Linear Algebra over a Ring/Projective and injective modules

Theorem (Baer's criterion):

A module over a commutative ring is injective if and only if for every ideal , every homomorphism of -modules may be extended to a homomorphism of -modules , ie. extended to a homomorphism that satisfies .

(On the condition of the axiom of choice.)

Proof: The "only if" part is obvious from the definition of injectivity. Conversely, assume that satisfies the extension property described in the theorem statement. Let now be -modules, let be a homomorphism and let be an injection. By the definition of injective modules, we have to prove that there exists a homomorphism that satisfies , ie. that extends wrt. the inclusion , as the algebraists say. Now the image is a submodule of that is isomorphic to by the injectivity of via , and hence precomposition of the inverse of with yields a homomorphism from to . Now we partially order all extensions of this isomorphism by the order that if and only if

  1. the domain of is contained within the domain of , and
  2. and coincide on the domain of .

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 with respect to that order. We claim that is defined on all of . Indeed, upon assuming otherwise, we may choose an element where is not defined. Let be the domain of definition of , so that . Define the ideal

of ; it may be the zero ideal. is defined on , and yields a homomorphism on . By assumption, this homomorphism may be extended to a homomorphism . We then define ; a submodule of that is strictly larger than , since it contains . On , we define the homomorphism

for and ,

which is a proper extension of , so that was not maximal, a contradiction. Hence, was defined on all of from the beginning and yields the desired extension of .