Ring Isomorphism

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November 1st 2008, 01:50 PM
vincisonfire

Ring Isomorphism

Let R and S be rings and let I ▹R, J ▹S be ideals. Prove that
$(R\times S)/(I\times J) \simeq (R/I) \times (S/J)$
I think
$(R\times S)/(I\times J)$ is of the form $(a,b) + (I,J) : a \in R, b \in S$ that is also of the form $(a + I, b + J) : a \in R, b \in S$ because of definition of addition.
This is the "cross set" $(R/I) \times (S/J)$ .
Something tells me it can't be that simple. Can you tell me why?
November 1st 2008, 05:18 PM
ThePerfectHacker

Quote:

Originally Posted by vincisonfire

Let R and S be rings and let I ▹R, J ▹S be ideals. Prove that
$(R\times S)/(I\times J) \simeq (R/I) \times (S/J)$
I think
$(R\times S)/(I\times J)$ is of the form $(a,b) + (I,J) : a \in R, b \in S$ that is also of the form $(a + I, b + J) : a \in R, b \in S$ because of definition of addition.
This is the "cross set" $(R/I) \times (S/J)$ .
Something tells me it can't be that simple. Can you tell me why?

1)First prove $I\times J\triangleleft R\times S$
2)Define $\phi : (R\times S)/(I\times J) \to (R/I)\times (S/J)$ by $\phi ((a,b)(I\times J)) = aI\times bJ$
3)Prove $\phi$ is well-defined.
4)Prove $\phi$ is one-to-one.
5)Prove $\phi$ is a ring homorphism.
November 2nd 2008, 02:57 AM
vincisonfire

Aren't $aI$ and $bJ$ always equal to zero in $<br /> R/I$ and $S/J$ .
The function maps everything to (0 , 0) = 0?