## Theorem

Let $A$ be a commutative ring with unity.

Let $A^\times$ be the group of units of $A$.

Let $\map {\operatorname {Jac} } A$ be the Jacobson radical of $A$.

Then:

$\map {\operatorname {Jac} } A = \set {a \in A: 1_A - a x \in A^\times \text{ for all } x \in A}$

where $1_A$ is the unity of $A$.

## Proof

Recall the definition of Jacobson radical of $A$:

$\map {\operatorname {Jac} } A$ is the intersection of all maximal ideals of $R$.

### $\subset$ Direction

Aiming for a contradiction, suppose that $x \in \map {\operatorname {Jac} } A$ such that there exists $y \in A$ such that $1_A - x y \notin A^\times$.

By definition $x$ is contained in all maximal ideals of $A$.

Let $\mathfrak m \subseteq A$ be one particular such maximal ideal of $A$.

Then $x \in \map {\operatorname {Jac} } A \subseteq \mathfrak m$ implies that $xy \in \mathfrak m$ and therefore $1_A \in \mathfrak m$.

But if $1_A \in \mathfrak m$ then from Ideal of Unit is Whole Ring: Corollary:

$\mathfrak m = R$

This contradicts the definition of $\mathfrak m$ as a maximal ideal of $A$.

$\map {\operatorname {Jac} } A \subseteq \set {a \in A: 1_A - a x \in A^\times \text{ for all } x \in A}$

$\Box$

### $\supset$ Direction

Now suppose that $x \notin \map {\operatorname {Jac} } A$.

Then $x \notin \mathfrak m$ for some maximal ideal $\mathfrak m$ of $A$.

Because $\mathfrak m$ is maximal, $x$ and $\mathfrak m$ generate $A$.

Therefore there exist $w \in \mathfrak m$ and $y \in A$ such that $w + x y = 1_A$.

Thus $1_A - x y \in \mathfrak m$, and $1_A - x y \notin A^\times$.

Hence:

$\map {\operatorname {Jac} } A \supseteq \set {a \in A: 1_A - a x \in A^\times \text{ for all } x \in A}$

$\Box$

The result follows by definition of set equality.

$\blacksquare$