**Proof.**
Proof of (1). It follows from (89.3.10.3) that (1)(a) and (1)(b) are equivalent. Also, if $A \to B$ is surjective, then (1)(a) and (1)(b) hold. Assume (1)(a). Since the kernel of $\text{d}_ A$ is the image of $H_1(L_{k/\Lambda })$ which also maps to $\mathfrak m_ B/\mathfrak m_ B^2$ we conclude that $\mathfrak m_ B/\mathfrak m_ B^2 \to \mathfrak m_ A/\mathfrak m_ A^2$ is surjective. Hence $B \to A$ is surjective by Lemma 89.3.5. This finishes the proof of (1).

Proof of (2). The equivalence of (2)(b) and (2)(c) is immediate from (89.3.10.3).

Assume (2)(b). Let $g : C \to B$ be a ring map in $\mathcal{C}_\Lambda $ such that $f \circ g$ is surjective. We conclude that $\mathfrak m_ C/\mathfrak m_ C^2 \to \mathfrak m_ A/\mathfrak m_ A^2$ is surjective by Lemma 89.3.5. Hence $\mathop{\mathrm{Im}}(\text{d}_ C) \to \mathop{\mathrm{Im}}(\text{d}_ A)$ is surjective and by the assumption we see that $\mathop{\mathrm{Im}}(\text{d}_ C) \to \mathop{\mathrm{Im}}(\text{d}_ B)$ is surjective. It follows that $C \to B$ is surjective by (1).

Assume (2)(a). Then $f$ is surjective and we see that $\Omega _{B/\Lambda } \otimes _ B k \to \Omega _{A/\Lambda } \otimes _ A k$ is surjective. Let $K$ be the kernel. Note that $K = \text{d}_ B(\mathop{\mathrm{Ker}}(\mathfrak m_ B/\mathfrak m_ B^2 \to \mathfrak m_ A/\mathfrak m_ A^2))$ by (89.3.10.3). Choose a splitting

\[ \Omega _{B/\Lambda } \otimes _ B k = \Omega _{A/\Lambda } \otimes _ A k \oplus K \]

of $k$-vector space. The map $\text{d} : B \to \Omega _{B/\Lambda }$ induces via the projection onto $K$ a map $D : B \to K$. Set $C = \{ b \in B \mid D(b) = 0\} $. The Leibniz rule shows that this is a $\Lambda $-subalgebra of $B$. Let $\overline{x} \in k$. Choose $x \in B$ mapping to $\overline{x}$. If $D(x) \not= 0$, then we can find an element $y \in \mathfrak m_ B$ such that $D(y) = D(x)$. Hence $x - y \in C$ is an element which maps to $\overline{x}$. Thus $C \to k$ is surjective and $C$ is an object of $\mathcal{C}_\Lambda $. Similarly, pick $\omega \in \mathop{\mathrm{Im}}(\text{d}_ A)$. We can find $x \in \mathfrak m_ B$ such that $\text{d}_ B(x)$ maps to $\omega $ by (1). If $D(x) \not= 0$, then we can find an element $y \in \mathfrak m_ B$ which maps to zero in $\mathfrak m_ A/\mathfrak m_ A^2$ such that $D(y) = D(x)$. Hence $z = x - y$ is an element of $\mathfrak m_ C$ whose image $\text{d}_ C(z) \in \Omega _{C/k} \otimes _ C k$ maps to $\omega $. Hence $\mathop{\mathrm{Im}}(\text{d}_ C) \to \mathop{\mathrm{Im}}(\text{d}_ A)$ is surjective. We conclude that $C \to A$ is surjective by (1). Hence $C \to B$ is surjective by assumption. Hence $D = 0$, i.e., $K = 0$, i.e., (2)(c) holds. This finishes the proof of (2).

Proof of (3). If $k'/k$ is separable, then $H_1(L_{k/\Lambda }) = \mathfrak m_\Lambda /\mathfrak m_\Lambda ^2 \otimes _{k'} k$, see Lemma 89.3.11. Hence $\mathop{\mathrm{Im}}(\text{d}_ A) = \mathfrak m_ A/(\mathfrak m_\Lambda A + \mathfrak m_ A^2)$ and similarly for $B$. Thus (3) follows from (2).

Proof of (4). A section $s$ of $f$ is not surjective (by definition a small extension has nontrivial kernel), hence $f$ is not essentially surjective. Conversely, assume $f$ is a small extension but not an essential surjection. Choose a ring map $C \to B$ in $\mathcal{C}_\Lambda $ which is not surjective, such that $C \to A$ is surjective. Let $C' \subset B$ be the image of $C \to B$. Then $C' \not= B$ but $C'$ surjects onto $A$. Since $f : B \to A$ is a small extension, $\text{length}_ C(B) = \text{length}_ C(A) + 1$. Thus $\text{length}_ C(C') \leq \text{length}_ C(A)$ since $C'$ is a proper subring of $B$. But $C' \to A$ is surjective, so in fact we must have $\text{length}_ C(C') = \text{length}_ C(A)$ and $C' \to A$ is an isomorphism which gives us our section.
$\square$

## Comments (1)

Comment #7731 by Mingchen on

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