Let R be a local ring and a,b in R such that (a)=(b). Prove that there exists a unit r of R such that a=rb.

We have a=rb and b=sa for some element r and s of R. Then a=rsa so a(1-rs)=0. Clearly if a=0 then b=0 and we can just take r=1. Suppose a≠0. If r∈R \ R* then, since R is local, r∈J(R) where J is the Jacobson radical of R. Then 1-rs ∈ R* so that a=0, contradiction. So r∈R* and we are done. □