= Solution
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 \b[$R/L$ 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 $\operatorname{Ann}_R(S)$ over all <simple modules>, where $\operatorname{Ann}$ 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 $d c=1$ 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 proved
$$
\boxed{J(R)=\{x\in R:1-axb\text{ is a unit for every }a,b\in R\}.}
$$
Finally let $R$ be <commutative> and $I$ an <ideal>. The set $S=1+I$ contains $1$, and
$$
(1+i)(1+j)=1+(i+j+ij)\in1+I,
$$
so 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 $r/t\in R_S$,
$$
1-\frac rt\frac is=\frac{ts-ri}{ts}.
$$
Both $ts$ and $ts-ri$ lie in $S$, since $ts\in1+I$ 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 $1-a(i/s)b$. Thus
$$
\boxed{I_S\subseteq J(R_S).}
$$
This 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.
Back to article page
