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Let R be a ring, S a subring of R and I an ideal of R. Prove that I ∩ S is an i...

Jun 4, 2024

Let R be a ring, S a subring of R and I an ideal of R. Prove that I ∩ S is an ideal of S.

Solution by Steps

step 1

To prove that

I \cap S

is an ideal of

S

, we need to show that it satisfies two conditions: (1) it is a subring of

S

, and (2) it is closed under multiplication by elements of

S

step 2

First, we show that

I \cap S

is a subring of

S

. Since

I

is an ideal of

R

, it is a subring of

R

. Therefore,

I \cap S

is a subset of

S

and inherits the ring operations from

R

step 3

To show that

I \cap S

is closed under addition and contains the additive identity, let

a, b \in I \cap S

. Since

a, b \in I

and

I

is an ideal,

a + b \in I

. Also, since

a, b \in S

and

S

is a subring,

a + b \in S

. Therefore,

a + b \in I \cap S

step 4

The additive identity

0

R

is in

I

because

I

is an ideal, and

0 \in S

because

S

is a subring. Therefore,

0 \in I \cap S

step 5

To show that

I \cap S

is closed under additive inverses, let

a \in I \cap S

. Since

a \in I

and

I

is an ideal,

-a \in I

. Also, since

a \in S

and

S

is a subring,

-a \in S

. Therefore,

-a \in I \cap S

step 6

Now, we show that

I \cap S

is closed under multiplication by elements of

S

. Let

a \in I \cap S

and

s \in S

. Since

a \in I

and

I

is an ideal,

sa \in I

. Also, since

a, s \in S

and

S

is a subring,

sa \in S

. Therefore,

sa \in I \cap S

step 7

Since

I \cap S

satisfies both conditions, it is an ideal of

S

Answer

I \cap S

is an ideal of

S

Key Concept

Ideal in a subring

Explanation

The intersection of an ideal of a ring with a subring is itself an ideal of the subring.