The Stacks project

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10.18 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 10.18.1. 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 10.16.2.

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. $\square$

Lemma 10.18.2. 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 10.16.2. $\square$


Comments (2)

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 to denote the Jacobson radical, since the same notation is very standard for the nilradical? Why not instead?

Comment #3746 by on

OK, I replaced "radical" by "Jacobson radical" in all places where appropriate and whenever we used the notation I have added text saying this denotes the Jacobson radical. See the corresponding changes here.


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