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One aspect of category theory that caught my eye is that it can give simultaneously prove the 1st isomorphism theorem for groups/rings/fields/vector spaces/... Yet whenever I look up works on category theory, it takes hundreds of pages to reach this theorem. Is it possible to prove this result using substantially less theory, and if so, is there any work which provides such a proof?

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But you don't have to read all these leading hundreds of pages to read the particular proof you want. . –  Regenbogen May 15 '10 at 14:23
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What is the first isomorphism theorem for fields? Also, I don't think categories are a framework to prove such things, from my point of view the isomorphism theorems are usually taken as axioms for a certain type of categories, say abelian categories, and then one proves that e.g. vector spaces form an abelian category. –  Xandi Tuni May 15 '10 at 14:36
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What specific category theory work reaches the first isomorphism theorem after hundreds of pages? –  S. Carnahan May 15 '10 at 15:06
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I'm having trouble understanding your motivation. You can simultaneously prove the first isomorphism theorem for groups, rings and modules just by understanding the proof in any one of the cases: the arguments are all the same. –  Pete L. Clark May 15 '10 at 15:14
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Perhaps instead of category theory, you should look at some basic book on universal algebra, for example, you can try part 3 of Cohn’s algebra. –  user2734 May 15 '10 at 16:11

4 Answers 4

up vote 4 down vote accepted

If you want a categorical proof that encompasses nonabelian groups and rings-without-identity, then you need to work with concepts such as "normal subobject" that are (I think) not really part of the category theory that most mathematicians routinely encounter. You may also find that a proof is not particularly enlightening at this level of generality. I think this might explain why a simultaneous generalization does not appear early in textbooks.

I'll consider the formulation of the first isomorphism theorem which states that for any homomorphism $f: A \to B$, $ker(f): K \to A$ is a normal subobject, and the image of $f$ in $B$ is isomorphic to the quotient of $A$ by $K$. The formalism is roughly as follows: We work in a category in which every morphism has a kernel and a cokernel. Then any morphism $f: A \to B$ factors into a composition of $coker(ker(f)): A \to A/K$, $m: A/K \to N$, and $ker(coker(f)): N \to B$, and the diagram is defined up to unique isomorphism by the universal properties of kernel and cokernel. For the first part of the theorem, one typically defines normal subobject to be a kernel (although there are alternative definitions involving congruences), so we reduce to the question of the image. Here we have a conflict of terminology, since the "image" is often defined in categories as the kernel of the cokernel, and the first isomorphism theorem interprets image set-theoretically. Given this alternative interpretation, the theorem amounts to the assertion that $m$ and $ker(coker(f))$ are monomorphisms. Here we encounter another problem, since as far as I know, $m$ does not need to be a monomorphism in general.

At this point, we need to check that the categories we like have kernels and cokernels. The kernel of a group homomorphism $f:A \to B$ is (the inclusion of) the preimage of the identity, and the cokernel of $f$ is given by taking the normal hull $N$ of $f(A)$ in $B$, and taking the quotient group $B/N$. We get our theorem because the maps of $f(A)$ into $N$ and $N$ into $B$ are inclusions, hence monomorphisms. A similar argument works for rings-without-identity. For abelian groups and vector spaces, the map $m: f(A) \to N$ is an isomorphism. For fields, there are no kernels, so the first isomorphism theorem doesn't make sense.

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On Abelian situation "Theory of categories" (BArry Mitchell) is easy and short.

In general algebras see: "Universal Algebra" by P. M. Cohn.

For a general categorical tractate about algebraic proprieties of algebraic categorical structures see: F. Borceux and D. Bourn, Malcev, Protomodular, Homological and Semi-Abelian Categories.

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Let $C$ be a finitary algebraic category. That is objects are sets with additional structure (given by some operations of certain arities, satisfying some equations) and morphisms are structure preserving maps.

Now the isomorphism theorem reads as follows: Let $f : A \to B$ a morphism and $\sim$ a congruence relation on $A$ such that $a \sim b \Rightarrow f(a)=f(b)$. Then $f$ factors uniquely through the projection $A \to A/\sim$ and a morphism $\overline{f} : A/\sim \to B$. The image of $f$ coincides with the image of $\overline{f}$. Besides, $\overline{f}$ is injective iff $a \sim b \Leftrightarrow f(a)=f(b)$.

The proof is straight forward and generalizes the isomorphism theorems for groups/rings/... by finding alternative descriptions of congruence relations.

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Try "Introduction to the Theory of Categories and Functors" by Bucur and Deleanu. They have a treatment of the isomorphism theorems in abelian categories beginning on page 101. They start by noting that the first isomorphism theorem follows from the definition of abelian category. Then they go on to prove the second and third theorems. Their discussion of abelian categories only begins on page 87, and they finish the third isomorphism theorem on page 111. If you already know the preliminaries, you might be able to get right to the heart of the matter without too much trouble.

Edit: I just stumbled onto a paper by Barros and Pombo, "A direct proof of Noether’s second isomorphism theorem for abelian categories". It's only 8 pages long. They reference Bucur & Deleanu, Freyd, and Grothendieck. You can find it at http://www.scielo.org.co/pdf/rcm/v43n1/v43n1a04.pdf.

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