Question

Let J be an ideal in R. Prove that I is an ideal, where $\mathit{I}\mathbf{=}\left\{a\in R/rt=0_{R}foreveryt\in J\right\}$.

Short answer

It is proved that Iis an ideal.

Step-by-step solution

Theorem statement

Let R be a ring and Iand S are non-empty subsets of the ring, then I is said to be the ideal if and only if it has the following two properties:

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

Satisfied property (i) of theorem

It is given that $0_{R}t=0_R$for all $t\in J$and $0_R$, so $I$is non-empty

Now, assume that $a,b\in I$and $t\in J$, then we have:

$$\begin{gathered} \left(a-b\right)t=at-bt \\ =0_R-0_R \\ =0_R \end{gathered}$$

Where $at-bt=0_R$because $a,b\in I$. As$t\in J$ was taken arbitrarily, we can conclude that $\left(a-b\right)t=0_R$, which implies that $a-b\in I$.

Hence, property (i) of the theorem is stated in step 1.

Satisfied property (ii) of theorem

Let $r\in R$and take any $t\in J$, then

$$\begin{gathered} \left(ra\right)t=r\left(at\right) \\ =r{0}_R \\ =0_R \end{gathered}$$

Where $\left(at\right)=0_R$because $a\in I$. Since $I\in J$was taken arbitrary, it can be concluded that$\left(ra\right)t=0_R$for all$t\in J$, so $ra\in I$.

Thus, property (ii) is also satisfied for the theorem stated in step 1.

Hence proved, Iis an ideal.