Question

Let $\mathit{R}$be a ring without identity. Let $\mathit{T}$be the ring with identity of Exercise 32 in Section 3.2. Show that $\mathit{R}$is isomorphic to the subring $\mathit{R}$of $\mathit{T}$. Thus, if $\mathit{R}$is identified with $\mathit{R}$, then $\mathit{R}$is a subring of a ring with identity.

Short answer

It is shown that $f$is an isomorphism and $R$is a subring of a ring with identity.

Step-by-step solution

Determine f is injective

Consider the give section 3.2, the ring with addition and multiplication is given by,

$\begin{array}{rcl}T& =& R\times \mathrm{\mathbb{Z}}\\ \left(r,m\right)+\left(s,n\right)& =& \left(r+s,m+n\right)\\ \left(r,m\right)\left(s,n\right)& =& \left(rs,ms,nr,mn\right)\end{array}$

And also $\overline{R}=\left\{\left(r,0\right)\overline{)r\in R}\right\}$is subring of $T$.

Let’s consider the given function, $f:R\to \overline{R}$and it’s given by $f\left(r\right)=\left(r,0\right)$.

If $r,s\in R$and role="math" localid="1648214333167" $f\left(r\right)=f\left(s\right)$then,

$\begin{array}{rcl}f\left(r\right)& =& f\left(s\right)\\ \left(r,0\right)& =& \left(s,0\right)\\ r& =& s\end{array}$

Therefore,$f$ is injective.

Determine R is a subring of a ring with identity

If $\left(r,0\right)$is an arbitrary element in $\overline{R}$.

$\left(r,0\right)=f\left(r\right)$for some $r\in R$that means that $f$is surjective.

Now, let’s consider that, $r,s\in R$then,

$$\begin{gathered} f\left(r+s\right)=\left(r+s,0\right)=\left(r+s,0+0\right)=\left(r,0\right)+\left(s,0\right)=f\left(r\right)+f\left(s\right) \\ f\left(rs\right)=\left(rs,0\right)=\left(rs+0_R,0·0\right)=\left(rs+0s+0r,0·0\right)=\left(r,0\right)\left(s,0\right)=f\left(r\right)f\left(s\right) \end{gathered}$$

Therefore, $f$is a homomorphism of rings.

Hence, $f$is an isomorphism and $R$is a subring of a ring with identity.