Modules are built to let rings act on abelian groups, but this flexibility comes with a cost: a module need not have coordinates. A [vector space](/page/Vector%20Space) over a field is controlled by a basis; every vector has a unique coordinate expansion, every linearly independent set can be extended to a basis, and every spanning set can be trimmed to a basis. Over a general ring, these familiar statements can fail. A free module is the part of module theory where coordinates still exist. The guiding question is not merely whether a module is generated by some set. Generation says that every element can be written using the generators. Freeness says that this expression is unique. That extra uniqueness is what lets matrices, rank, exact sequences, and tensor products behave in a controlled way. [example: A Generated Module with Non-Unique Coordinates] Let $R=\mathbb{Z}$ and regard $M=\mathbb{Z}/2\mathbb{Z}$ as a $\mathbb{Z}$-module, with scalar action $n\bar{a}=\overline{na}$. The element $\bar{1}$ generates $M$: if $m\in M$, then either $m=\bar{0}=0\bar{1}$ or $m=\bar{1}=1\bar{1}$. The coordinate expression using $\bar{1}$ is not unique. Indeed, \begin{align*} 1\bar{1}=\bar{1} \end{align*} and \begin{align*} 3\bar{1}=\bar{3}=\bar{1} \end{align*} because $3-1=2$ is divisible by $2$. Thus $1\bar{1}=3\bar{1}$ even though $1\ne 3$ in $\mathbb{Z}$. Equivalently, \begin{align*} 2\bar{1}=\bar{2}=\bar{0} \end{align*} so the nonzero coefficient $2\in\mathbb{Z}$ gives a relation on the proposed generator $\bar{1}$. Therefore $M$ is cyclic, but it is not free on $\{\bar{1}\}$. In fact $M$ is not free as a $\mathbb{Z}$-module. If $M$ had a basis $(e_i)_{i\in I}$ and $\bar{1}=\sum_{i\in I} a_i e_i$, then multiplying by $2$ would give \begin{align*} 0=2\bar{1}=\sum_{i\in I} (2a_i)e_i. \end{align*} Uniqueness of coordinates in a basis forces $2a_i=0$ for every $i$. Since $a_i\in\mathbb{Z}$, this implies $a_i=0$ for every $i$, so $\bar{1}=0$, a contradiction. The obstruction is torsion: $\bar{1}\ne 0$ but $2\bar{1}=0$. [/example] This example isolates the main obstruction. A finite generating set may still carry hidden relations. Free modules are exactly the modules for which the chosen generators have no hidden relations at all. ## Definition A [module](/page/Module) generalizes a vector space by allowing scalars from a ring rather than a field. The parent concept gives addition and scalar multiplication; the child concept of a free module asks for a coordinate system compatible with that module structure. [definition: Free Module] Let $R$ be a ring with identity, and let $M$ be a left $R$-module. The module $M$ is a free left $R$-module if there exists a set $I$ and elements $(e_i)_{i \in I}$ of $M$ such that for every $m \in M$ there is a unique family $(r_i)_{i \in I}$ with $r_i \in R$ and $r_i = 0_R$ for all but finitely many $i \in I$ satisfying \begin{align*} m = \sum_{i \in I} r_i e_i. \end{align*} [/definition] The finite-support condition is part of the definition: even when the basis set is infinite, each individual module element uses only finitely many basis elements. The unique object is the whole coefficient family, not a chosen finite index set; adding extra zero coefficients does not create a new coordinate expansion. To use freeness in calculations, we need to name the coordinate systems that witness it. Different choices of such systems can give different-looking coordinates for the same module element. ## Bases and Coordinate Models The definition of a free module asserts that some coordinate expansion exists, but it does not give a fixed coordinate system to use in later constructions. When we want to compare coordinates, define maps by specifying images of coordinate vectors, or speak about rank, we must keep track of the particular family that supplies those unique finite expansions. ### Bases A free module is useful only after we remember which family supplies its coordinates. The following definition isolates that family: it is the data that lets every element of $M$ be read uniquely as a finite coordinate vector, so later maps and matrix descriptions can be specified by what they do to these coordinate elements. [definition: Basis of a Free Module] Let $R$ be a ring with identity, and let $M$ be a left $R$-module. A family $(e_i)_{i \in I}$ of elements of $M$ is a basis of $M$ if for every $m \in M$ there is a unique family $(r_i)_{i \in I}$ with $r_i \in R$ and $r_i = 0_R$ for all but finitely many $i \in I$ satisfying \begin{align*} m = \sum_{i \in I} r_i e_i. \end{align*} [/definition] The word basis here means the same two conditions as in linear algebra: spanning and no relations. The difference is that coefficients now lie in $R$, and nonzero ring elements can behave in ways that nonzero field elements cannot. Finite coordinate systems are especially important because they are the setting where matrices, determinants over commutative rings, and finite presentations enter. Before rank becomes a numerical invariant, we first need a phrase for the concrete situation where a free module is presented with exactly $n$ coordinates. The invariant-basis issue is addressed later; at this stage the definition records the existence of an $n$-element basis. [definition: Free Module of Rank $n$] Let $R$ be a ring with identity and let $n \ge 0$ be an integer. A free left $R$-module of rank $n$ is a free left $R$-module admitting a basis indexed by $\{1, \ldots, n\}$, with this index set interpreted as empty