# Abstract Nonsense

## Injective Modules (Pt. I)

Point of Post: In this post we discuss the notion of injective modules and show that the category $R\text{-}\mathbf{Mod}$ has “enough injectives”.

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Motivation

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We are going to discuss now the “dual” notion to projective modules which, as one would expect, are just the modules one gets by dualizing the lifting axioms for projective modules. Of course, it should also follow then that we can dualize the other properties of projective modules to get other characterizations, $\text{Hom}_R(\bullet,M)$ is exact, etc.

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Our main use for injective modules, similar to the case of projective modules, is that injective modules are “nice” modules for which we can effectively approximate any other module (i.e. we can find some long exact sequence beginning with a given module and with the rest of the terms in the sequence being injective modules). This will be key for when we discuss the notion of derived functors.

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Injective Modules

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Let $R$ be any ring. We call a left $R$-module $I$ injective if whenever we have a diagram of the form

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$\begin{array}{ccccc} & & I & & \\ & & \big\uparrow & & \\ 0 & \to & M & \to & N\end{array}\quad\mathbf{(1)}$

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We can find an $R$-map $N\to I$ making the resulting diagram commute.

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What this roughly translates to is that every $R$-map from a submodule $M$ of a module $N$ into $I$ can be extended to an $R$-map $N\to I$. So, for example, $2\mathbb{Z}$ is not an injective $\mathbb{Z}$-module for if one tried to extend the identity map $2\mathbb{Z}\to 2\mathbb{Z}$ to a group map $f:\mathbb{Z}\to 2\mathbb{Z}$ one would be able to find $x\in 2\mathbb{Z}$ such that $2x=2$ (namely, $x=f(1)$) but this is impossible. So, one can’t extend the map $2\mathbb{Z}\to2\mathbb{Z}$ to a map $\mathbb{Z}\to2\mathbb{Z}$ and so $2\mathbb{Z}$ is not an injective $\mathbb{Z}$-module. For similar reasons $\mathbb{Z}$ is not an injective $\mathbb{Z}$-module since the identity map $\mathbb{Z}\to\mathbb{Z}$ cannot be extended to a map $\mathbb{Q}\to\mathbb{Z}$ since $5x=1$ isn’t solvable in $\mathbb{Z}$.

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Let us prove two quick things about injective modules:

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Theorem: Direct summands and products of injective modules are injective.

Proof: Suppose first that $M$ is any module and there exists another module $M'$ such that $M\oplus M'$ is injective. Suppose then that we have an injective map $i:L\to N$ and a map $f:L\to M$. We note then that we also have the map $\iota\circ f:L\to M\oplus M'$ (where $\iota$ is the usual inclusion $M\hookrightarrow M\oplus M'$). Since $M\oplus M'$ is injective there exists $g:N\to M\oplus M'$ such that $g\circ i=\iota\circ f$. Define then $\widetilde{g}:L\to M$ by $\pi\circ g$ where $\pi:M\oplus M'\to M$ is the usual projection. This is clearly an $R$-map and satisfies $\widetilde{g}\circ i=\pi\circ g\circ i=\pi\circ\iota\circ f=f$. Since everything was arbitrary it follows that $M$ is injective.

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The fact that the product of injectives is injective follows from the fact that the $\displaystyle \text{Hom}_R\left(M,\prod_{\alpha}I_\alpha\right)$ is naturally isomorphic to $\displaystyle \prod_\alpha\text{Hom}_R(M,I_\alpha)$. $\blacksquare$

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Of course, while this is the best definition in the sense that it captures what injective modules “do” it is somewhat transparent.

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Perhaps one of the clearest definitions of injective modules is given by the following:

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Theorem: Let $R$ be a ring. Then, a left $R$-module $I$ is injective if and only if every short exact sequence

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$0\to I\to M\to N\to 0$

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splits.

Proof: Let’s first assume that $I$ is injective and suppose that we have a sequence $0\to I\xrightarrow{f} M\to N\xrightarrow{g} 0$. Note that we then have the following commutative diagram

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$\begin{array}{ccccc} & & I & & \\ & & \big\uparrow_{\text{id}} & & \\ 0 & \to & I & \to & M\end{array}$

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so there exists an arrow $p:M\to I$ such that $\text{id}=p\circ f$. Thus, by the splitting lemma, we may conclude that our sequence splits.

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We will prove in just a little bit that every left $R$-module can be embedded into an injective $R$-module. So, suppose that $I$ satisfies the splitting property. We can find an injective module $M$ such that $I\hookrightarrow M$. But, because of $I$‘s splitting property this says that $I$ is a direct summand of $M$ and thus injective. $\blacksquare$

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It then becomes even more obvious, for example, that $2\mathbb{Z}$ is not an injective $\mathbb{Z}$-module since $0\to2\mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}\to0$ is not split.

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This also allows us to interpret injective modules as precisely the modules that always sit nicely inside any module containing them. More particularly:

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Theorem: Let $I$ be an injective module. Then, if $I\leqslant M$ then $I$ is a direct summand of $M$ (i.e. there exists $J\leqslant M$ with $M=I\oplus J$).

Proof: We know that $0\to I\to M\to M/I\to 0$ splits and so $M\cong I\oplus M/I$ and moreover $I\oplus 0$ maps isomorphically onto $I$. $\blacksquare$

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We now prove the equivalence that injective modules are just those that make the contaravariant Hom functor exact:

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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.