Question

If I and J are ideals in R, prove that$\mathit{I}\mathbf{\cap }\mathit{J}$ is an ideal.

Short answer

It is proved$I\cap J$ is an ideal.

Step-by-step solution

Theorem

Let R be a ring, and Ibea non-empty subset of the ring;then,I is said to be the ideal if and only if it has the following two properties:

(i) If $a,b\in I$, then $a-b\in I$.

(ii) If$r\in R$and $a\in I$, then $ra\in I$, and $ar\in I$.

Proof

It is given that I and J are ideals and both are non-empty, so by using the theorem stated in step 1,$I\cap J$is non-empty.

Now, assume that $a,b\in I\cap J$;then using the definition of intersection, we have$a,b\in I$and $a,b\in J$.

As I and J are ideals, so$a-b\in I$and $a-b\in J$, which implies $a-b\in I\cap J$. Thus,property (i) ofthe theorem is satisfied.

Next, assume that $r\in R$.

As$a\in I$ and I is an ideal, we have$ra\in I$and $ar\in I$.

Similarly,$ra\in J$and $ar\in J$.

By definition, we have$ra\in I\cap J$and $ar\in I\cap J$, so property (ii) of the theorem is satisfied.

Hence, it is proved$I\cap J$is an ideal.