when $n=0$. [/definition] For such a module, the standard model is $R^n$, whose elements are columns or tuples of ring elements; when $n=0$, this is the zero module $R^0=0$. A module that is free with the displayed $n$-element coordinate system is not merely generated by $n$ elements; it is isomorphic to this standard coordinate module. Over rings with invariant basis number, introduced below, the integer $n$ is independent of the chosen finite basis. [example: The Standard Free Module $R^n$] Let $R$ be a ring with identity, and write elements of $R^n$ as $n$-tuples. For each $i\in\{1,\ldots,n\}$, let $e_i$ be the tuple whose $i$th coordinate is $1_R$ and whose other coordinates are $0_R$. We show that $(e_i)_{i=1}^n$ is a basis of $R^n$. For any $x=(x_1,\ldots,x_n)\in R^n$, the $j$th coordinate of $\sum_{i=1}^n x_i e_i$ is \begin{align*} x_1(0_R)+\cdots+x_{j-1}(0_R)+x_j(1_R)+x_{j+1}(0_R)+\cdots+x_n(0_R)=x_j. \end{align*} Thus \begin{align*} x=\sum_{i=1}^n x_i e_i. \end{align*} So the standard coordinate vectors span $R^n$. Now suppose \begin{align*} \sum_{i=1}^n r_i e_i=0. \end{align*} The $j$th coordinate of the left side is \begin{align*} r_1(0_R)+\cdots+r_{j-1}(0_R)+r_j(1_R)+r_{j+1}(0_R)+\cdots+r_n(0_R)=r_j. \end{align*} The $j$th coordinate of the zero tuple is $0_R$, so $r_j=0_R$. Since this holds for every $j\in\{1,\ldots,n\}$, all coefficients are zero. Hence the standard coordinate vectors span $R^n$ and have no nonzero linear relation, so they form a basis. [/example] ### Coordinate Models The same idea works with infinitely many basis vectors, but the finite-support rule changes the ambient object. If an infinite basis is allowed, the natural coordinate module must remember that every element has only finitely many nonzero coordinates. This is the reason direct sums, rather than direct products, model free modules. [definition: Direct Sum of Copies of $R$] Let $R$ be a ring with identity and let $I$ be a set. The [direct sum](/page/Direct%20Sum) $\bigoplus_{i \in I} R$ is the left $R$-module consisting of all families $(r_i)_{i \in I}$ with $r_i \in R$ such that $r_i = 0_R$ for all but finitely many $i \in I$, with addition and scalar multiplication defined componentwise. [/definition] This construction is the coordinate model for every free module. The next theorem is needed because it upgrades the definition from an existential statement about a basis into a concrete isomorphism with a standard module. Once this isomorphism is available, calculations in an arbitrary free module can be transported to finite-support coordinate families. [quotetheorem:10013] This theorem turns the definition into a usable normal form. Once a basis has been chosen, every free module can be handled as a module of finite-support coordinate families. ## Coordinates and Universal Properties The coordinate expansion of a free module is more than a convenient notation. It tells us exactly how to define maps out of a free module: choose the images of the basis vectors, and extend by linearity. This is the algebraic mechanism behind presentations, matrices, and many constructions in [homological algebra](/page/Homological%20Algebra%20I%3A%20Complexes%20and%20Resolutions). ### Universal Mapping Property A basis gives unique coefficients, but uniqueness alone is cumbersome if we must repeatedly say "take the coefficient of this basis element." To make coordinates into reusable maps, we name the operation that extracts one coefficient from the finite expansion of an element. This is especially useful when checking formulas component by component: instead of comparing two finite sums directly, we can apply a coordinate extractor and compare the resulting scalar coefficients. The definition below packages that extractor as a map attached to a chosen basis, so later statements about homomorphisms and matrices can refer to coordinates without repeatedly unpacking the whole expansion. [definition: Coordinate Functional on a Free Module] Let $R$ be a ring with identity, let $M$ be a free left $R$-module with basis $(e_i)_{i \in I}$. For each $i \in I$, the $i$th coordinate functional with respect to this basis is the map \begin{align*} \pi_i: M \to R \end{align*} defined by sending each $m \in M$ to the coefficient $r_i \in R$ of $e_i$ in the unique expansion \begin{align*} m = \sum_{j \in I} r_j e_j \end{align*} where the coefficient family $(r_j)_{j \in I}$ has finite support. [/definition] Coordinates depend on the chosen basis, but the rule for building homomorphisms from coordinates does not. The possible obstruction would be a relation among basis elements: arbitrary chosen images might fail to respect that relation. A basis has no such hidden compatibility conditions, so prescribing images of its elements should determine exactly one homomorphism. [quotetheorem:10014] The result says that a [linear map](/page/Linear%20Map) out of a free module is determined by its values on a basis. This is the module-theoretic version of the familiar vector-space principle, but here it is often used as a definition-free construction tool. There is also a categorical way to state the same idea. For every set $I$, the construction $\bigoplus_{i \in I} R$ is the free left $R$-module on $I$: it is left adjoint to the forgetful functor from the category of left $R$-modules to the category of sets. In concrete terms, a set map from $I$ into the underlying set of a module $N$ extends uniquely to an $R$-[module homomorphism](/page/Module%20Homomorphism) from $\bigoplus_{i \in I} R$ to $N$. [example: Building a Homomorphism from Chosen Images] Let $R=\mathbb{Z}$ and let $M=\mathbb{Z}^3$ with standard basis $e_1,e_2,e_3$. Let $N=\mathbb{Z}/6\mathbb{Z}$, and choose \begin{align*} n_1=\bar{2}, \qquad n_2=\bar{3}, \qquad n_3=\bar{5}. \end{align*} By the *[Universal Property of Free Modules](/theorems/10014)*, there is a unique $\mathbb{Z}$-module homomorphism $f:\mathbb{Z}^3\to \mathbb{Z}/6\mathbb{Z}$ satisfying $f(e_1)=\bar{2}$, $f(e_2)=\bar{3}$, and $f(e_3)=\bar{5}$. Every $(a,b,c)\in\mathbb{Z}^3$ has the coordinate expansion \begin{align*} (a,b,c)=a e_1+b e_2+c e_3. \end{align*} Using $\mathbb{Z}$-linearity of $f$, this forces \begin{align*} f(a,b,c)=f(ae_1+be_2+ce_3)=a f(e_1)+b f(e_2)+c f(e_3). \end{align*} Substituting the chosen values gives \begin{align*} f(a,b,c)=a\bar{2}+b\bar{3}+c\bar{5}=\overline{2a+3b+5c}\in\mathbb{Z}/6\mathbb{Z}. \end{align*} In particular, \begin{align*} f(e_1)=f(1,0,0)=\overline{2}, \qquad f(e_2)=f(0,1,0)=\overline{3}, \qquad f(e_3)=f(0,0,1)=\overline{5}. \end{align*} The standard basis has no nonzero integer relation: if \begin{align*} \alpha e_1+\beta e_2+\gamma e_3=(0,0,0), \end{align*} then \begin{align*} (\alpha,\beta,\gamma)=(0,0,0), \end{align*} so $\alpha=\beta=\gamma=0$. Thus choosing the three images above imposes no compatibility condition beyond $\mathbb{Z}$-linearity, and the displayed formula is the unique homomorphism with those basis values. [/example] Maps into a free module are different: specifying coordinate functions into $R$ is enough only when those coordinate functions have finite support at each input. This distinction is why direct sums and direct products must not be confused. [remark: Maps Out and Maps In] If $M = \bigoplus_{i \in I} R$, then a homomorphism $M \to N$ is equivalent to a family of elements of $N$, one for each $i \in I$. A homomorphism $N \to M$ is equivalent to a family of homomorphisms $N \to R$ whose values have finite support for every fixed element of $N$. [/remark] ### Presentations by Generators and Relations The universal property also explains why free modules are the raw material for presenting arbitrary modules. Generators give a map from a free module, but the kernel of that map remembers all relations among the generators. A presentation is the formal device that records both the chosen generators and the relations at the same time. [definition: Presentation of a Module] Let $R$ be a ring with identity and let $M$ be a left $R$-module. A presentation of $M$ is an exact sequence of left $R$-modules \begin{align*} F_1 \xrightarrow{\varphi} F_0 \xrightarrow{\psi} M \to 0, \end{align*} where $F_0$ and $F_1$ are free left $R$-modules. [/definition] A presentation separates two tasks: first choose generators, then impose relations. Free modules appear because generators without relations are the starting point from which all modules can be quotiented. [example: Presenting $\mathbb{Z}/2\mathbb{Z}$] The cyclic $\mathbb{Z}$-module $\mathbb{Z}/2\mathbb{Z}$ is generated by $\bar{1}$, because for every $m\in \mathbb{Z}/2\mathbb{Z}$ either $m=\bar{0}=0\bar{1}$ or $m=\bar{1}=1\bar{1}$. Define \begin{align*} \psi:\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z},\qquad \psi(n)=\bar{n}. \end{align*} This is a $\mathbb{Z}$-module homomorphism: for $a,b,n\in\mathbb{Z}$, \begin{align*} \psi(a+b)=\overline{a+b}=\bar{a}+\bar{b}=\psi(a)+\psi(b). \end{align*} Also, \begin{align*} \psi(na)=\overline{na}=n\bar{a}=n\psi(a). \end{align*} The map is surjective because $\bar{0}=\psi(0)$ and $\bar{1}=\psi(1)$. Its kernel consists exactly of the even integers. Indeed, \begin{align*} n\in\ker(\psi)\Longleftrightarrow \bar{n}=\bar{0}\Longleftrightarrow 2\mid n. \end{align*} Therefore $\ker(\psi)=2\mathbb{Z}$. Let \begin{align*} \varphi:\mathbb{Z}\to\mathbb{Z},\qquad \varphi(n)=2n. \end{align*} Then $\operatorname{im}(\varphi)=2\mathbb{Z}=\ker(\psi)$, and \begin{align*} \psi(\varphi(n))=\psi(2n)=\overline{2n}=\bar{0} \end{align*} for every $n\in\mathbb{Z}$. Thus the sequence \begin{align*} \mathbb{Z} \xrightarrow{\times 2} \mathbb{Z} \xrightarrow{\psi} \mathbb{Z}/2\mathbb{Z} \to 0 \end{align*} is exact. Since both copies of $\mathbb{Z}$ are free $\mathbb{Z}$-modules with basis $1$, this is a presentation of $\mathbb{Z}/2\mathbb{Z}$. The second copy records the generator $\bar{1}=\psi(1)$, while the first copy records the relation \begin{align*} 2\bar{1}=\bar{2}=\bar{0}. \end{align*} [/example] ## Linear Independence, Spanning, and Relations In vector spaces, spanning and independence lead to bases through exchange arguments. For modules, the words still make sense, but the exchange arguments depend on the scalar ring. Free modules are the place where independence and spanning align into a basis, while nearby finitely generated modules may fail because of torsion or relations. The first step is to define independence without assuming a basis exists. This matters because a proposed coordinate system can fail in two different ways: it may miss elements, or it may contain relations. [Linear independence](/page/Linear%20Independence) isolates the second failure. [definition: Linearly Independent Family in a Module] Let $R$ be a ring with identity, let $M$ be a left $R$-module, and let $(m_i)_{i \in I}$ be a family of elements of $M$. The family is $R$-linearly independent if, for every finite subset $F \subset I$ and every choice of coefficients $r_i \in R$, the equality \begin{align*} \sum_{i \in F} r_i