[proofplan]
We prove both directions by contrapositive-style arguments involving maximal ideals. For the forward direction, we assume $x \in \operatorname{Jac}(R)$ (i.e., $x$ lies in every maximal ideal) and show that $1 - xy$ cannot lie in any maximal ideal, hence must be a unit. For the converse, we assume $x \notin \operatorname{Jac}(R)$, find a maximal ideal $\mathfrak{m}$ that misses $x$, and use maximality to produce $y \in R$ such that $1 - xy \in \mathfrak{m}$, which witnesses that $1 - xy$ is not a unit.
[/proofplan]
[step:Show that $x \in \operatorname{Jac}(R)$ implies $1 - xy$ is a unit for every $y \in R$]
Suppose $x \in \operatorname{Jac}(R)$ and let $y \in R$ be arbitrary. Assume for contradiction that $1 - xy$ is not a unit. Then the principal ideal $(1 - xy)$ is a proper ideal of $R$ (since it does not contain $1_R$). Every proper ideal of a ring is contained in some maximal ideal, so there exists a maximal ideal $\mathfrak{m}$ of $R$ with $1 - xy \in \mathfrak{m}$.
Since $x \in \operatorname{Jac}(R) = \bigcap_{\mathfrak{m}' \text{ maximal}} \mathfrak{m}'$, in particular $x \in \mathfrak{m}$. Then $xy \in \mathfrak{m}$ because $\mathfrak{m}$ is an ideal. We now have both $1 - xy \in \mathfrak{m}$ and $xy \in \mathfrak{m}$, so
\begin{align*}
1 = (1 - xy) + xy \in \mathfrak{m}.
\end{align*}
This implies $\mathfrak{m} = R$, contradicting the fact that $\mathfrak{m}$ is a maximal ideal (and therefore a proper ideal). Hence $1 - xy$ is a unit.
[guided]
Suppose $x \in \operatorname{Jac}(R)$ and let $y \in R$ be arbitrary. We want to show $1 - xy \in R^\times$.
Assume for contradiction that $1 - xy$ is not a unit. What does it mean for an element to fail to be a unit? It means the principal ideal $(1 - xy)$ does not equal $R$. Since $(1 - xy)$ is a proper ideal, it is contained in some maximal ideal $\mathfrak{m}$ of $R$ (every proper ideal sits inside a maximal ideal, by Zorn's lemma). In particular, $1 - xy \in \mathfrak{m}$.
Now we use the hypothesis $x \in \operatorname{Jac}(R)$. The Jacobson radical $\operatorname{Jac}(R)$ is defined as the intersection of all maximal ideals of $R$. Since $\mathfrak{m}$ is a maximal ideal, $x \in \operatorname{Jac}(R) \subseteq \mathfrak{m}$. Therefore $xy \in \mathfrak{m}$ (because $\mathfrak{m}$ is closed under multiplication by arbitrary ring elements).
We have two elements in $\mathfrak{m}$: $1 - xy$ and $xy$. Adding them:
\begin{align*}
(1 - xy) + xy = 1 \in \mathfrak{m}.
\end{align*}
But $1 \in \mathfrak{m}$ implies $\mathfrak{m} = R$, contradicting the fact that maximal ideals are proper. This contradiction shows that $1 - xy$ must be a unit.
[/guided]
[/step]
[step:Show that $x \notin \operatorname{Jac}(R)$ implies $1 - xy$ is not a unit for some $y \in R$]
Suppose $x \notin \operatorname{Jac}(R)$. Then there exists a maximal ideal $\mathfrak{m}$ of $R$ such that $x \notin \mathfrak{m}$. Since $\mathfrak{m}$ is maximal, the ideal $\mathfrak{m} + (x) = \mathfrak{m} + Rx$ strictly contains $\mathfrak{m}$ and therefore equals $R$.
In particular, $1 \in \mathfrak{m} + Rx$: there exist $t \in \mathfrak{m}$ and $y \in R$ such that $t + xy = 1$. Rearranging gives $1 - xy = t \in \mathfrak{m}$. Since $\mathfrak{m}$ is a proper ideal, $1 - xy$ is not a unit (a unit cannot belong to any proper ideal, because if $u \in \mathfrak{m}$ is a unit then $1 = u \cdot u^{-1} \in \mathfrak{m}$, forcing $\mathfrak{m} = R$).
[guided]
Now suppose $x \notin \operatorname{Jac}(R)$. We need to find some $y \in R$ such that $1 - xy$ is not a unit. The strategy is to find a maximal ideal containing $1 - xy$.
Since $x \notin \operatorname{Jac}(R) = \bigcap_{\mathfrak{m}' \text{ maximal}} \mathfrak{m}'$, there exists at least one maximal ideal $\mathfrak{m}$ of $R$ with $x \notin \mathfrak{m}$. Now we exploit maximality: $\mathfrak{m}$ is maximal, so the only ideal of $R$ that properly contains $\mathfrak{m}$ is $R$ itself. Since $x \notin \mathfrak{m}$, the ideal $\mathfrak{m} + Rx$ properly contains $\mathfrak{m}$ (it contains $x$ but $\mathfrak{m}$ does not). By maximality, $\mathfrak{m} + Rx = R$.
In particular, $1_R \in \mathfrak{m} + Rx$, so there exist $t \in \mathfrak{m}$ and $y \in R$ with $t + xy = 1$. Rearranging: $1 - xy = t$.
Since $t \in \mathfrak{m}$ and $\mathfrak{m}$ is a proper ideal, $t$ cannot be a unit. (If $t$ were a unit, then $t^{-1} \cdot t = 1 \in \mathfrak{m}$, forcing $\mathfrak{m} = R$, a contradiction.) Therefore $1 - xy$ is not a unit, which completes the proof.
[/guided]
[/step]
