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Assume $R$ is a nonzero unital ring and all its modules are unital. Order the proper left ideals by inclusion. A chain has an upper bound given by its union, which is still a proper left ideal: if it contained $1$, one member of the chain would contain $1$ and be all of $R$. The family is nonempty because it contains $0$. The Zorn lemma gives a maximal proper left ideal $L$. The nonzero quotient $R/L$ has no proper nonzero submodules, because its submodules correspond to left ideals between $L$ and $R$. Thus $R/L$ is a simple left module. The nonzero-ring qualification is necessary: the zero ring has no nonzero unital simple module.

Define the Jacobson radical $J(R)$ as the intersection of all maximal left ideals. It is equivalently the intersection of the ideals ${\mathrm{Ann}}_R(S)$ over all simple modules, where $\mathrm{Ann}$ is the annihilator of a module. To verify this equivalence, for a nonzero vector $v$ in a simple module $S$, the surjection $R\to S$, $r\mapsto rv$, has a maximal left-ideal kernel. Thus $x\in J(R)$ implies $xv=0$ for every $v$ in every simple module. Conversely, if $x$ annihilates all simple modules, it annihilates $1+L$ in every $R/L$, so $x\in L$. An annihilator of a module is a two-sided ideal; this description therefore proves that $J(R)$ is a two-sided ideal.

For the forward implication of the unit criterion for the Jacobson radical, first let $y\in J(R)$. If $R(1-y)$ were a proper left ideal, it would lie in a maximal one $L$. Both $y$ and $1-y$ would belong to $L$, an impossibility. Hence $c(1-y)=1$ for some $c\in R$. A left inverse alone does not yet establish that $1-y$ is a unit in a general noncommutative ring. However, $c=1+cy$, with $cy\in J(R)$, so the same argument gives $dc=1$ for some $d$. Multiplying $c(1-y)=1$ by $d$ on the left gives $d=1-y$. Therefore $(1-y)c=dc=1$ as well. This proves that $1-y$ is a unit.

For $x\in J(R)$, its two-sided-ideal property gives $axb\in J(R)$ for all $a,b$, so $1-axb$ is a unit. Conversely, if $x\notin J(R)$, choose a maximal left ideal $L$ with $x\notin L$. Then $L+Rx=R$, so $l+ax=1$ for some $l\in L$, $a\in R$. Hence $1-ax=l\in L$ is not a unit: if it were invertible, multiplying it on the left by its inverse would put $1$ in $L$. Taking $b=1$ contradicts the given unit property. We have provedFinally let $R$ be commutative and $I$ an ideal. The set $S=1+I$ contains $1$, andso it is a multiplicative subset. In the localization of a module $I_S\subseteq R_S$, any element is $i/s$ with $i\in I$ and $s\in S$. For any $r/t\in R_S$,Both $ts$ and $ts-ri$ lie in $S$, since $ts\in 1+I$ and $ri\in I$. The fraction is consequently a unit in $R_S$. In a commutative ring, the product of any two coefficients is again a single coefficient, so the proved unit criterion for the Jacobson radical applies to all expressions $1-a(i/s)b$. ThusThis proves localization away from one plus an ideal lies in the Jacobson radical. If $I=R$, then $0\in S$, $R_S=0$, and the inclusion holds trivially.