Lemma 10.112.2. Suppose that $R \to S$ is a ring map with the going up property, see Definition 10.41.1. If $\mathfrak q \subset S$ is a maximal ideal. Then the inverse image of $\mathfrak q$ in $R$ is a maximal ideal too.
Proof. Trivial. $\square$
Lemma 10.112.2. Suppose that $R \to S$ is a ring map with the going up property, see Definition 10.41.1. If $\mathfrak q \subset S$ is a maximal ideal. Then the inverse image of $\mathfrak q$ in $R$ is a maximal ideal too.
Proof. Trivial. $\square$
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