Question

If $\mathit{f}\mathbf{:}\mathit{R}\mathbf{\to }\mathit{S}$is an isomorphism of rings, which of the following properties

are preserved by this isomorphism? Justify your answers.

(a) $\mathit{a}\mathbf{\in }\mathit{R}$is a zero divisor.

(b) $\mathit{a}\mathbf{\in }\mathit{R}$is idempotent.*

(c) $\mathit{R}$is an integral domain.

Short answer

It is proved that:

(a) $a\in R$is a zero divisor.

(b)role="math" localid="1648059459505" $R$ is an idempotent.

(c) $R$and $S$are integral domains.

Step-by-step solution

Determine a∈R is a zero divisor

(a)

Let’s consider $a$is an arbitrary element of ring $R$and function $f:R\to S$is an isomorphism of rings.

$a\in R$is zero divisor,

$\Rightarrow a\ne 0_{R}and\exists b\in R,b\ne 0_{R}sothatab=0_R$

$$\begin{gathered} \Rightarrow 0_S\stackrel{Theorem3.10}{=}f\left(0_R\right)=f\left(ab\right)=f\left(a\right)f\left(b\right) \\ \left[f\left(a\right)\ne 0_{S}andf\left(b\right)\ne 0_S\left(a\ne 0\_b\ne 0\_R,f\left(0\_R\right)=0\_S\right)\right] \end{gathered}$$

and $f$is an injection.

$\Rightarrow f\left(a\right)$is a zero divisor

Hence, $a\in R$is a zero divisor.

Determine a∈Ris an idempotent*

(b)

If $a^2=a$then,$a\in R$is called idempotent.

$a\in R$is idempotent$\Rightarrow a^2=a$.

$\Rightarrow f\left(a\right)=f\left(a^2\right)=f\left(aa\right)=f\left(a\right)f\left(a\right)=f{\left(a\right)}^2$

$\Rightarrow f\left(a\right)$is idempotent

Therefore, $R$is an idempotent.

Hence, the property of being idempotent is isomorphism.

Determine R  is an integral domain

(c)

Let’s consider that $R$is an integral domain and $f:R\to S$is an isomorphism of ring.

Then, $R$is a commutative ring with identity.

As $f$is a surjection then, the arbitrary elements $c$and $d$in $S$sheet exists $a,b\in R$so that $f\left(a\right)-c$and $f\left(b\right)-d$.

Then, it can be:

$cd=f\left(a\right)f\left(b\right)=f\left(ab\right)=f\left(b\right)f\left(a\right)=dc$

The multiplication condition satisfies so $S$is a commutative ring.

Using Theorem 3.10, $S$is ring with identity$1_s=f\left(1_R\right)$

Assume that, $cd=0_s$then,

$$\begin{gathered} cd=0_s\Rightarrow f\left(a\right)f\left(b\right)=0_s\Rightarrow f\left(ab\right)=0_s \\ \Rightarrow ab=0_s \\ \Rightarrow a=0_{R}orb=0_s\Rightarrow f\left(a\right)=f\left(0_R\right)orf\left(b\right)=f\left(0_R\right) \\ \Rightarrow c=0_{s}ord=0_s \end{gathered}$$

Hence, $S$is an integral domain.

Therefore, $R$and $S$are integral domains and$f:R\to S$ is an isomorphism.