ProductOfInjectiveModulesIsInjective 2009-02-21 09:50:13 2009-02-21 09:54:15 Theorem injective module % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \textbf{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. \textit{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 $f'_i:B\to Q_i$. Then we have $$f':B\to Q;$$ $$f'(b)=\big( f'_i (b)\big)_{i\in I}.$$ It is easy to check, that if $a\in A$, then $f'(a)=f(a)$, so $f'$ is an extension of $f$. Thus $Q$ is injective. $\square$ \textbf{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 $f'(b)$ need not be an element of the direct sum, more precisely, it is possible that $f'_i(b)\neq 0$ for infinetly many $i\in I$. Nevertheless products are always injective.