m_i = 0 \end{align*} implies $r_i = 0_R$ for every $i \in F$. [/definition] Independence detects relations, but it does not guarantee that the family reaches the whole module. To speak about coordinates, we also need the opposite condition: every element must be built from the proposed family using finitely many coefficients. [definition: Spanning Family in a Module] Let $R$ be a ring with identity, let $M$ be a left $R$-module, and let $(m_i)_{i \in I}$ be a family of elements of $M$. The family spans $M$ if every $m \in M$ can be written in the form \begin{align*} m = \sum_{i \in F} r_i m_i, \end{align*} where $F \subset I$ is finite and $r_i \in R$ for each $i \in F$. [/definition] Spanning alone gives generators, but a spanning expression may not be unique. Independence alone rules out nonzero finite relations, but it does not ensure every element is reached. The obstruction to being a basis is therefore exactly one of these two failures, and the characterization below separates the existence and uniqueness parts of coordinate expansion. [quotetheorem:10015] This characterization is useful because many failures of freeness can be diagnosed by finding either too few generators or a nonzero relation. [example: An Ideal That Is Not Free as a Module Over Its Ring] Let $k$ be a field and let \begin{align*} R = k[x,y]/(x,y)^2. \end{align*} Write $\bar{x}$ and $\bar{y}$ for the residue classes of $x$ and $y$, and let $I=(\bar{x},\bar{y})$. Since $(x,y)^2=(x^2,xy,y^2)$, we have \begin{align*} \bar{x}^2=\bar{0},\qquad \bar{x}\bar{y}=\bar{0},\qquad \bar{y}^2=\bar{0}. \end{align*} The element $\bar{x}$ is not zero in $R$, because every element of $(x,y)^2$ has zero degree-one part, while $x$ has degree-one part $x$. Thus $I\ne 0$. The ideal $I$ is generated by $\bar{x}$ and $\bar{y}$ as an $R$-module. More explicitly, every element of $R$ has the form \begin{align*} \lambda+\mu\bar{x}+\nu\bar{y} \end{align*} with $\lambda,\mu,\nu\in k$, so \begin{align*} (\lambda+\mu\bar{x}+\nu\bar{y})\bar{x}=\lambda\bar{x}+\mu\bar{x}^2+\nu\bar{y}\bar{x}=\lambda\bar{x}. \end{align*} Similarly, \begin{align*} (\lambda+\mu\bar{x}+\nu\bar{y})\bar{y}=\lambda\bar{y}+\mu\bar{x}\bar{y}+\nu\bar{y}^2=\lambda\bar{y}. \end{align*} Hence elements of $I$ are exactly the $k$-linear combinations $\alpha\bar{x}+\beta\bar{y}$. Now take two elements $a=\alpha\bar{x}+\beta\bar{y}$ and $b=\gamma\bar{x}+\delta\bar{y}$ of $I$. Their product is \begin{align*} ab=\alpha\gamma\bar{x}^2+\alpha\delta\bar{x}\bar{y}+\beta\gamma\bar{y}\bar{x}+\beta\delta\bar{y}^2=\bar{0}. \end{align*} Thus every element of $I$ annihilates every element of $I$. Suppose, for contradiction, that $I$ were a free $R$-module with basis $(e_j)_{j\in J}$. The basis cannot be empty, because an empty basis spans only the zero module, while $\bar{x}\in I$ is nonzero. Choose $j\in J$. Since $e_j\in I$ and $\bar{x}\in I$, the calculation above gives \begin{align*} \bar{x}e_j=0. \end{align*} But $\bar{x}\ne 0$ in $R$, so this is a nonzero coefficient relation on the single basis vector $e_j$, contradicting linear independence. Therefore $I$ is not free as an $R$-module. Since $I$ is an ideal of $R$, it is a submodule of the free rank-one module $R$, so this shows that submodules of free modules over a general ring need not be free. [/example] The failure in this example is one of the main differences between modules over fields and modules over arbitrary rings. It raises the natural repair question: which rings are controlled enough that submodules of free modules remain free? Principal ideal domains give one of the most important positive answers. [quotetheorem:10016] This theorem explains why modules over $\mathbb{Z}$ behave much closer to vector spaces than modules over arbitrary rings. It is also the structural reason finitely generated abelian groups can be studied using matrices over $\mathbb{Z}$. ## Rank and Invariance For a vector space, dimension is the number of basis vectors, independent of the chosen basis. For free modules, the corresponding number is called rank. The subtlety is that over unusual rings, different bases of a free module can have different cardinalities unless the ring satisfies an invariance condition. Most rings encountered in commutative algebra have the desired property, but the distinction matters conceptually. The finite-rank case is the most common setting, and it is the one in which matrices are used directly. To define rank responsibly, we must first acknowledge that the number of basis elements should not depend on the chosen basis. The definition below packages that independence into the notation. [definition: Rank of a Finite Free Module] Let $R$ be a ring with identity, and let $M$ be a free left $R$-module. If $M$ admits a basis with $n \ge 0$ elements, and every basis of $M$ has $n$ elements, then the rank of $M$ is $n$, written $\operatorname{rank}_R(M)=n$. [/definition] This definition includes the invariance requirement rather than hiding it. The next concept names the property of a ring that guarantees finite free modules have a well-defined rank. Without it, the notation $\operatorname{rank}_R(R^n)=n$ could be misleading. [definition: Invariant Basis Number] A ring $R$ with identity has invariant basis number if, whenever $m,n \ge 0$ are integers and $R^m \cong R^n$ as left $R$-modules, then $m=n$. [/definition] Invariant basis number is the ring-theoretic condition that makes rank a reliable numerical invariant. Without it, an isomorphism between