Question

Let S be a subring of a ring$R$ with identity.

(a) If $S$has an identity, show by example that$1_R$ may not be the same as$1_R$ .

1. If both $RandS$are integral domains, prove that $1_S=1_R$?

Short answer

(a) It is proved that if $S$has an identity, then $1_R$ may not be equal to $1_S$.

(b) Hence, when both rings are integral domains, then $1_R=1_S$.

Step-by-step solution

Property of Rings:

If any ringis designated as $R$,such that $a,b\in R$, then:

$a=b\iff a-b=0$

Proof:

It is given that $S$has an identity which is a subring of $R$. Let there be a set:

$S=\left\{0,2,4,6,8\right\}$.

In which, we have:

$r=2,s=5andK=1$

Thus, the set is given by:

$S=\left\{0,r,2r,.........,\left(s-1\right)r\right\}$

But, the identity of the subring is:$Ks+1=6\ne 1$

Hence it is proved that if $S$ has an identity, then$1_R$ may not be equal to $1_s$.

Proof:

Let $RandS$be integral domains where $a\in S$and$a\ne 0$. Then, we have:

$$\begin{gathered} a\in Randa\in S \\ \Rightarrow a{1}_R=a{1}_S=a \\ \Rightarrow a{1}_R-a{1}_S=0_R \\ \Rightarrow a\left(1_R-1_S\right)=0_R \\ \Rightarrow 1_R-1_S=0_R \\ \Rightarrow 1_R=0_R+1_S \\ \Rightarrow 1_R=1_S \end{gathered}$$

Hence, when both rings are integral domains, then $1_R=1_S$.