The Stacks project

Proof. If $f \in \text{rad}(R)$, then $f \in \mathfrak m$ for all maximal ideals $\mathfrak m$ of $R$. Hence $1 + f \not\in \mathfrak m$ for all maximal ideals $\mathfrak m$ of $R$. Thus the closed subset $V(1 + f)$ of $\mathop{\mathrm{Spec}}(R)$ is empty. This implies that $1 + f$ is a unit, see Lemma 10.17.2.

Conversely, assume that $1 + f$ is a unit for all $f \in I$. If $\mathfrak m$ is a maximal ideal and $I \not\subset \mathfrak m$, then $I + \mathfrak m = R$. Hence $1 = f + g$ for some $g \in \mathfrak m$ and $f \in I$. Then $g = 1 + (-f)$ is not a unit, contradiction.

For the final statement let $f \in R$ map to a unit in $R/I$. Then we can find $g \in R$ mapping to the multiplicative inverse of $f \bmod I$. Then $fg = 1 \bmod I$. Hence $fg$ is a unit of $R$ by (2) which implies that $f$ is a unit. $\square$

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. $\square$