Question

Let $R$be a commutative ring with identity and $b\in R$ . Let $T$ be the subring of all multiples of $b$ (as in Exercise 8). If $u$ is a unit in $R$ and $u\in T$, prove that $T=R$.

Short answer

Hence, it is proved that $T=R$if $u$ is a unit in $R$.

Step-by-step solution

Property of Rings: 

If any ringis designated as$R$,such that , then, if the ring is an integral domain, we have:

role="math" localid="1646828965859" $a·0=0$

Cancellation Rule: 

The given subring can be expressed as:

$T=\left\{rb|r\in R\right\}$

Where the given ring $R$is commutative with identity, and let it have an element as: $b\in R$

Let the subring $T$have a unit element as: $u\in T$.

Then, there exists a unit element $u^{\text{'}}=u^{\text{'}}b$such that:

$u=u^{\text{'}}b$

Therefore, we have:

$T\subseteq R·········\left\{1\right\}$

Now, for any element y,we have:

$\begin{array}{rcl}y& =& 1_{R}y=u{u}^{-1}y\\ u{u}^{-1}yb& =& \left(u{u}^{-1}y\right)b\\ & \Rightarrow & u^{-1}\in R\end{array}$

This implies:

$R\subseteq T···············\left\{2\right\}$

From 1 and 2, we get:

$T=R$

Hence it is proved that $T=R$ if $u$ is a unit in $R$.