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\begin{document}

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\section*{free module}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{algebra}{}\paragraph*{{Algebra}}\label{algebra}

[[!include higher algebra - contents]]

\hypertarget{homological_algebra}{}\paragraph*{{Homological algebra}}\label{homological_algebra}

[[!include homological algebra - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{definition}{Definition}\dotfill \pageref*{definition} \linebreak
\noindent\hyperlink{general}{General}\dotfill \pageref*{general} \linebreak
\noindent\hyperlink{over_rings}{Over rings}\dotfill \pageref*{over_rings} \linebreak
\noindent\hyperlink{properties}{Properties}\dotfill \pageref*{properties} \linebreak
\noindent\hyperlink{as_a_monoidal_functor}{As a monoidal functor}\dotfill \pageref*{as_a_monoidal_functor} \linebreak
\noindent\hyperlink{SubmodulesOfFreeModules}{Submodules of free modules}\dotfill \pageref*{SubmodulesOfFreeModules} \linebreak
\noindent\hyperlink{over_a_field_vector_spaces}{Over a field: vector spaces}\dotfill \pageref*{over_a_field_vector_spaces} \linebreak
\noindent\hyperlink{related_concepts}{Related concepts}\dotfill \pageref*{related_concepts} \linebreak
\noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

A \emph{free module} over some [[ring]] $R$ is [[free construction|freely generated]] on a [[set]] of [[basis]] elements.

Under the \href{module#RelationToVectorBundlesInIntroduction}{interpretation of modules as generalized vector bundles} a \emph{free module} corresponds to a \emph{trivial} bundle.

\hypertarget{definition}{}\subsection*{{Definition}}\label{definition}

\hypertarget{general}{}\subsubsection*{{General}}\label{general}

Let $C$ be a [[monoidal category]], and $Alg(C)$ the [[category]] of [[monoids]] in $C$ and for $A \in Alg(C)$ let $A$[[Mod]]$(C)$ be the category of $A$-[[modules]] in $C$.

There is the evident [[forgetful functor]] $U : A Mod(C) \to C$ that sends each module $(N,\rho)$ to its underlying object $N \in C$.

\begin{defn}
\label{}\hypertarget{}{}
The [[left adjoint]] $C \to A Mod(C)$ is the corresponding [[free construction]]. The modules in the image of this functor are \emph{free modules}.

\end{defn}
\hypertarget{over_rings}{}\subsubsection*{{Over rings}}\label{over_rings}

Let $R$ be a ring. We discuss free [[modules]] over $R$.

\begin{prop}
\label{}\hypertarget{}{}
For $R \in$ [[Ring]] a [[ring]] and $S \in$ [[Set]], the free $R$-module on $S$ is isomorphic to the ${\vert S\vert}$-fold [[direct sum]] of $R$ with itself

\begin{displaymath}
R^{(S)}\simeq \oplus_{s \in S} R
  \,.
\end{displaymath}
\end{prop}
\hypertarget{properties}{}\subsection*{{Properties}}\label{properties}

\hypertarget{as_a_monoidal_functor}{}\subsubsection*{{As a monoidal functor}}\label{as_a_monoidal_functor}

Let $R$ be a commutative ring, and let $R\{X\}$ denote the free $R$-module on a set $X$.

\begin{prop}
\label{}\hypertarget{}{}
The free $R$-module functor is strong monoidal with respect to the Cartesian monoidal structure on sets, and the tensor product of $R$-modules.

\end{prop}
In other words, the free module construction turns set-theoretic products into tensor products. Thus, it preserves algebraic objects (such as [[monoid object]]s, [[Hopf monoid]] objects, etc.) and their homomorphisms. In particular, if $M$ is a monoid in the category of sets (and hence a bimonoid with the canonical comonoid structure) then $R\{M\}$ is a bimonoid object in $R \mathsf{Mod}$, which is precisely a $K$-bialgebra. A group $G$ in the category of sets is a Hopf monoid, and hence $R\{G\}$ is a Hopf algebra --- this is precisely the [[group algebra]] of $G$.

\hypertarget{SubmodulesOfFreeModules}{}\subsubsection*{{Submodules of free modules}}\label{SubmodulesOfFreeModules}

Let $R$ be a commutative [[ring]].

\begin{prop}
\label{submod}\hypertarget{submod}{}
Assuming the [[axiom of choice]], the following are equivalent

\begin{enumerate}%
\item every [[submodule]] of a free $R$-module is itself free;


\item every [[ideal]] in $R$ is a free $R$-module;


\item $R$ is a [[principal ideal domain]].



\end{enumerate}
\end{prop}
\begin{proof}
(See also \hyperlink{Rotman}{Rotman, pages 650-651}.) Condition 1. immediately implies condition 2., since ideals of $R$ are the same as submodules of $R$ seen as an $R$-module. Now assume condition 2. holds, and suppose $x \in R$ is any nonzero element. Let $\lambda_x$ denote multiplication by $x$ (as an $R$-module map). We have a sequence of surjective $R$-module maps

\begin{displaymath}
R \stackrel{\lambda_x}{\to} (x) \cong \oplus_J R \stackrel{\nabla}{\to} R
\end{displaymath}
(where $\nabla$ is the [[codiagonal]] map); by the [[Yoneda lemma]], the composite map $R \to R$ is of the form $\lambda_r$, where $r \in R$ is the value of the composite at $1 \in R$. Since $\lambda_r$ is surjective, we have $\lambda_r(s) = r s = 1$ for some $s$, so that $r$ is invertible. Hence $\lambda_r$ is invertible, and this implies $\lambda_x$ is monic. Therefore $R$ is a domain. From that, we infer that if $f$ and $g$ belong to a basis of an ideal $I$, then

\begin{displaymath}
0 \neq f g \in R\cdot f \cap R \cdot g
\end{displaymath}
whence $f$ and $g$ are not linearly independent, so $f = g$ and $I$ as an $R$-module is generated by a single element, i.e., $R$ is a principal ideal domain.

That condition 3. implies condition 1. is proved \href{http://ncatlab.org/nlab/show/principal+ideal+domain#free}{here}.

\end{proof}
\begin{cor}
\label{}\hypertarget{}{}
Assuming the [[axiom of choice]], over a ring $R$ which is a [[principal ideal domain]], every [[module]] has a [[projective resolution]] of length 1.

\end{cor}
See at \href{projective+resolution#Lenght1ResolutionsOfAbelianGroups}{projective resolution -- Resolutions of length 1} for more.

\hypertarget{over_a_field_vector_spaces}{}\subsubsection*{{Over a field: vector spaces}}\label{over_a_field_vector_spaces}

Assuming the [[axiom of choice]], if $R = k$ is a [[field]] then every $R$-[[module]] is free: it is $k$-[[vector space]] and by the \emph{[[basis theorem]]} every such has a [[basis of a vector space|basis]].

\hypertarget{related_concepts}{}\subsection*{{Related concepts}}\label{related_concepts}

\begin{itemize}%
\item [[locally free module]]


\item [[finitely generated module]], [[finitely presented module]]


\item \textbf{free module} $\Rightarrow$ [[projective module]] $\Rightarrow$ [[flat module]] $\Rightarrow$ [[torsion-free module]]


\item [[basis of a free module]]


\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]]


\item flat object, [[flat resolution]]

\begin{itemize}%
\item [[flat module]]

\end{itemize}


\end{itemize}
\hypertarget{references}{}\subsection*{{References}}\label{references}

\begin{itemize}%
\item Rotman \emph{Advanced Modern Algebra}, pp. 650--651

\end{itemize}
[[!redirects free modules]]



\end{document}