product of injective modules is injective

Proposition. Let $R$ be a ring and ${\{Q_i\}}_{i\in I}$ a family of injective $R$-modules. Then the product

$$Q=\prod _{i\in I}{Q}_i$$

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 $f_i^{\prime }:B\to Q_i$. Then we have

$$f^{\prime }:B\to Q;$$

$$f^{\prime }(b)={\left(f_i^{\prime }(b)\right)}_{i\in I}.$$

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. $\mathrm{\square }$

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 $f_i^{\prime }(b)\ne 0$ for infinetly many $i\in I$. Nevertheless products are always injective.

Title	product of injective modules is injective
Canonical name	ProductOfInjectiveModulesIsInjective
Date of creation	2013-03-22 18:50:17
Last modified on	2013-03-22 18:50:17
Owner	joking (16130)
Last modified by	joking (16130)
Numerical id	5
Author	joking (16130)
Entry type	Theorem
Classification	msc 16D50