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\newtheorem{prop}{Proposition} \newtheorem{cor}{Corollary} \newtheorem*{utheorem}{Theorem} \newtheorem*{ulemma}{Lemma} \newtheorem*{uprop}{Proposition} \newtheorem*{ucor}{Corollary} \theoremstyle{definition} \newtheorem{defn}{Definition} \newtheorem{example}{Example} \newtheorem*{udefn}{Definition} \newtheorem*{uexample}{Example} \theoremstyle{remark} \newtheorem{remark}{Remark} \newtheorem{note}{Note} \newtheorem*{uremark}{Remark} \newtheorem*{unote}{Note} %------------------------------------------------------------------- \begin{document} %------------------------------------------------------------------- \section*{projective module} \hypertarget{context}{}\subsubsection*{{Context}}\label{context} \hypertarget{homological_algebra}{}\paragraph*{{Homological algebra}}\label{homological_algebra} [[!include homological algebra - contents]] \hypertarget{contents}{}\section*{{Contents}}\label{contents} \noindent\hyperlink{definition}{Definition}\dotfill \pageref*{definition} \linebreak \noindent\hyperlink{properties}{Properties}\dotfill \pageref*{properties} \linebreak \noindent\hyperlink{existence_of_enough_projective_modules}{Existence of enough projective modules}\dotfill \pageref*{existence_of_enough_projective_modules} \linebreak \noindent\hyperlink{explicit_characterizations}{Explicit characterizations}\dotfill \pageref*{explicit_characterizations} \linebreak \noindent\hyperlink{RelationToProjectiveResolution}{Relation to projective resolutions of chain complexes}\dotfill \pageref*{RelationToProjectiveResolution} \linebreak \noindent\hyperlink{examples}{Examples}\dotfill \pageref*{examples} \linebreak \noindent\hyperlink{related_concepts}{Related concepts}\dotfill \pageref*{related_concepts} \linebreak \noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak \hypertarget{definition}{}\subsection*{{Definition}}\label{definition} \begin{defn} \label{ProjectiveModule}\hypertarget{ProjectiveModule}{} For $R$ a [[ring]], a \emph{projective $R$-module} is a [[projective object]] in the [[category]] $R$[[Mod]]. Hence an $R$-module $N$ is \emph{projective} precisely if for all [[diagrams]] of $R$-[[module]] [[homomorphisms]] of the form \begin{displaymath} \itexarray{ && A \\ & & \downarrow^{\mathrlap{epi}} \\ N &\underset{f}{\longrightarrow}& B } \end{displaymath} there exists a [[lift]], hence a morphism $N \overset{\phi}{\to} A$ making a [[commuting diagram]] of the form \begin{displaymath} \itexarray{ && A \\ & {}^{\mathllap{\phi}}\nearrow & \downarrow^{\mathrlap{epi}} \\ N &\underset{f}{\longrightarrow}& B } \,. \end{displaymath} \end{defn} \begin{prop} \label{NProjectiveIFFHomNExact}\hypertarget{NProjectiveIFFHomNExact}{} An $R$-module $N$ is projective (def. \ref{ProjectiveModule}) precisely if the [[hom functor]] \begin{displaymath} Hom_{R Mod}(N, - ) : R Mod \to Ab \end{displaymath} (out of $N$) is an [[exact functor]]. \end{prop} \begin{proof} The hom-functor in question is a [[left exact functor]] for all $N$, hence we need to show that it is a [[right exact functor]] precisely if $N$ is projective. That $Hom_R(N,-)$ is right exact means equivalently that for \begin{displaymath} 0 \to A \overset{i}{\longrightarrow} B \overset{p}{\longrightarrow} C \to 0 \,, \end{displaymath} any [[short exact sequence]], hence for $p$ any [[epimorphism]] and $i$ its [[kernel]] inclusion, then $Hom_R(N,p)$ is an epimorphism, hence that for any element $f \in