Exercise 6.2

Let \(R\) be a ring, and suppose \(1\) is a zero-divisor. That is, there exists a \(b \in R\) with \(b \neq 0\) such that \(1 \cdot b = 0\) or \(b \cdot 1 = 0\). But \(1 \cdot b = b \cdot 1 = b\) since \(1\) is the multiplicative identity. So \(b = 0\), but this is a contradiction. Therefore \(1\) cannot be a zero-divisor.