two different finite powers of the same ring would make the phrase "the number of basis elements" ambiguous. Nonzero commutative rings avoid this obstruction, which is why rank can be used over them with the same numerical confidence as dimension over fields. [quotetheorem:8630] [example: Rank over $\mathbb{Z}$] The $\mathbb{Z}$-module $\mathbb{Z}^4$ has standard basis $(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)$, so it is free of rank $4$. Consider the subgroup \begin{align*} N = \{(a,b,c,d) \in \mathbb{Z}^4 : a+b+c+d=0\}. \end{align*} Define \begin{align*} v_1=(1,-1,0,0),\qquad v_2=(1,0,-1,0),\qquad v_3=(1,0,0,-1). \end{align*} Each $v_i$ lies in $N$, since \begin{align*} 1+(-1)+0+0=0,\qquad 1+0+(-1)+0=0,\qquad 1+0+0+(-1)=0. \end{align*} We show that $v_1,v_2,v_3$ form a basis of $N$. Let $(a,b,c,d)\in N$. Then \begin{align*} a+b+c+d=0. \end{align*} Subtracting $b+c+d$ from both sides gives \begin{align*} a=-b-c-d. \end{align*} Now \begin{align*} -bv_1=(-b,b,0,0),\qquad -cv_2=(-c,0,c,0),\qquad -dv_3=(-d,0,0,d). \end{align*} Adding these three vectors coordinatewise gives \begin{align*} -bv_1-cv_2-dv_3=(-b-c-d,b,c,d). \end{align*} Since $a=-b-c-d$, this becomes \begin{align*} -bv_1-cv_2-dv_3=(a,b,c,d). \end{align*} Thus $v_1,v_2,v_3$ span $N$. It remains to check uniqueness of the coefficients. Suppose \begin{align*} \alpha v_1+\beta v_2+\gamma v_3=(0,0,0,0) \end{align*} with $\alpha,\beta,\gamma\in\mathbb{Z}$. Expanding the left side coordinatewise, \begin{align*} \alpha(1,-1,0,0)+\beta(1,0,-1,0)+\gamma(1,0,0,-1)=(\alpha+\beta+\gamma,-\alpha,-\beta,-\gamma). \end{align*} Hence \begin{align*} (\alpha+\beta+\gamma,-\alpha,-\beta,-\gamma)=(0,0,0,0). \end{align*} Equality of the second, third, and fourth coordinates gives \begin{align*} -\alpha=0,\qquad -\beta=0,\qquad -\gamma=0. \end{align*} Therefore $\alpha=\beta=\gamma=0$. So there is no nonzero integer relation among $v_1,v_2,v_3$. The spanning calculation gives existence of an expression in terms of $v_1,v_2,v_3$, and the relation calculation gives uniqueness: if two expressions are equal, subtracting them gives a relation, so the corresponding coefficients must be equal. Therefore $N$ is a free $\mathbb{Z}$-module with basis $v_1,v_2,v_3$, and hence $N$ has rank $3$. [/example] Rank should behave additively when two independent coordinate systems are placed side by side, but this only makes sense after rank is known to be independent of the chosen basis. For a direct sum, the natural candidate basis is obtained by putting the two bases in separate summands; the point is to ensure that this construction gives the correct numerical invariant. This additivity result is the next structural test for rank: if rank is to function as a size invariant for free modules, it must convert a decomposed module into the sum of the sizes of its independent pieces. The theorem records that compatibility and gives a rule that will later mirror how matrices and exact sequences track separate coordinate blocks. [quotetheorem:10017] The theorem matches the geometric intuition that placing coordinate systems side by side adds the number of coordinates. It also anticipates the behaviour of bases under exact sequences when splitting is available. ## Matrices and Homomorphisms Free modules over a commutative ring are where matrices act in the familiar column-vector way. A homomorphism $R^n \to R^m$ is encoded by an $m \times n$ matrix with entries in $R$, and composition of homomorphisms becomes ordinary matrix multiplication. This is an extra commutativity assumption, not a feature of arbitrary left modules: over noncommutative rings, ordinary $Ax$ column multiplication is not compatible with every left-module convention unless the side of scalar action and the matrix action are chosen with care. For that reason, this section states the matrix results in the commutative setting. To use this bridge, we first fix the matrix convention. The convention is needed because a matrix has two indexing directions: columns will record images of domain basis vectors, and rows will record coordinates in the codomain basis. [definition: Matrix of a Homomorphism Between Finite Free Modules] Let $R$ be a commutative ring with identity, let $M$ and $N$ be finite free $R$-modules with ordered bases $(e_j)_{j=1}^n$ and $(f_i)_{i=1}^m$, and let $\varphi: M \to N$ be an $R$-module homomorphism. The matrix of $\varphi$ with respect to these bases is the matrix $A=(a_{ij}) \in R^{m \times n}$ determined by \begin{align*} \varphi(e_j)=\sum_{i=1}^m a_{ij}f_i \end{align*} for each $j \in \{1,\ldots,n\}$. [/definition] The $j$th column records the image of the $j$th basis vector. The possible concern is that matrix data might either miss a homomorphism or encode extra choices beyond the homomorphism itself. Because a map out of a free module is determined by basis images, the column convention should give a one-to-one translation between homomorphisms and matrices. The formal correspondence is needed before matrix notation can be used as more than shorthand. It says that choosing a matrix is exactly the same data as choosing an $R$-linear map between the two finite free modules with their ordered bases, so later computations with entries and columns genuinely describe homomorphisms rather than an auxiliary encoding. The remaining question is existence and uniqueness in both directions: given a homomorphism, its columns are forced by the images of the basis vectors, while given an arbitrary