## Third Ring Isomorphism Theorem

**Point of Post: **In this post we prove the third ring isomorphism theorem which takes the form .

*Motivation*

Similar to the second ring isomorphism theorem the third isomorphism theorem has an almost immediate proof from the first ring isomorphism theorem (which, of course, both really should harken back to the proofs of the first and second group isomorphism theorems since the ring isomorphism theorems are generalizations (for abelian groups at least) if we define the trivial product on given by for all ). Thus, just like the second ring isomorphism theorem the real meat of the theorem, the real thing that need be considered is precisely how the question “is ?’ naturally comes up in our study of rings. So, suppose for a second that we have some ring and two -ideals and with . We first note that and moreover that is an ideal of since it’s the image of under the canonical projection and we know that ideals are preserved under epimorphisms. So, it makes sense to consider the quotient ring . That said, the elements of this quotient ring are, for lack of a better word, horrific. I mean, the general element looks like . A wish would be that there was some nice way to describe this ‘double quotient ring’ as something nicer, something less unwieldy. The third ring isomorphism theorem grants us this wish by telling us that the ‘s “cancel” in the sense that . To see the intuition of this theorem let’s look at the case when . We have by previous comment that the ideals of are all of the form for some . So, suppose we had two -ideals, say and . We note that evidently if and only if .We then note that what ‘looks like’ (by definition) the set . So, now let’s list the elements of . They look like:

We note then that the first of these are distinct since

if . That said, for elements beyond the first listed we see the elements are of the form for in which case it is equal to since . In particular we see that . Thus, from this we are able to conclude that as a set .Moreover, it’s easy to extract from this that for we have that

and similarly

Thus, is an isomorphism, and since the conclusion follows.

This example should give you an intuition then why the full theorem is true, with the intuitive idea that what makes this theorem true is precisely the reason why –once the first hits, the second isn’t affected.

*Third Ring Isomorphism Theorem*

We now actually prove the third ring isomorphism theorem. The proof is much, much simpler than the above fully-worked example for why if . Most of this is masked because we’ll use the first isomorphism theorem

**Theorem: ***Let be a ring with ideals and with . Then, *

**Proof: **Define by . It’s not evident that this map is well-defined–to see this suppose that then and so by definition , from where well-definedness follows. We now claim that is a morphism. To see this let be arbitrary, we see then that

and

Evidently is epic (an epimorphism) and by definition we see that if and only if , or in oher words if and only if . The conclusion then follows from the first ring isomorphism theorem.

**References:**

1. Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. *Basic Abstract Algebra*. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

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