Abstract Nonsense

Crushing one theorem at a time

First Ring Isomorphism Theorem


Point of Post: In this post we give a proof of the first ring isomorphism theorem namely that R/\ker f\cong\text{im }f. 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).

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Abstract Nonsense Proof

We now use the universal characterization of quotient rings to prove the first ring isomorphism theorem:

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Theorem: Let R and R' be rings and f:R\to R' a ring morphism. Then, R/\ker f\cong\text{im }f.

Proof: We note firstly that \text{im }f is a ring with an epimorphism f:R\twoheadrightarrow\text{im }f. Then, let S be any ring and g:R\to S a ring morphism with \ker f\subseteq \ker g. Define \eta:\text{im }f\to S to be \eta(f(x))=g(x). To see this is well-defined we merely note that if f(x)=f(y) then x=y+a with a\in\ker f\subseteq \ker g and so g(x)=g(y+a)=g(y)+g(a)=g(y). Evidently this is a morphism since

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\eta(f(x)+f(y))=\eta(f(x+y))=g(x+y)=g(x)+g(y)=\eta(f(x))+\eta(f(y))

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and

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\eta(f(x)f(y))=\eta(f(xy))=g(xy)=g(x)g(y)=\eta(f(x))\eta(f(y))

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since evidently g=\eta\circ f we may conclude from the universal characterization of quotient rings that \text{im }f\cong R/\ker f. \blacksquare

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Geometric Idea

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 f:R\to S is a morphism, and above each s\in S there sits a ‘string’ attached to it. Perhaps think about it as looking like the following:

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(ignoring the labels). We call these strings the fibers of f, namely the fiber above s is the set f^{-1}(s). An interesting question is, if we take s,s'\in S and is there a way to describe the fiber f^{-1}(s+s') in terms of f^{-1}(s) and f^{-1}(s').  It’s pretty easy to see then that f^{-1}(s+s')=f^{-1}(s)+f^{-1}(s'), namely that the fiber above s+s' is the setwise sum of the fibers of s and s'. It’s similarly easy to see that the fiber above ss' is the setwise product of f^{-1}(ss'). Thus, we see that there seems to be a natural bijection from the collection \mathcal{F} of fibers to \text{im }f given by f^{-1}(s)\mapsto s. 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 \mathcal{F}\to \text{im }f p for projection. The above shows that if we define addition and multiplication on \mathcal{F} by setwise addition and product then p is a ‘morphism’ (in the sense that p(f^{-1}(s)+f^{-1}(s'))=s+s' and p(f^{-1}(s)f^{-1}(s'))=ss'. Moreover, since each fiber lies above a different point we see that p 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 f(x)=s then f^{-1}(s)=x+\ker f. We then see that the setwise product of f^{-1}(f(x)) and f^{-1}(f(y)) is equal to x+y+\ker f and similarly f^{-1}(f(x))f^{-1}(f(y)) is equal to xy+\ker f and so the above theorem says that the map R/\ker f=\left\{x+\ker f:x\in R\right\}=\mathcal{F}\to\text{im }f given by x+\ker f\mapsto f(x) is an isomorphism which is precisely what the first ring isomorphism theorem says.

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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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June 27, 2011 - Posted by | Algebra, Ring Theory | , , , ,

13 Comments »

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