Question

Let$\mathit{I}$ be an ideal in commutative ring$\mathit{R}$ and let $\mathit{J}\mathbf{=}\left\{r\in R\overline{)r^n\in Iforsomepositiveintegern}\right\}$.

Prove that$\mathit{J}$ is an ideal that contains $\mathit{I}$. [Hint: You will need the Binomial Theorem from Appendix E. Exercise 30 is the case when $\mathit{I}\mathbf{=}\left(0_R\right)$.]

Short answer

It is proved that J is an ideal that contains I.

Step-by-step solution

Write the conditions for theorem 6.1

A non-empty subset$I$ of a ring$R$ is an ideal if and only if it follows the below-listed properties:

1. If$a,b\in I$then$a-b\in I$
2. If $r\in R$and $a\in I$, then $ra\in I$and$ar\in I$

Show that J is an ideal that contains I

If$a\in I$ then $a=a^{-1}\in J$, so $I\underline{\subset J}$. To verify$J$ is an ideal we show that the conditions for theorem 6.1 are satisfied.

1. If $r,s\in J$with $r^m,s^n\in I$for $m,n\in Z_{+}$then${\left(r-s\right)}^{m+n}=\sum _{i=0}^{m+n}\left(\begin{array}{c}m+n\\ i\end{array}\right){\left(-1\right)}^i{r}^{m+n-i}{s}^i\in I$

Since the range $0\le i\le n$we have $r^{m+n-i}=r^m{r}^{n-i}\in I$and in the range $n+1\le i\le m+n$we have $s^i=s^n{s}^{i-n}\in I$. Therefore, $r-s\in J$.

1. If $r\in J$with $r^m\in I$for $m\in Z_{+}$and $s\in R$then, ${\left(rs\right)}^m={\left(sr\right)}^m=s^m{r}^m\in I$

So, $rs=sr\in J$.

Hence, the given statement is proved.