Lemma 10.128.4. Suppose that $R \to S$ is a local homomorphism of local rings. Denote $\mathfrak m$ the maximal ideal of $R$. Let $u : M \to N$ be a map of $S$-modules. Assume

1. $S$ is essentially of finite presentation over $R$,

2. $M$, $N$ are finitely presented over $S$,

3. $N$ is flat over $R$, and

4. $\overline{u} : M/\mathfrak mM \to N/\mathfrak mN$ is injective.

Then $u$ is injective, and $N/u(M)$ is flat over $R$.

Proof. By Lemma 10.127.13 and its proof we can find a system $R_\lambda \to S_\lambda$ of local ring maps together with maps of $S_\lambda$-modules $u_\lambda : M_\lambda \to N_\lambda$ satisfying the conclusions (1) – (6) for both $N$ and $M$ of that lemma and such that the colimit of the maps $u_\lambda$ is $u$. By Lemma 10.128.3 we may assume that $N_\lambda$ is flat over $R_\lambda$ for all sufficiently large $\lambda$. Denote $\mathfrak m_\lambda \subset R_\lambda$ the maximal ideal and $\kappa _\lambda = R_\lambda / \mathfrak m_\lambda$, resp. $\kappa = R/\mathfrak m$ the residue fields.

Consider the map

$\Psi _\lambda : M_\lambda /\mathfrak m_\lambda M_\lambda \otimes _{\kappa _\lambda } \kappa \longrightarrow M/\mathfrak m M.$

Since $S_\lambda /\mathfrak m_\lambda S_\lambda$ is essentially of finite type over the field $\kappa _\lambda$ we see that the tensor product $S_\lambda /\mathfrak m_\lambda S_\lambda \otimes _{\kappa _\lambda } \kappa$ is essentially of finite type over $\kappa$. Hence it is a Noetherian ring and we conclude the kernel of $\Psi _\lambda$ is finitely generated. Since $M/\mathfrak m M$ is the colimit of the system $M_\lambda /\mathfrak m_\lambda M_\lambda$ and $\kappa$ is the colimit of the fields $\kappa _\lambda$ there exists a $\lambda ' > \lambda$ such that the kernel of $\Psi _\lambda$ is generated by the kernel of

$\Psi _{\lambda , \lambda '} : M_\lambda /\mathfrak m_\lambda M_\lambda \otimes _{\kappa _\lambda } \kappa _{\lambda '} \longrightarrow M_{\lambda '}/\mathfrak m_{\lambda '} M_{\lambda '}.$

By construction there exists a multiplicative subset $W \subset S_\lambda \otimes _{R_\lambda } R_{\lambda '}$ such that $S_{\lambda '} = W^{-1}(S_\lambda \otimes _{R_\lambda } R_{\lambda '})$ and

$W^{-1}(M_\lambda /\mathfrak m_\lambda M_\lambda \otimes _{\kappa _\lambda } \kappa _{\lambda '}) = M_{\lambda '}/\mathfrak m_{\lambda '} M_{\lambda '}.$

Now suppose that $x$ is an element of the kernel of

$\Psi _{\lambda '} : M_{\lambda '}/\mathfrak m_{\lambda '} M_{\lambda '} \otimes _{\kappa _{\lambda '}} \kappa \longrightarrow M/\mathfrak m M.$

Then for some $w \in W$ we have $wx \in M_\lambda /\mathfrak m_\lambda M_\lambda \otimes \kappa$. Hence $wx \in \mathop{\mathrm{Ker}}(\Psi _\lambda )$. Hence $wx$ is a linear combination of elements in the kernel of $\Psi _{\lambda , \lambda '}$. Hence $wx = 0$ in $M_{\lambda '}/\mathfrak m_{\lambda '} M_{\lambda '} \otimes _{\kappa _{\lambda '}} \kappa$, hence $x = 0$ because $w$ is invertible in $S_{\lambda '}$. We conclude that the kernel of $\Psi _{\lambda '}$ is zero for all sufficiently large $\lambda '$!

By the result of the preceding paragraph we may assume that the kernel of $\Psi _\lambda$ is zero for all $\lambda$ sufficiently large, which implies that the map $M_\lambda /\mathfrak m_\lambda M_\lambda \to M/\mathfrak m M$ is injective. Combined with $\overline{u}$ being injective this formally implies that also $\overline{u_\lambda } : M_\lambda /\mathfrak m_\lambda M_\lambda \to N_\lambda /\mathfrak m_\lambda N_\lambda$ is injective. By Lemma 10.99.1 we conclude that (for all sufficiently large $\lambda$) the map $u_\lambda$ is injective and that $N_\lambda /u_\lambda (M_\lambda )$ is flat over $R_\lambda$. The lemma follows. $\square$

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