Question

Prove that the intersection of two ideals of a ring is again an ideal.​

Answer 1

Answer:

Step-by-step explanation:

Let A, B both be Ideals of a ring R.

Suppose I≡A∩B.

Since A and B are both Ideals of a ring R, A and B are both Subrings of a ring R. In particular, we have that (A,+),(A∖{0},⋅),(B,+),(B∖{0},⋅) are Abelian.

Now, Suppose x1,x2∈I.

I'm not entirely sure how I can justify x1+(−x2)∈I. Might be overthinking this but I might have to use the fact that I is the intersection.

Answer 2

Note :

Ring : A non empty set R equipped with two binary operations called addition and multiplication denoted by ( + ) and ( • ) is said to be a ring if the following properties holds :

1. (R,+) is an abelian group .
2. (R,•) is a semi-group .
3. (R,+,•) holds distribute law .

• a•(b + c) = a•b + a•c
• (b + c)•a = b•a + c•a

Ideal : A non empty subset U of ring R is said to be an ideal (two sided ideal) of R if :

1. a , b ∈ U → a - b ∈ U and
2. a ∈ U , r ∈ R → ar ∈ U and ra ∈ U

Solution :

To prove :

Intersection of two ideals of a ring is again an ideal .

Proof :

(Please refer to the attachment)