matrix, those columns should assemble into one well-defined homomorphism. The theorem below records that this two-way passage is exact. [quotetheorem:10018] This is the point at which free modules justify the use of matrices outside vector spaces. The coefficients no longer have to lie in a field, so row reduction may lose some of its power, but matrix notation remains meaningful. [example: A Homomorphism over $\mathbb{Z}$] Consider the homomorphism $\varphi: \mathbb{Z}^2 \to \mathbb{Z}^3$ whose first standard basis vector maps to $(2,0,4)$ and whose second standard basis vector maps to $(1,3,-1)$. Write the standard basis of $\mathbb{Z}^2$ as $e_1=(1,0)$ and $e_2=(0,1)$. Every $(a,b)\in\mathbb{Z}^2$ has the coordinate expansion \begin{align*} (a,b)=a e_1+b e_2. \end{align*} Using $\mathbb{Z}$-linearity of $\varphi$, \begin{align*} \varphi(a,b)=\varphi(ae_1+be_2)=a\varphi(e_1)+b\varphi(e_2). \end{align*} Substituting $\varphi(e_1)=(2,0,4)$ and $\varphi(e_2)=(1,3,-1)$ gives \begin{align*} \varphi(a,b)=a(2,0,4)+b(1,3,-1). \end{align*} Scalar multiplication in $\mathbb{Z}^3$ is coordinatewise, so \begin{align*} a(2,0,4)=(2a,0,4a) \end{align*} and \begin{align*} b(1,3,-1)=(b,3b,-b). \end{align*} Adding these coordinatewise gives \begin{align*} \varphi(a,b)=(2a+b,3b,4a-b). \end{align*} The image consists exactly of all integer linear combinations of the two displayed column vectors: \begin{align*} \operatorname{im}(\varphi)=\{a(2,0,4)+b(1,3,-1):a,b\in\mathbb{Z}\}. \end{align*} Thus $\operatorname{im}(\varphi)$ is the submodule of $\mathbb{Z}^3$ generated by $(2,0,4)$ and $(1,3,-1)$. The kernel is obtained by setting the three coordinates of $\varphi(a,b)$ equal to the three coordinates of $(0,0,0)$. Hence \begin{align*} (a,b)\in\ker(\varphi)\Longleftrightarrow (2a+b,3b,4a-b)=(0,0,0). \end{align*} Coordinate equality in $\mathbb{Z}^3$ gives \begin{align*} (a,b)\in\ker(\varphi)\Longleftrightarrow 2a+b=0,\ 3b=0,\ 4a-b=0. \end{align*} Since $3b=0$ in $\mathbb{Z}$ implies $b=0$, the equation $2a+b=0$ then gives $2a=0$, hence $a=0$. Therefore the third equation is also satisfied, and \begin{align*} \ker(\varphi)=\{(0,0)\}. \end{align*} So this homomorphism embeds $\mathbb{Z}^2$ into $\mathbb{Z}^3$ as the submodule generated by $(2,0,4)$ and $(1,3,-1)$. [/example] Changing bases changes the matrix but not the homomorphism. To know which matrices can arise from basis changes, we need the module-theoretic version of invertibility. Over a commutative ring, an invertible change of coordinates is exactly an invertible matrix over that ring. [quotetheorem:10019] The theorem is the correct replacement for the vector-space basis-change story. The entries lie in $R$, and invertibility happens inside the matrix ring over $R$. ## Free, Projective, and Flat Modules Free modules are rarely the endpoint of module theory. They are the easiest modules from which to build maps, but many naturally occurring modules are not free. The next layer consists of modules that behave like direct summands of free modules. ### Projective Modules A direct summand keeps enough of the coordinate structure of a free module to retain many exactness properties. The key test is whether maps out of the module can be lifted across surjections. This test leads to projective modules. [definition: Projective Module] Let $R$ be a ring with identity. A left $R$-module $P$ is projective if, for every surjective homomorphism $\pi: M \to N$ of left $R$-modules and every homomorphism $f: P \to N$, there exists a homomorphism $\widetilde{f}: P \to M$ such that \begin{align*} \pi \circ \widetilde{f}=f. \end{align*} [/definition] Projective modules are the modules whose maps lift through surjections. Free modules have this property because maps out of them are determined by basis elements, and each basis image can be lifted separately. The next theorem formalizes this first link between freeness and homological algebra. [quotetheorem:4201] The theorem is one of the first reasons free modules matter in [homological algebra](/page/Homological%20Algebra%20I%3A%20Complexes%20and%20Resolutions): projective resolutions are built from modules that lift maps well, and free modules provide the most concrete supply. ### Flat Modules The [tensor product](/page/Tensor%20Product) gives a second way in which free modules behave well. Tensoring can destroy injections, so we need a condition that says a module preserves injections under tensor product. This condition is flatness, and free modules satisfy it because tensoring with a direct sum of copies of $R$ is especially controlled. [definition: Flat Module] Let $R$ be a commutative ring with identity. An $R$-module $M$ is flat if the functor from the category of $R$-modules to the category of $R$-modules given on objects by \begin{align*} N \mapsto N \otimes_R M \end{align*} sends injective homomorphisms of $R$-modules to injective homomorphisms. [/definition] Flatness is motivated by tensor products rather than by lifting. The obstruction it rules out is that tensoring might collapse distinct elements and turn an injection into a non-injection. Free modules avoid this problem because tensoring with a direct sum of copies of the base ring reduces to taking a corresponding direct sum of the original module. [quotetheorem:10020] The route from free to flat usually passes through the fact that tensoring with a direct sum of copies of $R$ is a direct sum of copies of the original module. Thus free modules preserve the relevant injectivity. [example: