Lemma 10.18.3. Let $\varphi : R \to S$ be a ring map. Assume $R$ and $S$ are local rings. The following are equivalent:

1. $\varphi$ is a local ring map,

2. $\varphi (\mathfrak m_ R) \subset \mathfrak m_ S$, and

3. $\varphi ^{-1}(\mathfrak m_ S) = \mathfrak m_ R$.

4. For any $x \in R$, if $\varphi (x)$ is invertible in $S$, then $x$ is invertible in $R$.

Proof. Conditions (1) and (2) are equivalent by definition. If (3) holds then (2) holds. Conversely, if (2) holds, then $\varphi ^{-1}(\mathfrak m_ S)$ is a prime ideal containing the maximal ideal $\mathfrak m_ R$, hence $\varphi ^{-1}(\mathfrak m_ S) = \mathfrak m_ R$. Finally, (4) is the contrapositive of (2) by Lemma 10.18.2. $\square$

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