Question

Let I be an ideal in R. Prove that K is an ideal, where $\mathit{K}\mathbf{=}\left\{a\in R/ra=Iforeveryr\in R\right\}$.

Short answer

It is proved that K is 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 ideal if and only if it has the following two 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$ar\in I$ .

Satisfied property (i) of theorem

It is given that $r{0}_R=0_R\in I$for all $r\in R$and $0_R\in K$, so $K$is non-empty.

Now, assume that $a,b\in K$and $s\in R$, then we have:

Because I is closed under subtraction, it is known that$s\in R$ was taken arbitrary, from which we can conclude that $\left(a-b\right)\in K$.

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

Satisfied property (ii) of theorem

Let $r\in R$and take any $s\in R$, then

$s\left(ra\right)=\left(sr\right)a\in I$

Where $\left(sr\right)a\in I$because $s\in R$. Since was taken arbitrary, from which it can be concluded that $s\left(ra\right)\in I$for all $s\in R$, so$ra\in K$ .

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

Hence, it is proved that K is an ideal.