## Theorem

Let $A$ be a commutative ring with unity.

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

Let $\operatorname{Jac} \left({A}\right)$ be the Jacobson radical of $A$.

Then:

$\operatorname{Jac} \left({A}\right) = \left\{ {a \in A: 1_A - a x \in A^\times \text{ for all } x \in A}\right\}$

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

## Proof

First suppose that $1_A - x y \notin A^\times$.

Then it is contained in some maximal ideal $\mathfrak m \subseteq A$.

Then $x \in \operatorname{Jac} \left({A}\right) \subseteq \mathfrak m$ implies that $y \in \mathfrak m$ and therefore $1_A \in \mathfrak m$.

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

$\mathfrak m = R$

So by definition $\mathfrak m$ would not be maximal.

This shows that:

$\operatorname{Jac} \left({A}\right) \subseteq \left\{{a \in A: 1_A - a x \in A^\times \text{ for all } x \in A}\right\}$

Now suppose that $x \notin \mathfrak m$ for some maximal ideal $\mathfrak m$ of $A$.

Since $\mathfrak m$ 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 $1_A - x y \notin A^\times$.

This completes the proof.

$\blacksquare$