Abstract Nonsense

Crushing one theorem at a time

The Module Isomorphism Theorems


Point of Post: In this post we discuss and prove the four big module isomorphism theorems.

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Motivation

As always, when discussing a new algebraic theory it is always important, after one defines subobjects and quotient objects to discuss the isomorphism theorems. While this all might seem quite vague there is actually a rigorous home for such statements in the subject of universal algebra–there one can phrase the isomorphism theorems in their full generality with congruences. Regardless in this post we discuss the four module isomorphism theorems which have their analogues in ring theory. In particular, the four isomorphism theorems we are going to discuss have the same feel and motivation as the first, second, third, and fourth ring isomorphism theorems with “ring” replaced by “left R-module” and “ideal” replaced by “submodule”.

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The Isomorphism Theorems

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We introduce a bit of notation. For a R-morphism f:M\to N we call the space M/\ker f the coimage of f and denote it \text{coim }f. The first isomorphism theorem amounts to the statement that for left R-modules one has that \text{im }f\cong\text{coim }f as left R-modules. Indeed, by the universal characterization of quotient modules it will suffice for us to prove that \text{im }f has the same universal mapping property as f. In other words, we have the epimorphism f:M\twoheadrightarrow\text{im }f\leqslant N and we’d like to show that every time there is a map g:M\to L with \ker f\subseteq\ker g then there exists a unique map \zeta:\text{im }f\to L with g=\zeta\circ f. But, it’s clear how to do this, namely defining \zeta(f(x))=g(x)–since \ker f\subseteq\ker g this map is well-defined. Moreover, since f is epi (so that \zeta\circ f=\eta\circ f implies \zeta=\eta) we have that \zeta is the unique map with this property. From this we may conclude that \text{im }f satisfies the universal characterization of quotient modules and so we may conclude the following:

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Theorem (First Module Isomorphism Theorem): Let R be a ring and M and N two left R-modules. Then, given any R-morphism f:M\to N one has that \text{im }f\cong\text{coim }f, canonically.

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Theorem(Second Module Isomorphism Theorem): Let R be a ring and let M be a left R-module with L\leqslant N\leqslant M. Then, N/L is a submodule of M/L and (M/L)/(N/L)\cong M/N.

Proof: Evidently N/L\leqslant M/L since it’s the image of N under the R-morphism \pi. To see that this isomorphism is true, define M/L\to M/N by x+L\mapsto x+N. Note that since L\subseteq N this is well-defined for any two representatives of x+L differ by a L-element and in particular, differ only be a N-element and so will have equal N-cosets. Clearly this map is a R-epimorphism and has kernel equal to \left\{x+L:x\in N\right\}=N/L. The rest follows from the first isomorphism theorem. \blacksquare

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Theorem(Third Module Isomorphism Theorem): Let R be a ring and M a left R-module. Then, if N,L\leqslant M one has that \displaystyle N/(N\cap L)\cong (N+L)/L.

Proof: Define f:N\to (N+L)/L by n\mapsto n+L. We claim that that f is an epi. Indeed, evidently f is a morphism and if (n+\ell)+L\in (N+L)/L then we evidently see that f(n)=n+L=n+L+(0+L)=(n+L)+(\ell+L)=(n+\ell)+L from where epiness follows. Clearly though we have that \ker f=\left\{n\in N:n\in L\right\}=N\cap L and so the rest follows from the first isomorphism theorem. \blacksquare

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Theorem(Fourth Module Isomorphism Theorem): Let R be a ring and M a left R-module and let N\leqslant M. Then, the map L\mapsto L/N is an isomorphism of posets, between the submodules of R containing N and the submodules of M/N.

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References:

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

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. Module Theory. Clarendon, 1990. Print.

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November 11, 2011 - Posted by | Algebra, Module Theory, Ring Theory | , , , , , , , ,

5 Comments »

  1. […] Proof: We merely define a map by sending to . This is evidently a -epimorphism and since evidently its kernel is we may conclude by the first isomorphism theorem. […]

    Pingback by Product of Modules « Abstract Nonsense | November 11, 2011 | Reply

  2. […] is well-defined since it’s well-defined on a basis). Clearly is an epi and and so by the first isomorphism theorem , but since is an integral domain it’s trivial that and so . From this point we set to be […]

    Pingback by Submodules of Free Modules Need Not be Free Unless Ring is a PID (pt. II) « Abstract Nonsense | November 21, 2011 | Reply

  3. […] it descends to an epimorphism . To finish we prove that from where the rest will follow from the first isomorphism theorem. Indeed, suppose that is in then we see […]

    Pingback by Using Partial Exactness to Compute Things (Pt. I) « Abstract Nonsense | January 20, 2012 | Reply

  4. […] The fact that can be -embedded into an -module is clear by the first isomorphism theorem. Suppose now that is some left -module and we have some -map . We then consider the map given by […]

    Pingback by Extension of Scalars and Change of Ring (Pt. I) « Abstract Nonsense | January 24, 2012 | Reply

  5. […] for -algebras (which, if you aren’t aware of is just a combination of the first ring and module isomorphism […]

    Pingback by Algebraic Extensions (Pt. II) « Abstract Nonsense | March 25, 2012 | Reply


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