A Projective Module That Need Not Be Free] Let $R=R_1\times R_2$, where $R_1$ and $R_2$ are nonzero commutative rings with identity, and let \begin{align*} P=R_1\times 0\subset R. \end{align*} Set \begin{align*} Q=0\times R_2. \end{align*} Every element $(a,b)\in R$ decomposes as \begin{align*} (a,b)=(a,0)+(0,b), \end{align*} with $(a,0)\in P$ and $(0,b)\in Q$. If $(a,0)\in P\cap Q$, then also $(a,0)=(0,b)$ for some $b\in R_2$, so equality of coordinates gives $a=0$ and $b=0$. Hence $P\cap Q=\{0\}$, and therefore \begin{align*} R=P\oplus Q. \end{align*} We verify projectivity from the lifting definition. Let $\pi:M\to N$ be a surjective $R$-module homomorphism, and let $f:P\to N$ be an $R$-module homomorphism. Write \begin{align*} e=(1_{R_1},0)\in P. \end{align*} Since $\pi$ is surjective, choose $m\in M$ such that \begin{align*} \pi(m)=f(e). \end{align*} For $(a,0)\in P$, define \begin{align*} \widetilde{f}(a,0)=(a,0)m. \end{align*} This is $R$-linear because scalar multiplication in $M$ is $R$-linear. Also, \begin{align*} (\pi\circ \widetilde{f})(a,0)=\pi((a,0)m)=(a,0)\pi(m)=(a,0)f(e). \end{align*} Since $(a,0)=(a,0)e$ and $f$ is $R$-linear, \begin{align*} (a,0)f(e)=f((a,0)e)=f(a,0). \end{align*} Thus $\pi\circ \widetilde{f}=f$, so $P$ is projective. Now set \begin{align*} s=(0,1_{R_2})\in R. \end{align*} For every $(a,0)\in P$, \begin{align*} s(a,0)=(0,1_{R_2})(a,0)=(0a,1_{R_2}0)=(0,0). \end{align*} Thus $s$ annihilates all of $P$. The element $s$ is nonzero because $1_{R_2}\ne 0$. Suppose, for contradiction, that $P$ were a free $R$-module with basis $(b_i)_{i\in I}$. Since $R_1$ is nonzero, choose $a\in R_1$ with $a\ne 0$; then $(a,0)\in P$ is nonzero, so $P\ne 0$ and the basis is not empty. Choose $j\in I$. Because $b_j\in P$, the annihilation calculation gives \begin{align*} s b_j=0. \end{align*} This is a linear relation on the single basis element $b_j$ with coefficient $s\ne 0$, contradicting linear independence of a basis. Therefore $P$ is projective but not free. Projective modules can therefore extend beyond free modules. [/example] The example is a useful warning: freeness is a coordinate condition, not just a lifting condition. A projective module may look locally like a coordinate module while failing to have one global basis. ## Free Resolutions and Presentations Many modules are not free, but they can be approximated by free modules. A presentation gives the first approximation; a free resolution continues recording relations among relations. This is the starting point for derived functors such as $\operatorname{Tor}_n^R(M,N)$ and $\operatorname{Ext}_R^n(M,N)$. The definition is naturally expressed as an exact sequence. It is needed because a single presentation records only generators and first relations, while longer homological constructions need the higher relations as well. [definition: Free Resolution] Let $R$ be a ring with identity and let $M$ be a left $R$-module. A free resolution of $M$ is an exact sequence of left $R$-modules \begin{align*} \cdots \to F_2 \to F_1 \to F_0 \to M \to 0, \end{align*} where each $F_i$ is a free left $R$-module. [/definition] Free resolutions exist in great generality because [every module is a quotient of a free module](/theorems/4552). The obstruction is that a module may have relations, torsion, or other features preventing it from being free itself; the repair is to start with formal generators in a free module and map them onto the given module, leaving the relations in the kernel. [quotetheorem:4552] This theorem is not saying that every module is free. It says that every module can be reached from a free module by imposing relations. Iterating the same idea on successive kernels gives a full resolution. [quotetheorem:4557] The construction repeatedly takes kernels and covers them by free modules. Each step records the next layer of relations. [example: A Free Resolution of $\mathbb{Z}/2\mathbb{Z}$] Let $\iota:0\to\mathbb{Z}$ be the zero map from the zero module, let \begin{align*} \mu:\mathbb{Z}\to\mathbb{Z},\qquad \mu(n)=2n, \end{align*} and let \begin{align*} \pi:\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z},\qquad \pi(n)=\bar{n}. \end{align*} We verify that \begin{align*} 0 \to \mathbb{Z} \xrightarrow{\mu} \mathbb{Z} \xrightarrow{\pi} \mathbb{Z}/2\mathbb{Z} \to 0 \end{align*} is exact. First, $\mu$ is injective: if $\mu(n)=0$, then $2n=0$ in $\mathbb{Z}$, so $n=0$. Hence $\operatorname{im}(\iota)=\{0\}=\ker(\mu)$. Next, \begin{align*} (\pi\circ\mu)(n)=\pi(2n)=\overline{2n}=\bar{0} \end{align*} for every $n\in\mathbb{Z}$, so $\operatorname{im}(\mu)\subseteq\ker(\pi)$. Conversely, if $n\in\ker(\pi)$, then \begin{align*} \bar{n}=\bar{0} \end{align*} in $\mathbb{Z}/2\mathbb{Z}$, which means $2\mid n$. Thus $n=2k$ for some $k\in\mathbb{Z}$, and then \begin{align*} n=2k=\mu(k). \end{align*} Therefore $\ker(\pi)\subseteq\operatorname{im}(\mu)$, so $\ker(\pi)=\operatorname{im}(\mu)$. Finally, $\pi$ is surjective because every element of $\mathbb{Z}/2\mathbb{Z}$ is either $\bar{0}=\pi(0)$ or $\bar{1}=\pi(1)$. Since both copies of $\mathbb{Z}$ are free $\mathbb{Z}$-modules with basis $1$, this exact sequence is a free resolution of $\mathbb{Z}/2\mathbb{Z}$. The middle equality $\ker(\pi)=2\mathbb{Z}$ records the relation on the generator $\bar{1}$: \begin{align*} 2\bar{1}=\bar{2}=\bar{0}. \end{align*} [/example] Free resolutions turn non-free modules into complexes of free modules. Homological algebra then studies how much information