Lemma 10.19.2. Let $\varphi : R \to S$ be a ring map such that the induced map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is surjective. Then an element $x \in R$ is a unit if and only if $\varphi (x) \in S$ is a unit.

**Proof.**
If $x$ is a unit, then so is $\varphi (x)$. Conversely, if $\varphi (x)$ is a unit, then $\varphi (x) \not\in \mathfrak q$ for all $\mathfrak q \in \mathop{\mathrm{Spec}}(S)$. Hence $x \not\in \varphi ^{-1}(\mathfrak q) = \mathop{\mathrm{Spec}}(\varphi )(\mathfrak q)$ for all $\mathfrak q \in \mathop{\mathrm{Spec}}(S)$. Since $\mathop{\mathrm{Spec}}(\varphi )$ is surjective we conclude that $x$ is a unit by part (17) of Lemma 10.17.2.
$\square$

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