let R be a ring. i need help proving that the intersection of arbitrary non-empty collection of ideals in R is an ideal of R. and i dont want to assume that the collection is countable.iwas able to prove it when i just had 2 ideals I,J of R. not sure how to do it when the collection is uncountable. thx will give 24fg to whoever can help me out

Let (R,+,x) a ring.
Let I a non-empty set (countable or not) and let Ji ( i ∈ I ) a collection of ideals of R.
I assume here we're talking about bilateral ideals (but this doesn't change anything in the end).

Let J = ∩ Ji, for i ∈ I.

We want to prove that J is an ideal of R, ie :
- J is an additive subgroub of R,
- for every j in J, for every r in R, jr and rj are in J.

For every i in I, Ji is an additive subgroub of R, hence 0 lies in Ji for every i in I : 0 lies in J.
Let x, y in J. Hence x and y belong to every Ji, thus x-y is in every Ji, since they all are subgroubs of (R,+).
This shows that J is an additive subgroub of R.

Now let j in J, and r in R.
Since j is in J, it is in every Ji. Since every Ji is an ideal, jr and rj are in Ji, for every i in I : jr and rj are in J.

So, J is an ideal of R.

this is similar to wat i was thinking of doing but wasnt sure if it deals with uncountable collection of ideals