Abstract Nonsense

Crushing one theorem at a time

Injective Modules (Pt. II)


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

Theorem: Let I be a left R-module. Then, I is injective if and only if \text{Hom}_R(\bullet,I) is exact.

Proof: Suppose first that I is injective and let 0\to M\xrightarrow{\alpha} N\xrightarrow{\beta} L\to 0 be arbitrary. Applying \text{Hom}_R(\bullet,I) a priori only gives us the exact sequence 0\to \text{Hom}_R(L,I)\xrightarrow{\beta^\ast}\text{Hom}_R(N,I)\xrightarrow{\alpha^\ast}\text{Hom}_R(M,I). So, what we really need to check is that \alpha^\ast is injective. So, let f:M\to I be arbitrary. We then have the following sequence

\text{ }

\begin{array}{ccccc} & & I & & \\ & & \big\uparrow _{f} & & \\ 0 & \to & M & \xrightarrow{\alpha} & N\end{array}

\text{ }

By injectivity of I we get a map g:N\to I such that f=g\circ\alpha or f=\alpha^\ast(g). Since f was arbitrary it follows that \alpha^\ast is surjective and since our sequence was arbitrary it follows that \text{Hom}_R(\bullet,M) is exact.

\text{ }

Conversely, suppose that \text{Hom}_R(\bullet,I) is exact and that we have a diagram of the form

\text{ }

\begin{array}{ccccc} & & I & & \\ & & \big\uparrow_{f} & & \\ 0 & \to & M & \xrightarrow{\alpha} & N \end{array}

\text{ }

Since \text{Hom}_R(\bullet,I) is exact we may apply it to the sequence 0\to M\xrightarrow{\alpha}N to get the exact sequence \text{Hom}_R(N,I)\xrightarrow{\alpha^\ast}\text{Hom}_R(M,I)\to 0.  Now, since this is surjective there exists g\in\text{Hom}_R(N,I) such that \alpha^\ast(g)=f or f=g\circ\alpha. The conclusion follows. \blacksquare

\text{ }

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 I is injective, instead of having to check that whenever we have M\to I and M\leqslant N that we can extend to get a map N\to I we only have to check that any map \mathfrak{a}\to I (where \mathfrak{a} is an ideal of R) we can extend to get a map R\to I.

\text{ }

Theorem(Baer’s Criterion): A left R-module I is injective if and only if for every ideal \mathfrak{a} of R and every map f:\mathfrak{a}\to I, we can find an extension \widetilde{f}:R\to I.

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

\text{ }

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

\text{ }

\begin{array}{ccccc} & & I & & \\ & & \big\uparrow_{f} & & \\ 0 & \to & M & \xrightarrow{\alpha} & N\end{array}

\text{ }

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

\text{ }

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

\text{ }

So, let P be the poset of all ordered pairs (L,g) such that M\subseteq L\subseteq N and g extends f and with (L,g)\leqslant (L',g') if L\subseteq L' and g' extends g. Suppose now that \{(L_i,g_i)\} is a chain in P. Since the union of a chain of submodules is a submodule we have that L=\displaystyle \bigcup_{i}L_i is a submodule of N containing M. Define g:L\to I by g(\ell)=g_i(\ell) for any i with \ell\in L_i–this is well-defined precisely because we have a chain. This is clearly an R-map and extends each g_i. Thus, (L,g) is an upper bound for \{(L_i,g_i)\}.

\text{ }

By Zorn’s lemma we get a maximal element (L,g) of P. We claim that L=N. Indeed, suppose not, then there exists x\in N-L. Define \mathfrak{a}=\left\{r\in R:rx\in L\right\}, this is clearly an ideal of R. Define then h:\mathfrak{a}\to I by h(r)=g(rx). We can then, by hypothesis, extend this to a map \widetilde{h}:R\to I. We then define j:L+\langle x\rangle\to I by j(\ell+rx)=g(\ell)+r\widetilde{h}(1). To see that this is well-defined we note that if \ell+rx=\ell'+r'x then \ell-\ell'=(r'-r)x and so r'-r\in\mathfrak{a}. We see then that

\text{ }

g(\ell-\ell')=g((r'-r)x)=h(r'-r)=\widetilde{h}(r'-r)=(r'-r)\widetilde{h}(1)

\text{ }

from where well-definedness follows. Clearly j extends g and since L+\langle x\rangle properly contains L this contradicts maximality. Thus, L=N and the conclusion follows. \blacksquare

\text{ }

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

\text{ }

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

Proof: By Baer’s criterion it suffices to show that we can always extend maps f:\mathfrak{a}\to k to maps \widetilde{f}:R\to k. But, this is easy. For if r\in R is arbitrary define \displaystyle \widetilde{f}(r)=f(ra)a^{-1} for any a\in\mathfrak{a}. This is well-defined for if b is also in \mathfrak{b} then bf(ra)=f(arb)=af(rb) so that f(ra)a^{-1}=f(rb)b^{-1}. This is clearly an R-map that extends f. The conclusion follows. \blacksquare

\text{ }

Corollary: \mathbb{Q} is an injective \mathbb{Z}-module.

\text{ }

\text{ }

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.

Advertisements

April 28, 2012 - Posted by | Algebra, Module Theory, Ring Theory | , , , , , , ,

No comments yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: