In this problem we know that the set formed ideals where the set is defined as I

Question

In this problem we know that the set formed ideals where the set is defined as
I+J to be {i+j: i E I and j E I}..
and since the cosets of ideals form rings, we have three rings: R/(I + J),
I/(I intersection J), (I + J)/J.

Prove that the latter two rings, I/(I intersection J) and (I + J)/J are isomorphic. (“Isomorphic”means there is a homomorphism from one ring to the other that is one-to-one and onto.)

Here are some hints....
(a) Start by working your way through a concrete example. Let R = Z, I be the multiples of 10 and J be the multiples of 14. Figure out what I + J, I intersection J, and (I + J)/J look like.
(b) Then create a homomorphism from I to (I + J)/J. Aft?er part a, this should be easy.
(c) What is the kernel of the homomorphism you created? Use one of the theorems we covered today to complete the proof.

Thanks for the help!

Explanation / Answer

Alright, so we're pretty much going to construct the only natural map from I to (I+J)/J. First obviously I is in I+J since we have all elements of the form i + 0 since 0 is in J. Next, theres the quotient map from I+J to (I+J)/J just taking an element of I+J to its coset in the quotient. To create our map from I to (I+J)/J, we are just going to compose these two, that is first we send x in I to itself in I+J, and then we send x in I+J to its coset x + J in (I+J)/J. Since inclusion and the quotient map are both homomorphisms, this composition, call it f: I --> (I+J)/J is automatically a well defined homomorphism. Next, we show it's surjective. Let a + J be an element of (I+J)/J. Then a is in I+J so a = x + y where x is in I and y is in J. However, since y is in J, the coset of y, y + J is just 0 + J = J itself, so a + J = x + y + J = x + 0 + J = x + J. Thus, since x is in I, and by definition, f(x) = x + J, then f(x) = a + J so for every a + J in (I + J)/J there exists an x in I with f(x) = a + J. Therefore, f is surjective. Next, we find the kernel of f. Suppose f(x) = 0 + J. Then x + J = 0 + J = J so x is in J. Therefore, x is in I and x is in J so x is in I intersect J. Thus ker f is a subset of I intersect J. Conversely, suppose x is in I intersect J, then f(x) = x + J = 0 + J = J because x is in J so x is in the ker f. Therefore I intersect J is in the ker f. Thus, ker f = I intersect J. By first isomorphism theorem, I/ker f = im f but im f = (I + J)/J because f is surjective. Therefore, I/ker f = I/(I intersect J) = (I + J)/J where by equals in that last step i mean isomorphic to.

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