1. ## Ideals

I need some help on this problem as well:
Let R and S be rings and let I be an ideal of R and J be an ideal of S.
Prove that:
(R x S)/(I x J) is isomorphic with (R/I) x (S/J).

Thank you.

2. Originally Posted by DaisukeMatsuzaka
I need some help on this problem as well:
Let R and S be rings and let I be an ideal of R and J be an ideal of S.
Prove that:
(R x S)/(I x J) is isomorphic with (R/I) x (S/J).

Thank you.
Did you consider the map,
$\phi: (R\times S)/(I\times J)\to R/I \times S/J$
Well defined as,
$\phi((r,s)(I\times J))=rI\times sJ$

3. Thanks for the response. I really don't know what the second half means. Could you elaborate?

4. Originally Posted by DaisukeMatsuzaka
Thanks for the response. I really don't know what the second half means. Could you elaborate?
Yes.

How do we know that this map is well-defined?
Independent from the choices we can choose for $r,s$?

Well, let us assume there are two ways to expess an element in $(R\times S)/(I\times J)$
$(r_1,s_1)(I\times J)$ and $(r_2,s_2)(I\times J)$
Then the function maps them into,
$r_1I\times s_1J$, $r_2I\times s_2J$ respectively.
But,
$r_1I=r_2I$ and $s_1J=s_2J$ because the conditon we had said that,
$(r_1,s_1)(I\times J)=(r_2,s_2)(I\times J)$
Thus, the map is well-defined.

The function is a group homomorphism,
$\phi((r_1,s_1)(I\times J)\cdot (r_2,s_2)(I\times J))$= $\phi((r_1r_2,s_1s_2)(I\times J))=r_1r_2I\times s_1s_2J=(r_1I\times s_1J)(r_2I\times s_2J)$ $=\phi((r_1,s_1)(I\times J))\cdot \phi((r_2,s_2)(I\times J))$

The function is one-to-one,
$\phi((r_1,s_1)(I\times J))=\phi((r_2,s_2)(I\times J))$
$r_1I\times s_1J=r_2I\times s_2J$
Thus,
$r_1,r_2$ are both in same coset of $I$ and,
$s_1,s_2$ are both in same coset of $J$.

The function is onto,
That result is trivial.
Any, $rI\times sJ$ can be obtained from mapping the element,
$(r,s)(I\times J)$