Hom_R(N,C)$, \begin{displaymath} \itexarray{ && B \\ && \downarrow^{\mathrlap{p}} \\ N &\underset{f}{\longrightarrow}& C } \end{displaymath} there exists $\phi \colon N \to B$ such that $f = Hom_R(N,p)(\phi) \coloneqq p \circ \phi$, hence that \begin{displaymath} \itexarray{ && B \\ &{}^{\mathllap{\phi}}\nearrow& \downarrow^{\mathrlap{p}} \\ N &\underset{f}{\longrightarrow}& C } \,. \end{displaymath} This is manifestly the condition that $N$ is projective. \end{proof} \hypertarget{properties}{}\subsection*{{Properties}}\label{properties} \hypertarget{existence_of_enough_projective_modules}{}\subsubsection*{{Existence of enough projective modules}}\label{existence_of_enough_projective_modules} \begin{lemma} \label{FreeModulesAreProjective}\hypertarget{FreeModulesAreProjective}{} Assuming the [[axiom of choice]], a [[free module]] $N \simeq R^{(S)}$ is projective. \end{lemma} \begin{proof} Explicitly: if $S \in Set$ and $F(S) = R^{(S)}$ is the [[free module]] on $S$, then a module homomorphism $F(S) \to N$ is specified equivalently by a [[function]] $f : S \to U(N)$ from $S$ to the underlying set of $N$, which can be thought of as specifying the images of the unit elements in $R^{(S)} \simeq \oplus_{s \in S} R$ of the ${\vert S\vert}$ copies of $R$. Accordingly then for $\tilde N \to N$ an epimorphism, the underlying function $U(\tilde N) \to U(N)$ is an epimorphism, and the [[axiom of choice]] in [[Set]] says that we have all lifts $\tilde f$ in \begin{displaymath} \itexarray{ && U(\tilde N) \\ & {}^{\tilde f} \nearrow & \downarrow \\ S &\stackrel{f}{\to}& U(N) } \,. \end{displaymath} By [[adjunction]] these are equivalently lifts of module homomorphisms \begin{displaymath} \itexarray{ && \tilde N \\ & \nearrow & \downarrow \\ R^{(S)} &\stackrel{}{\to}& N } \,. \end{displaymath} \end{proof} \begin{prop} \label{FreeForgetfulCounitEpimorphism}\hypertarget{FreeForgetfulCounitEpimorphism}{} Assuming the [[axiom of choice]], the category $R$[[Mod]] has \emph{\href{projective%20object#EnoughProjectives}{enough projectives}}: for every $R$-[[module]] $N$ there exists an [[epimorphism]] $\tilde N \to N$ where $\tilde N$ is a projective module. \end{prop} \begin{proof} Let $F(U(N))$ be the [[free module]] on the [[set]] $U(N)$ underlying $N$. By lemma \ref{FreeModulesAreProjective} this is a projective module. The [[counit of an adjunction|counit]] \begin{displaymath} \epsilon : F(U(N)) \to N \end{displaymath} of the [[free functor|free]]/[[forgetful functor|forgetful]]-[[adjunction]] $(F \dashv U)$ is an [[epimorphism]]. \end{proof} Actually, the full axiom of choice is not necessary here; it is enough to have the [[presentation axiom]], which states the [[category of sets]] has enough projectives (whereas the axiom of choice states that \emph{every} set is projective). Then we can replace $U(N)$ above by a projective set $P \twoheadrightarrow U(N)$, giving an epimorphism $F(P) \twoheadrightarrow F(U(N)) \twoheadrightarrow N$ (and $F(P)$ is projective). \hypertarget{explicit_characterizations}{}\subsubsection*{{Explicit characterizations}}\label{explicit_characterizations} We discuss the more explicit characterization of projective modules as [[direct sum|direct summands]] of [[free modules]]. \begin{lemma} \label{DirectSummandOfFreeIsProjective}\hypertarget{DirectSummandOfFreeIsProjective}{} If $N \in R Mod$ is a [[direct sum|direct summand]] of a [[free module]], hence if there is $N' \in R Mod$ and $S \in Set$ such that \begin{displaymath} R^{(S)} \simeq N \oplus N' \,, \end{displaymath} then $N$ is a projective module. \end{lemma} \begin{proof} Let $\tilde K \to K$ be a surjective homomorphism of modules and $f : N \to K$ a homomorphism. We need to show that there is a lift $\tilde f$ in \begin{displaymath} \itexarray{ && \tilde K \\ & {}^{\mathllap{\tilde f}}\nearrow & \downarrow \\ N &\stackrel{f}{\to}& K } \,. \end{displaymath} By definition of [[direct sum]] we can factor the [[identity]] on $N$ as \begin{displaymath} id_N : N \to N \oplus N' \to N \,. \end{displaymath} Since $N \oplus N'$ is free by assumption, and hence projective by lemma \ref{FreeModulesAreProjective}, there is a lift $\hat f$ in \begin{displaymath} \itexarray{ && && \tilde K \\ && & {}^{\mathllap{\hat f}}\nearrow & \downarrow \\ N &\to& N \oplus N' &\to& K } \,. \end{displaymath} Hence $\tilde f : N \to N \oplus N' \stackrel{\hat f}{\to} \tilde K$ is a lift of $f$. \end{proof} \begin{prop} \label{ProjectiveIsPreciselyDirectSummandOfFreeModule}\hypertarget{ProjectiveIsPreciselyDirectSummandOfFreeModule}{} An $R$-module $N$ is projective precisely if it is the [[direct summand]] of a [[free module]]. \end{prop} \begin{proof} By lemma \ref{DirectSummandOfFreeIsProjective} if $N$ is a direct summand then it is projective. So we need to show the converse. Let $F(U(N))$ be the [[free module]] on the [[set]] $U(N)$ underlying $N$ as in the proof of prop. \ref{FreeForgetfulCounitEpimorphism}. The [[counit of an adjunction|counit]] \begin{displaymath} \epsilon : F(U(N)) \to N \end{displaymath} of the [[free functor|free]]/[[forgetful functor|forgetful]]-[[adjunction]] $(F \dashv U)$ is an [[epimorphism]]. Thefore if $N$ is projective, there is a [[section]] $s$ of $\epsilon$. This exhibits $N$ as a direct summand of $F(U(N))$. \end{proof} This proposition is often stated more explicitly as the existence of a [[dual basis]], see there. In some cases this can be further strengthened: \begin{prop} \label{}\hypertarget{}{} If the [[ring]] $R$ is a [[principal ideal domain]] (in particular $R = \mathbb{Z}$ the [[integers]]), then every projective $R$-module is free. \end{prop} The details are discussed at \emph{\href{http://ncatlab.org/nlab/show/principal+ideal+domain#StructureTheoryOfModules}{pid - Structure theory of modules}}. \begin{prop} \label{}\hypertarget{}{} For an $R$-module $P$, the following statements are equivalent: \begin{enumerate}% \item $P$ is finite locally free in that there exists a partition $1 = \sum_i f_i \in R$ such that the localized modules $P[f_i^{-1}]$ are finite free modules over $R[f_i^{-1}]$. \item $P$ is finitely generated and projective. \item $P$ is a [[dualizable object]] in the category of $R$-modules (equipped with the tensor product as monoidal structure). \item There exist elements $x_1,\ldots,x_n \in P$ and linear forms $\vartheta_1,\ldots,\vartheta_n \in Hom(P,R)$ such that $x = \sum_i \vartheta_i(x) x_i$ for all $x \in P$. \end{enumerate} \end{prop} \begin{proof} The equivalence of 2., 3., and 4. is mostly formal. For the equivalence with 1., see \href{http://math.stackexchange.com/questions/16814/finitely-generated-projective-modules-are-locally-free}{this math.SE discussion} for good references. Note that the equivalences are true without assuming that $R$ is [[Noetherian ring|Noetherian]] or that $P$ satisfies some finiteness condition. \end{proof} \hypertarget{RelationToProjectiveResolution}{}\subsubsection*{{Relation to projective