## Submodules of Free Modules Need Not be Free Unless Ring is a PID (pt. II)

**Point of Post: **This is a continuation of this post.

Let be a free module with basis and . By the well-ordering theorem we can make into a well-ordered set , we then define a new set for some and extend to by declaring that is larger than every element of . Evidently then is a well-ordered set where every element has a successor . We note the obvious fact that if and only if , and moreover that . Now, define by sending to itself and every other to and extending by linearity (this map 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 equal to . We make the similar observations that implies , , and .

The basic idea from here is simple. Note that which, by the third isomorphism theorem is isomorphic to , but note that since and since we may conclude that is isomorphic to a submodule of . But, since is a PID every submodule is of the form . If then since is an integral domain and if then . From this we may conclude that is a free -module of rank or . That said, we have an obvious short exact sequence

but since is free, we know this sequence splits and so there exists some backmap and we know that and since is free we know that where if and if . We then let .

We claim that is a basis of . To see first that is linearly independent, suppose that with each in . Suppose for a second that one of the coefficients in this sum was nonzero, we could then select (by the well-ordering) a largest one, say so that , but this clearly implies that which is contradictory to assumption. Thus, there does not exist an index for which and so for all . The linear independence of follows.

To see that spans we note that if we can prove that the set spans then we’ll be done since we’d clearly have then that

Since our set is well-ordered we can prove this by transfinite induction. Suppose then that for all and let . We split this into two cases, depending on whether is a successor in or not. If for some then we know that . We see then that for some , and by the induction hypothesis is in and so evidently . Suppose now that is not the immediate successor of anything. Since we know that can be written as . Choose the largest index, , such that then evidently . That said, by assumption we have that and since by assumption we have that . But, note then that . But, by the induction and so . Either way, we see that and so the induction is complete. It thus follows that is a basis.

In fact, it’s clear from the above that:

**Corollary: ***Let be a PID and a free module. Then, any submodules is free and .*

**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.

[4] Hungerford, Thomas W. *Algebra*. New York: Holt, Rinehart and Winston, 1974. Print.

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