- Let \(R\) be a ring with identity \(1 \ne 0\), and let \(I \trianglelefteq R\) be an ideal of \(R\). Prove that if \(I\) contains a unit \(u \in R^\times\), then \(I = R\).
- Prove that the group of units \(R^\times\) need not be closed under addition, and provide concrete examples in standard rings.
Solution
If an ideal \(I\trianglelefteq R\) contains a unit \(u\), then it also contains \[ u^{-1}u=1. \] Hence for every \(r\in R\), \[ r=r\cdot1\in I, \] so \[ \boxed{I=R}. \] The same argument works for one-sided ideals using the appropriate side of the inverse.
The unit group \(R^\times\) is a group under multiplication, not generally under addition. For example, \[ \mathbb Z^\times=\{\pm1\}, \] but \[ 1+1=2\notin\mathbb Z^\times. \] More universally, in any nonzero unital ring, if \(u\) is a unit then so is \(-u\), while \[ u+(-u)=0 \] is not a unit. Thus \(R^\times\) need not be additively closed.