survives after applying functors such as tensor product and $\operatorname{Hom}_R(-,N)$. ## Failure Modes and Ring Dependence The phrase free module sounds close to vector space, but the scalar ring controls much of the behaviour. Over fields, every module is free. Over principal ideal domains, submodules of free modules are free. Over general rings, finite generation, torsion-freeness, projectivity, and freeness are separate properties. A common false expectation is that a submodule of a free module should be free. The earlier ideal example shows this can fail. Another false expectation is that torsion-free should imply free, so we first name the property that removes torsion. [definition: Torsion-Free Module over an Integral Domain] Let $R$ be an [integral domain](/page/Integral%20Domain), and let $M$ be an $R$-module. The module $M$ is torsion-free if, whenever $r \in R$ and $m \in M$ satisfy $rm=0$, with $r \ne 0_R$, then $m=0$. [/definition] Torsion-freeness removes the most visible obstruction to freeness, but it does not supply a basis. The missing ingredient is the existence of enough independent generators. [example: A Torsion-Free Module That Is Not Free] Regard $\mathbb{Q}$ as a $\mathbb{Z}$-module by ordinary integer multiplication. It is torsion-free: if $nq=0$ with $n\ne 0$, write $q=\frac{a}{b}$ with $a,b\in\mathbb{Z}$ and $b\ne 0$. Then \begin{align*} nq=n\frac{a}{b}=\frac{na}{b}=0. \end{align*} Since $\frac{na}{b}=0$ in $\mathbb{Q}$, we have $na=0$ in $\mathbb{Z}$. Because $n\ne 0$ and $\mathbb{Z}$ has no zero divisors, $a=0$, so $q=0$. We show that $\mathbb{Q}$ is not a free $\mathbb{Z}$-module. Suppose, for contradiction, that $\mathbb{Q}$ has a basis $(e_i)_{i\in I}$. Since $1\ne 0$ in $\mathbb{Q}$, its coordinate expansion has at least one nonzero coefficient: \begin{align*} 1=\sum_{i\in F} a_i e_i \end{align*} for some finite subset $F\subset I$, with $a_i\in\mathbb{Z}$ and with $a_j\ne 0$ for at least one $j\in F$. Choose a [prime number](/page/Prime%20Number) $p$ that does not divide $a_j$; such a prime exists because only finitely many primes divide the nonzero integer $a_j$. In $\mathbb{Q}$, the element $\frac{1}{p}$ satisfies \begin{align*} p\cdot \frac{1}{p}=1. \end{align*} Expand $\frac{1}{p}$ in the supposed basis: \begin{align*} \frac{1}{p}=\sum_{i\in G} b_i e_i \end{align*} with $G\subset I$ finite and $b_i\in\mathbb{Z}$. Multiplying by $p$ gives \begin{align*} 1=p\cdot \frac{1}{p}=p\sum_{i\in G} b_i e_i=\sum_{i\in G} (pb_i)e_i. \end{align*} The basis expansion of $1$ is unique, so the coefficient of $e_j$ in this last expression must equal the coefficient of $e_j$ in the earlier expression. Hence \begin{align*} a_j=pb_j \end{align*} if $j\in G$, and $a_j=0$ if $j\notin G$. The second case contradicts $a_j\ne 0$, while the first case says that $p$ divides $a_j$, contradicting the choice of $p$. Therefore $\mathbb{Q}$ is not free as a $\mathbb{Z}$-module. Thus $\mathbb{Q}$ has no integer torsion, but it still has no $\mathbb{Z}$-basis; torsion-free does not imply free. [/example] This example explains why module theory cannot be reduced to checking torsion. At the opposite end, fields remove every obstruction to freeness because scalar division restores the usual basis theory. The next theorem identifies vector spaces as the special case where every module has coordinates. [quotetheorem:10021] This theorem identifies vector spaces as the best-behaved special case of module theory. It also explains why many module-theoretic failures are invisible in ordinary linear algebra. ## Beyond and Connected Topics Free modules sit at the entrance to several larger theories. In commutative algebra, finite free modules are the coordinate models used to define finite presentations, projective modules, flatness, and local rank. The next layer is local freeness: a module may fail to be free over $R$ but become free after localizing at every prime ideal $\mathfrak{p} \in \operatorname{Spec}(R)$. In homological algebra, free modules provide the most concrete resolutions. Applying $\operatorname{Hom}_R(-,N)$ or $-\otimes_R N$ to a free resolution produces complexes whose homology defines $\operatorname{Ext}_R^n(M,N)$ and $\operatorname{Tor}_n^R(M,N)$. This is the natural continuation in [Homological Algebra I: Complexes and Resolutions](/page/Homological%20Algebra%20I%3A%20Complexes%20and%20Resolutions). In algebraic geometry, finite locally free modules correspond to vector bundles on affine schemes. The failure of a projective module to be globally free becomes geometric: a [vector bundle](/page/Vector%20Bundle) can be locally a product while having no global frame. In undergraduate algebra, free abelian groups and modules over polynomial rings provide the computational setting for Smith normal form, presentations, and classification theorems. The relevant background is developed in [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules) and extended in [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). ## References Androma, [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules). Androma, [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). Androma, [Homological Algebra I: Complexes and Resolutions](/page/Homological%20Algebra%20I%3A%20Complexes%20and%20Resolutions). Androma, [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). Atiyah and Macdonald, *Introduction to Commutative Algebra* (1969). Lang, *Algebra* (2002). Rotman, *An Introduction to Homological Algebra* (2009).