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.
Yes.
How do we know that this map is well-defined?
Independent from the choices we can choose for $\displaystyle r,s$?
Well, let us assume there are two ways to expess an element in $\displaystyle (R\times S)/(I\times J)$
$\displaystyle (r_1,s_1)(I\times J)$ and $\displaystyle (r_2,s_2)(I\times J)$
Then the function maps them into,
$\displaystyle r_1I\times s_1J$, $\displaystyle r_2I\times s_2J$ respectively.
But,
$\displaystyle r_1I=r_2I$ and $\displaystyle s_1J=s_2J$ because the conditon we had said that,
$\displaystyle (r_1,s_1)(I\times J)=(r_2,s_2)(I\times J)$
Thus, the map is well-defined.
The function is a group homomorphism,
$\displaystyle \phi((r_1,s_1)(I\times J)\cdot (r_2,s_2)(I\times J))$=$\displaystyle \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)$$\displaystyle =\phi((r_1,s_1)(I\times J))\cdot \phi((r_2,s_2)(I\times J))$
The function is one-to-one,
$\displaystyle \phi((r_1,s_1)(I\times J))=\phi((r_2,s_2)(I\times J))$
$\displaystyle r_1I\times s_1J=r_2I\times s_2J$
Thus,
$\displaystyle r_1,r_2$ are both in same coset of $\displaystyle I$ and,
$\displaystyle s_1,s_2$ are both in same coset of $\displaystyle J$.
The function is onto,
That result is trivial.
Any, $\displaystyle rI\times sJ$ can be obtained from mapping the element,
$\displaystyle (r,s)(I\times J)$