**Proof.**
For any ring map $f : A \to B$ in $\mathcal{C}_\Lambda $ we have $f(\mathfrak m_ A) \subset \mathfrak m_ B$ for example because $\mathfrak m_ A$, $\mathfrak m_ B$ is the set of nilpotent elements of $A$, $B$. Suppose $f$ is surjective. Let $y \in \mathfrak m_ B$. Choose $x \in A$ with $f(x) = y$. Since $f$ induces an isomorphism $A/\mathfrak m_ A \to B/\mathfrak m_ B$ we see that $x \in \mathfrak m_ A$. Hence the induced map $\mathfrak m_ A/\mathfrak m_ A^2 \to \mathfrak m_ B/\mathfrak m_ B^2$ is surjective. In this way we see that (1) implies (2).

It is clear that (2) implies (3). The map $A \to B$ gives rise to a canonical commutative diagram

\[ \xymatrix{ \mathfrak m_\Lambda /\mathfrak m_\Lambda ^2 \otimes _{k'} k \ar[r] \ar[d] & \mathfrak m_ A/\mathfrak m_ A^2 \ar[r] \ar[d] & \mathfrak m_ A/(\mathfrak m_\Lambda A + \mathfrak m_ A^2) \ar[r] \ar[d] & 0 \\ \mathfrak m_\Lambda /\mathfrak m_\Lambda ^2 \otimes _{k'} k \ar[r] & \mathfrak m_ B/\mathfrak m_ B^2 \ar[r] & \mathfrak m_ B/(\mathfrak m_\Lambda B + \mathfrak m_ B^2) \ar[r] & 0 } \]

with exact rows. Hence if (3) holds, then so does (2).

Assume (2). To show that $A \to B$ is surjective it suffices by Nakayama's lemma (Algebra, Lemma 10.20.1) to show that $A/\mathfrak m_ A \to B/\mathfrak m_ AB$ is surjective. (Note that $\mathfrak m_ A$ is a nilpotent ideal.) As $k = A/\mathfrak m_ A = B/\mathfrak m_ B$ it suffices to show that $\mathfrak m_ AB \to \mathfrak m_ B$ is surjective. Applying Nakayama's lemma once more we see that it suffices to see that $\mathfrak m_ AB/\mathfrak m_ A\mathfrak m_ B \to \mathfrak m_ B/\mathfrak m_ B^2$ is surjective which is what we assumed.
$\square$

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