# Linear Algebra over a Ring/Projective and injective modules

Theorem (Baer's criterion):

A module ${\displaystyle M}$ over a commutative ring ${\displaystyle R}$ is injective if and only if for every ideal ${\displaystyle I\leq R}$, every homomorphism ${\displaystyle \varphi :I\to M}$ of ${\displaystyle R}$-modules may be extended to a homomorphism of ${\displaystyle R}$-modules ${\displaystyle \Phi :R\to M}$, ie. extended to a homomorphism ${\displaystyle \Phi :R\to M}$ that satisfies ${\displaystyle \Phi \upharpoonright _{I}=\varphi }$.

(On the condition of the axiom of choice.)

Proof: The "only if" part is obvious from the definition of injectivity. Conversely, assume that ${\displaystyle M}$ satisfies the extension property described in the theorem statement. Let now ${\displaystyle K,L}$ be ${\displaystyle R}$-modules, let ${\displaystyle \psi :K\to M}$ be a homomorphism and let ${\displaystyle \iota :K\to L}$ be an injection. By the definition of injective modules, we have to prove that there exists a homomorphism ${\displaystyle \Psi :L\to M}$ that satisfies ${\displaystyle \Psi \circ \iota =\psi }$, ie. that extends ${\displaystyle \psi }$ wrt. the inclusion ${\displaystyle \iota :K\to L}$, as the algebraists say. Now the image ${\displaystyle \operatorname {Im} \iota :=L'}$ is a submodule of ${\displaystyle L}$ that is isomorphic to ${\displaystyle K}$ by the injectivity of ${\displaystyle \iota }$ via ${\displaystyle \iota }$, and hence precomposition of the inverse of ${\displaystyle \iota }$ with ${\displaystyle \psi }$ yields a homomorphism from ${\displaystyle L'}$ to ${\displaystyle M}$. Now we partially order all extensions of this isomorphism by the order that ${\displaystyle \alpha \leq \beta }$ if and only if

1. the domain of ${\displaystyle \alpha }$ is contained within the domain of ${\displaystyle \beta }$, and
2. ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ coincide on the domain of ${\displaystyle \alpha }$.

By Zorn's lemma, which applies since each chain with respect to this order has an upper bound (namely the function that assigns to an element within the union of the respective domains the value that any of the respective homomorphisms assumes on it), there exists a maximal homomorphism ${\displaystyle \Psi }$ with respect to that order. We claim that ${\displaystyle \Psi }$ is defined on all of ${\displaystyle L}$. Indeed, upon assuming otherwise, we may choose an element ${\displaystyle l\in L}$ where ${\displaystyle \Psi }$ is not defined. Let ${\displaystyle L''\leq L}$ be the domain of definition of ${\displaystyle \Psi }$, so that ${\displaystyle l\notin L''}$. Define the ideal

${\displaystyle I:=\{r\in R|rl\in L''\}}$

of ${\displaystyle R}$; it may be the zero ideal. ${\displaystyle \Psi }$ is defined on ${\displaystyle I}$, and yields a homomorphism on ${\displaystyle I}$. By assumption, this homomorphism may be extended to a homomorphism ${\displaystyle f:R\to M}$. We then define ${\displaystyle L''':=L''+\langle l\rangle }$; a submodule of ${\displaystyle L}$ that is strictly larger than ${\displaystyle L''}$, since it contains ${\displaystyle l}$. On ${\displaystyle L'''}$, we define the homomorphism

${\displaystyle F(x+sl):=\Psi (x)+sf(l)}$ for ${\displaystyle x\in L''}$ and ${\displaystyle s\in R}$,

which is a proper extension of ${\displaystyle \Psi }$, so that ${\displaystyle \Psi }$ was not maximal, a contradiction. Hence, ${\displaystyle \Psi }$ was defined on all of ${\displaystyle L}$ from the beginning and yields the desired extension of ${\displaystyle \psi }$. ${\displaystyle \Box }$