10.19 The Jacobson radical of a ring

We recall that the Jacobson radical $\text{rad}(R)$ of a ring $R$ is the intersection of all maximal ideals of $R$. If $R$ is local then $\text{rad}(R)$ is the maximal ideal of $R$.

Lemma 10.19.1. Let $R$ be a ring with Jacobson radical $\text{rad}(R)$. Let $I \subset R$ be an ideal. The following are equivalent

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

2. 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 10.17.2.

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. $\square$

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. $\square$

Comment #3650 by Brian Conrad on

Isn't it potentially confusing (especially for someone who follows links into this section without beginning from the start) to write ${\rm{rad}}(R)$ to denote the Jacobson radical, since the same notation is very standard for the nilradical? Why not ${\rm{jrad}}(R)$ instead?

Comment #3746 by on

OK, I replaced "radical" by "Jacobson radical" in all places where appropriate and whenever we used the notation $\text{rad}(R)$ I have added text saying this denotes the Jacobson radical. See the corresponding changes here.

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