## Injective Modules (Pt. II)

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

**Theorem:** *Let be a left -module. Then, is injective if and only if is exact.*

**Proof: **Suppose first that is injective and let be arbitrary. Applying a priori only gives us the exact sequence . So, what we really need to check is that is injective. So, let be arbitrary. We then have the following sequence

By injectivity of we get a map such that or . Since was arbitrary it follows that is surjective and since our sequence was arbitrary it follows that is exact.

Conversely, suppose that is exact and that we have a diagram of the form

Since is exact we may apply it to the sequence to get the exact sequence . Now, since this is surjective there exists such that or . The conclusion follows.

What we would now like to prove is the big-deal theory about injective modules–the characterization that will actually allow us to actually find non-trivial injective modules. What it says is that if we want to check that is injective, instead of having to check that whenever we have and that we can extend to get a map we only have to check that any map (where is an ideal of ) we can extend to get a map .

**Theorem(Baer’s Criterion): ***A* *left -module is injective if and only if for every ideal of and every map , we can find an extension .*

**Proof: **It’s clear that if is injective then we can extend any ideal sourced map to an -sourced map.

Conversely, suppose that we can always extend ideal sourced maps to -sourced maps and suppose that we have the following diagram

We may assume, without loss of generality, that is inclusion.

As is standard in extension arguments we shall apply Zorn’s lemma.

So, let be the poset of all ordered pairs such that and extends and with if and extends . Suppose now that is a chain in . Since the union of a chain of submodules is a submodule we have that is a submodule of containing . Define by for any with –this is well-defined precisely because we have a chain. This is clearly an -map and extends each . Thus, is an upper bound for .

By Zorn’s lemma we get a maximal element of . We claim that . Indeed, suppose not, then there exists . Define , this is clearly an ideal of . Define then by . We can then, by hypothesis, extend this to a map . We then define by . To see that this is well-defined we note that if then and so . We see then that

from where well-definedness follows. Clearly extends and since properly contains this contradicts maximality. Thus, and the conclusion follows.

So, as promised, we can actually produce an example of a non-trivial injective -module for some .

**Theorem:** *Let be an integral domain with field of fractions , then is an injective -module.*

**Proof: **By Baer’s criterion it suffices to show that we can always extend maps to maps . But, this is easy. For if is arbitrary define for any . This is well-defined for if is also in then so that . This is clearly an -map that extends . The conclusion follows.

**Corollary: *** is an injective -module.*

**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] Lang, Serge. *Algebra*. Reading, MA: Addison-Wesley Pub., 1965. Print.

[5] Grillet, Pierre A. *Abstract Algebra*. New York: Springer, 2007. Print.

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