# product of injective modules is injective

Proof. Let $B$ be an arbitrary $R$-module, $A\subseteq B$ a submodule  and $f:A\to Q$ a homomorphism          . It is enough to show that $f$ can be extended to $B$. For $i\in I$ denote by $\pi_{i}:Q\to Q_{i}$ the projection. Since $Q_{i}$ is injective for any $i$, then the homomorphism $\pi_{i}\circ f:A\to Q_{i}$ can be extended to $f^{\prime}_{i}:B\to Q_{i}$. Then we have

 $f^{\prime}:B\to Q;$
 $f^{\prime}(b)=\big{(}f^{\prime}_{i}(b)\big{)}_{i\in I}.$

It is easy to check, that if $a\in A$, then $f^{\prime}(a)=f(a)$, so $f^{\prime}$ is an extension   of $f$. Thus $Q$ is injective. $\square$

Remark. Unfortunetly direct sum    of injective modules  need not be injective. Indeed, there is a theorem which states that direct sums of injective modules are injective if and only if ring $R$ is Noetherian  . Note that the proof presented above cannot be used for direct sums, because $f^{\prime}(b)$ need not be an element of the direct sum, more precisely, it is possible that $f^{\prime}_{i}(b)\neq 0$ for infinetly many $i\in I$. Nevertheless products are always injective.

Title product of injective modules is injective ProductOfInjectiveModulesIsInjective 2013-03-22 18:50:17 2013-03-22 18:50:17 joking (16130) joking (16130) 5 joking (16130) Theorem msc 16D50