First Ring Isomorphism Theorem
Point of Post: In this post we give a proof of the first ring isomorphism theorem namely that . This will follow very easily from the universal characterization of quotient rings. Thus, most of the post will be motivating this theorem in a more geometric sense and then giving a different (more intuitive proof).
Abstract Nonsense Proof
We now use the universal characterization of quotient rings to prove the first ring isomorphism theorem:
Theorem: Let and be rings and a ring morphism. Then, .
Proof: We note firstly that is a ring with an epimorphism . Then, let be any ring and a ring morphism with . Define to be . To see this is well-defined we merely note that if then with and so . Evidently this is a morphism since
since evidently we may conclude from the universal characterization of quotient rings that .
I may have played up the geometric ‘proof’ which I should now be discussing, perhaps a better way to put it is that there is a nice geometric way to see what’s going on, and once you see this the proof to the first ring isomorphism theorem is immediate. Namely, imagine that is a morphism, and above each there sits a ‘string’ attached to it. Perhaps think about it as looking like the following:
(ignoring the labels). We call these strings the fibers of , namely the fiber above is the set . An interesting question is, if we take and is there a way to describe the fiber in terms of and . It’s pretty easy to see then that , namely that the fiber above is the setwise sum of the fibers of and . It’s similarly easy to see that the fiber above is the setwise product of . Thus, we see that there seems to be a natural bijection from the collection of fibers to given by . Using our picture we see this as taking each of the fibers and ‘projecting it down’ to the point it lies over. Let’s call this map for projection. The above shows that if we define addition and multiplication on by setwise addition and product then is a ‘morphism’ (in the sense that and . Moreover, since each fiber lies above a different point we see that is a bijection. So, what does this have to do with the first ring isomorphism theorem? It all comes nicely together when we note that if then . We then see that the setwise product of and is equal to and similarly is equal to and so the above theorem says that the map given by is an isomorphism which is precisely what the first ring isomorphism theorem says.
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.