resolutions of chain complexes}}\label{RelationToProjectiveResolution} \begin{defn} \label{}\hypertarget{}{} For $N \in R Mod$ a \textbf{[[projective resolution]]} of $N$ is a [[chain complex]] $(Q N)_\bullet \in Ch_\bullet(R Mod)$ equipped with a [[chain map]] \begin{displaymath} Q N \to N \end{displaymath} (with $N$ regarded as a complex concentrated in degree 0) such that \begin{enumerate}% \item this morphism is a [[quasi-isomorphism]] (this is what makes it a [[resolution]]), which is equivalent to \begin{displaymath} \cdots \to (Q N)_1 \to (Q N)_0 \to N \end{displaymath} being an [[exact sequence]]; \item all whose entries $(Q N)_n$ are projective modules. \end{enumerate} \end{defn} \begin{remark} \label{}\hypertarget{}{} This means precisely that $Q N \to N$ is an [[cofibrant resolution]] with respect to the standard [[model structure on chain complexes]] (see \href{model%20structure%20on%20chain%20complexes#StandardQuillenOnBounded}{here}) for which the fibrations are the positive-degreewise epimorphisms. Notice that in this model structure every object is fibrant, so that cofibrant resolutions are the only resolutions that need to be considered. \end{remark} \begin{prop} \label{}\hypertarget{}{} Every $R$-module has a projective resolution. \end{prop} See at \emph{[[projective resolution]]}. \hypertarget{examples}{}\subsection*{{Examples}}\label{examples} \begin{prop} \label{ModuleOverAFieldIsProjective}\hypertarget{ModuleOverAFieldIsProjective}{} Assuming the [[axiom of choice]], then by the [[basis theorem]] every module over a [[field]] is a [[free module]] and hence in particular every module over a field is a projective module (by prop. \ref{DirectSummandOfFreeIsProjective}). \end{prop} \begin{prop} \label{}\hypertarget{}{} If $R$ is the [[integers]] $\mathbb{Z}$, or a [[field]] $k$, or a [[division ring]], then every projective $R$-module is already a free $R$-module. \end{prop} \hypertarget{related_concepts}{}\subsection*{{Related concepts}}\label{related_concepts} \begin{itemize}% \item [[projective object]], [[projective presentation]], [[projective cover]], [[projective resolution]] \item [[injective object]], [[injective presentation]], [[injective envelope]], [[injective resolution]] \begin{itemize}% \item [[injective module]] \end{itemize} \item [[free object]], [[free resolution]] \begin{itemize}% \item [[free module]], [[locally free module]] \end{itemize} \item flat object, [[flat resolution]] \begin{itemize}% \item [[flat module]] \end{itemize} \item [[free module]] $\Rightarrow$ \textbf{projective module} $\Rightarrow$ [[flat module]] $\Rightarrow$ [[torsion-free module]] \item [[Serre-Swan theorem]] \item [[Quillen-Suslin theorem]] \end{itemize} \hypertarget{references}{}\subsection*{{References}}\label{references} Lecture notes include \begin{itemize}% \item [[Charles Weibel]], \emph{[[An Introduction to Homological Algebra]]}, section 2.2 \item \emph{Projective modules, Presentations and resolutions} (\href{http://livetoad.org/Courses/Documents/8875/Notes/presentations_and_resolutions.pdf}{pdf}) \item Thomas Lam, chapter 6 (\href{http://www.math.lsa.umich.edu/~tfylam/Math221/6.pdf}{pdf}) \end{itemize} Original articles include \begin{itemize}% \item Irving Kaplansky, \emph{Projective modules}, The Annals of Mathematics Second Series, Vol. 68, No. 2 (Sep., 1958), pp. 372-377 (\href{http://www.jstor.org/stable/1970252}{JSTOR}) \end{itemize} [[!redirects projective modules]] \end{document}