e., $\mathcal{C}_{X/S} = 0$.
$\square$

Lemma 37.8.8. Let $f : X \to S$ be a morphism of schemes. Assume $X$ and $S$ are affine. Then $f$ is formally étale if and only if $\mathcal{O}_ S(S) \to \mathcal{O}_ X(X)$ is a formally étale ring map.

**Proof.**
This is immediate from the definitions (Definition 37.8.1 and Algebra, Definition 10.150.1) by the equivalence of categories of rings and affine schemes, see Schemes, Lemma 26.6.5.
$\square$

Lemma 37.8.9. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

The morphism $f$ is étale, and

the morphism $f$ is locally of finite presentation and formally étale.

**Proof.**
Assume $f$ is étale. An étale morphism is locally of finite presentation, flat and unramified, see Morphisms, Section 29.36. Hence $f$ is locally of finite presentation and formally étale, see Lemma 37.8.7.

Conversely, suppose that $f$ is locally of finite presentation and formally étale. Being étale is local in the Zariski topology on $X$ and $S$, see Morphisms, Lemma 29.36.2. By Lemma 37.8.5 we can cover $X$ by affine opens $U$ which map into affine opens $V$ such that $U \to V$ is formally étale (and of finite presentation, see Morphisms, Lemma 29.21.2). By Lemma 37.8.8 we see that the ring maps $\mathcal{O}(V) \to \mathcal{O}(U)$ are formally étale (and of finite presentation). We win by Algebra, Lemma 10.150.2. (We will give another proof of this implication when we discuss formally smooth morphisms.)
$\square$

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