# Abstract Nonsense

## 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.

June 27, 2011 -

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