Lemma 10.112.3. Suppose that $R \to S$ is a ring map such that $S$ is integral over $R$. Then $\dim (R) \geq \dim (S)$, and every closed point of $\mathop{\mathrm{Spec}}(S)$ maps to a closed point of $\mathop{\mathrm{Spec}}(R)$.
Lemma 10.112.3. Suppose that $R \to S$ is a ring map such that $S$ is integral over $R$. Then $\dim (R) \geq \dim (S)$, and every closed point of $\mathop{\mathrm{Spec}}(S)$ maps to a closed point of $\mathop{\mathrm{Spec}}(R)$.
Proof. Immediate from Lemmas 10.36.20 and 10.112.2 and the definitions. $\square$
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