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


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.1 Exercise 7.1.3

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.


Solution: 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 $-1$.


Linearity

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