Let \(R\) be a ring with identity and let \(J(R)\) be its Jacobson radical. Prove that \[ x\in J(R)\iff 1-xr\in R^\times\quad\text{for every }r\in R. \]
Solution
We use the description \[ J(R)=\bigcap_{M\text{ maximal left ideal}}M. \]
1. If \(j\in J(R)\), then \(1-j\) is a unit.
If the left ideal \(R(1-j)\) were proper, it would lie in a maximal left ideal \(M\). Since \(j\in J(R)\subseteq M\), we would have \[ 1=(1-j)+j\in M, \] a contradiction. Hence \(R(1-j)=R\), so some \(u\in R\) satisfies \[ u(1-j)=1. \] Then \[ 1-u=-uj\in J(R) \] because \(J(R)\) is a two-sided ideal. Applying the same argument to \(1-u\in J(R)\) shows that \(u=1-(1-u)\) has a left inverse \(v\), so \(vu=1\). If \(a=1-j\), then \(ua=1\) and \[ v=v(ua)=(vu)a=a. \] Thus \(au=1\) as well, and \(a=1-j\) is a unit.
2. If \(x\in J(R)\), then \(1-xr\) is a unit for every \(r\in R\).
Since \(J(R)\) is a two-sided ideal, \(xr\in J(R)\). Apply 1 to \(j=xr\).
3. Conversely, suppose \(1-xr\) is a unit for every \(r\in R\). Then \(x\in J(R)\).
We first use the elementary identity: if \(1-ab\) is a unit, then so is \(1-ba\), with inverse \[ (1-ba)^{-1}=1+b(1-ab)^{-1}a. \] Hence the hypothesis also gives that \(1-rx\) is a unit for every \(r\in R\).
Let \(M\) be a maximal left ideal. If \(x\notin M\), then maximality gives \[ M+Rx=R. \] Thus \(1=m+rx\) for some \(m\in M\) and \(r\in R\), so \[ m=1-rx. \] But \(1-rx\) is a unit, impossible for an element of the proper left ideal \(M\). Therefore \(x\in M\) for every maximal left ideal \(M\), hence \(x\in J(R)\).
Thus \[ x\in J(R)\iff 1-xR\subseteq R^\times. \]