# Abstract Nonsense

## 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

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

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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.

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

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

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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. $\blacksquare$

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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 $R\to I$.

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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.

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

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

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

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

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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 $\{(L_i,g_i)\}$.

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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

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

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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. $\blacksquare$

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

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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. $\blacksquare$

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

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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.