Ideal and Factor Ring: Proving AxB is an Ideal of RxS

[hsong9]

Homework Statement

a) If A is an ideal of R and B is an ideal of S.
Show that A x B is an ideal of R x S.

b) Show that every ideal C of RxS has the form C = AxB as in(a)
[hint: A = { a in R | (a,0) in C}]

The Attempt at a Solution

a)Since A and B are ideal of R and S, aR and Ra are subsets of A, bS and Sb are subsets of B.
Let (a,b) in AxB and (r,s) in RxS, (a,b)(r,s) = (ar,bs) in AxB since ar in A and bs in B.

b) Let A = { a in R | (a,0) in C} and
B = { b in S | (0,b) in C}
We need to show that AxB = <(a,0),(0,b)>.
my idea is correct?

 

[Billy Bob]

(a) looks good if you already have a theorem that A x B is a ring. If you don't have that theorem, then don't forget to do that routine step.

For (b), you have to show A\times B\subset C and C\subset A\times B

I regard A\times B=\langle(a,0),(0,b)\rangle\subset C as the easier direction because you only need to use the fact that C is a ring. Really there's nothing to show.

To prove C\subset A\times B, I think you need to use the facts that C is an ideal and furthermore that R and S have identities. (Are you allowed to assume R and S have identities? I hope so.)