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Prove (I + J)/J is isomorphic to I(R/J) as R modules

oblixps

Member
May 20, 2012
38
Let R be a commutative ring and I, J be ideals of R. Show that (I + J)/J is isomorphic to I(R/J) as R modules.

I am having trouble coming up with the explicit isomorphism. For I(R/J) I know any element can be expressed as i(r + J) = ir + J by definition of the action of R on R/J.

As for (I + J)/J, any element can be expressed as i + j + J = i + J so i was thinking of mapping i + J to ir + J but the problem is that this map doesn't seem to be 1 - 1.

But I am having trouble coming up with any other sensible maps besides this one. Can someone offer a hint on how to proceed?

Thanks!
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
Let R be a commutative ring and I, J be ideals of R. Show that (I + J)/J is isomorphic to I(R/J) as R modules.

I am having trouble coming up with the explicit isomorphism. For I(R/J) I know any element can be expressed as i(r + J) = ir + J by definition of the action of R on R/J.

As for (I + J)/J, any element can be expressed as i + j + J = i + J so i was thinking of mapping i + J to ir + J but the problem is that this map doesn't seem to be 1 - 1.

But I am having trouble coming up with any other sensible maps besides this one. Can someone offer a hint on how to proceed?

Thanks!
It looks as though your definition of a commutative ring includes the requirement that the ring should have an identity element, otherwise this result is not true.

I think you would do best to consider the reverse of the mapping that you suggest. If $i\in I$ and $r\in R$ then $ir\in I$ (because $I$ is an ideal). The map that takes the element $i(r+J) \in I(R/J)$ to the "same" element $(ir)+J \in (I+J)/J$ is a well-defined 1–1 homomorphism (check that!). It is also an onto map, because we can take $r=1$ (the identity element of $R$) to see that $i+J \in (I+J)/J$ is the image of $i(1+J)\in I(R/J)$.
 

oblixps

Member
May 20, 2012
38
ah yes I should have mentioned that my definition of rings include 1.

thank you for your answer!

i was having trouble showing that the "reverse" of the map i originally suggested was onto but now i understand.