## The Module Isomorphism Theorems

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

*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 -module” and “ideal” replaced by “submodule”.

*The Isomorphism Theorems*

We introduce a bit of notation. For a -morphism we call the space the *coimage of *and denote it . The first isomorphism theorem amounts to the statement that for left -modules one has that as left -modules. Indeed, by the universal characterization of quotient modules it will suffice for us to prove that has the same universal mapping property as . In other words, we have the epimorphism and we’d like to show that every time there is a map with then there exists a unique map with . But, it’s clear how to do this, namely defining –since this map is well-defined. Moreover, since is epi (so that implies ) we have that is the unique map with this property. From this we may conclude that satisfies the universal characterization of quotient modules and so we may conclude the following:

**Theorem (First Module Isomorphism Theorem): ***Let be a ring and and two left -modules. Then, given any -morphism one has that , canonically.*

**Theorem(Second Module Isomorphism Theorem): ***Let be a ring and let be a left -module with . Then, is a submodule of and .*

**Proof: **Evidently since it’s the image of under the -morphism . To see that this isomorphism is true, define by . Note that since this is well-defined for any two representatives of differ by a -element and in particular, differ only be a -element and so will have equal -cosets. Clearly this map is a -epimorphism and has kernel equal to . The rest follows from the first isomorphism theorem.

**Theorem(Third Module Isomorphism Theorem): ***Let be a ring and a left -module. Then, if one has that .*

**Proof: **Define by . We claim that that is an epi. Indeed, evidently is a morphism and if then we evidently see that from where epiness follows. Clearly though we have that and so the rest follows from the first isomorphism theorem.

**Theorem(Fourth Module Isomorphism Theorem): ***Let be a ring and a left -module and let . Then, the map is an isomorphism of posets, between the submodules of containing and the submodules of .*

**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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