Question

If I is an ideal in R and S is a subring of R, prove that $\mathit{I}\mathbf{\cap }\mathit{S}$is an ideal in S.

Short answer

It is proved $I\cap S$is an idealin S.

Step-by-step solution

Theorem

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

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

Proof

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

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

As I and S are ideals, $a-b\in I$ and $a-b\in S$, which implies that$a-b\in I\cap S$, so property (i) of the theorem is satisfied.

Next, assume that$s\in S$ .

As $a\in I$ and I is an ideal, we have $sa\in I$and$as\in I$.

By definition, we have $sa\in I\cap S$and $as\in I\cap S$, so property (ii) of the theorem is satisfied.

Hence, proved that $I\cap S$ is an idea in S.