In a subring containing the identity, units are units in the superring

Let R be a ring with identity and let S \subseteq R be a subring containing 1. Prove that if u is a unit in S then u is also a unit in R. Show by example that the converse is false.


Since 1_R \in S, S has a 1 and we have 1_S = 1_R. If u is a unit in S, then there exists an element v \in S such that uv = vu = 1_S = 1_R. Since v is in R, u is a unit in R as well.

The integers \mathbb{Z} and the rationals \mathbb{Q} are rings, with \mathbb{Z} \subseteq \mathbb{Q} a subring, \mathbb{Q} has a 1 (namely 1), and 1 \in \mathbb{Z}. Now 2 \in \mathbb{Q} is a unit since 2 \cdot \frac{1}{2} = 1. However, by Example 1 in the text, the only units in \mathbb{Z} are 1 and -1.

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