Suppose \(R\) is a commutative ring (with 1), \(I\) is a proper ideal in \(R\), and \(a \in R\). Suppose \((a) + I = R\). Prove that \(a+I\) is a unit (i.e., invertible) in the quotient ring \(R/I\).
(For half credit: Prove that if \(a\) and \(n\) are relatively prime integers, then \(a+n\mathbb{Z}\) is a unit in \(\mathbb{Z}/n\mathbb{Z}\).)
Solution
Because \((a)+I=R\), we have \(1\in(a)+I\). Hence there exist \(r\in R\) and \(i\in I\) such that \[ ra+i=1. \] Passing to the quotient gives \[ (r+I)(a+I)=ra+I=1+I. \] Thus \(r+I\) is an inverse of \(a+I\), so \(a+I\) is a unit in \(R/I\).
For \(R=\mathbb Z\) and \(I=n\mathbb Z\), the hypothesis \((a)+n\mathbb Z=\mathbb Z\) is exactly \(\gcd(a,n)=1\), recovering the familiar statement for \(\mathbb Z/n\mathbb Z\).