Lemma 10.18.1. Let $R$ be a ring with Jacobson radical $\text{rad}(R)$. Let $I \subset R$ be an ideal. The following are equivalent

$I \subset \text{rad}(R)$, and

every element of $1 + I$ is a unit in $R$.

In this case every element of $R$ which maps to a unit of $R/I$ is a unit.

**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 10.16.2.

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.
$\square$

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