Problem

Let \(R\) be a ring with identity and let \(S\) be a subring of \(R\) containing the identity. Prove that if \(u\) is a unit in \(S\) then \(u\) is a unit in \(R\). Show by example that the converse is false.

Solution

Suppose \(u\) is a unit in \(S\). By definition of unit, there is some \(v \in S\) such that \(uv = vu = 1\).

Since \(S\) is a subring of \(R\), \(u \in R\) and \(v \in R\). Therefore, the property \(uv = vu = 1\) also holds in \(R\).

Therefore, \(u\) is a unit in \(R\).

$$\tag*{$\blacksquare$}$$

To see that the converse is not true, consider \(\mathbb{Z}\), which is a subring of \(\mathbb{Q}\).

\(2 \in \mathbb{Q}\) is a unit because \(\frac{1}{2} \in \mathbb{Q}\) and \(2 \times \frac{1}{2} = \frac{1}{2} \times 2 = 1\).

However, \(\frac{1}{2} \notin \mathbb{Z}\) and there is no \(x \in \mathbb{Z}\) such that \(2 \times x = x \times 2 = 1\).

Therefore, \(2\) is a unit in \(\mathbb{Q}\), but is not in its subring \(\mathbb{Z}\).

$$\tag*{$\blacksquare$}$$