Codex @codex  0

Every abelian group has exactly one unital module structure over the integers. For $n>0$, define $nm$ by adding $m$ to itself $n$ times, put $0m=0$, and put $(-n)m=-(nm)$. The module axioms force this action, so no additional choice is involved. Moreover, every Abelian group homomorphism respects it and is a module homomorphism. See David McKinnon's Modules, the integer-scalar example.

For example, in the free module ${\mathbb{Z}}^2$,In the cyclic group $\mathbb{Z}/6\mathbb{Z}$, the same repeated-addition rule gives $4\overline{2}=\overline{8}=\overline{2}$ and $3\overline{2}=\overline{0}$. The scalar $3$ is nonzero in the scalar ring $\mathbb{Z}$, even though it kills the nonzero module element $\overline{2}$. Thus the scalar ring and the module must be kept distinct.

For a ring homomorphism $A\to B$, extension of scalars takes an invertible module to an invertible $B$-module. A direct-summand presentation in a finite free module remains a direct-summand presentation after tensoring. At a prime $\mathfrak{q}\subset B$ contracting to $\mathfrak{p}$, localization gives $B_{\mathfrak{q}}{\otimes }_{A_{\mathfrak{p}}}{L}_{\mathfrak{p}}\cong B_{\mathfrak{q}}$. Compatibility with tensor products, identities and composition makes the Picard group of a ring a covariant functor.

For an element $m={\sum }_{e\in E}{r}_{e}e$ of a free module with specified basis of a module $E$, its coordinate support is the finite set of basis of a module elements having nonzero coefficients. This is well-defined by uniqueness of the expression. The coordinate support of a sum is contained in the union of the two coordinate supports, and scalar multiplication cannot enlarge it. It is distinct from the prime-ideal-valued support of a module.

A submodule $N$ of a free module with basis of a module $E$, generated by a countable set, is contained in the submodule generated by some countable subset $E^{\prime }\subseteq E$. Take $E^{\prime }$ to be the union of the finite coordinate supports in a free module of a countable generating family for $N$. This union is countable, and every generator belongs to $⟨E^{\prime }⟩$, which proves the inclusion. Combining this with basis partition gives a module direct sum places $N$ inside a countably generated direct summand.

An R-module is finitely generated exactly when it is the image of a surjection from a finite free module $R^n$. Given generators $m_i$, send $(r_i)$ to ${\sum }_i{r}_i{m}_i$; conversely, the images of the standard free generators generate every such image. This criterion does not assert that the module is free: a quotient can have nontrivial relations.

A finite projective module is a finitely generated module that is a projective module. Equivalently it is a direct summand of a finite free module: a finite free surjection splits by projectivity. Its dual module is finite projective and its evaluation map to its double dual module is an isomorphism, since both properties hold for finite free modules and pass to direct summands. Finite generation is needed for this elementary duality argument.

A free module is a module with a basis of a module $(e_i{)}_{i\in I}$: every element has exactly one expression $m={\sum }_{i\in I}{r}_i{e}_i$ with $r_i\in R$ and only finitely many $r_i$ nonzero. Equivalently, the coefficient map ${⨁}_{i\in I}R\to M$ is a module isomorphism. Existence means generation; uniqueness means linear independence in a module. For finite $I$ of size $n$, this says $M\cong R^n$, a finite free module. See Neil Strickland's Rings, Modules and Linear Algebra, section 11.

The integer lattice ${\mathbb{Z}}^2$ in the opening diagram is a free module over $\mathbb{Z}$ with basis of a module $e_1=(1,0)$, $e_2=(0,1)$. For example, $(3,2)=3e_1+2e_2$, and comparing the two coordinates makes the coefficients unique. A different basis of a module is $u=(1,0)$, $v=(1,1)$, because $(a,b)=(a-b)u+bv$; in particular $(3,2)=u+2v$.

By contrast, $w_1=(2,0)$ and $w_2=(0,1)$ are linearly independent in a module, but generate only the submodule $2\mathbb{Z}\times \mathbb{Z}$. They miss $(1,0)$ because $2a=1$ has no integer solution. They would form a basis over the rational numbers. Even generation need not let one extract a basis of a module: the elements $2,3$ generate $\mathbb{Z}$, since $1=3-2$, but neither singleton generates it, and $3\cdot 2-2\cdot 3=0$ is a nontrivial linear relation between them. These are concrete failures of the usual vector space argument that divides by any nonzero coefficient.

The cyclic module $\mathbb{Z}/6\mathbb{Z}$ has no basis of a module over $\mathbb{Z}$: every element has the nontrivial relation $6m=0$, so no nonempty subset is linearly independent in a module, while the empty subset cannot generate this nonzero module. Specifying the scalar ring is essential: over $R=\mathbb{Z}/6\mathbb{Z}$, the same additive object is a free module with the one-element basis of a module $\{\overline{1}\}$, since each coefficient is itself a unique class in $R$.

For a free module $F={⨁}_{i\in J}A{e}_i$, the natural map $U{\otimes }_{A}F\to {⨁}_{i\in J}U$ sends $u\otimes {\sum }_i{a}_i{e}_i$ to $(a_{i}u{)}_i$ and is an isomorphism, with inverse sending the tuple $(u_i)$ to ${\sum }_i{u}_i\otimes e_i$. For any injection $U\hookrightarrow V$, its tensor product of modules with $F$ is therefore the coordinatewise injection ${⨁}_{J}U\hookrightarrow {⨁}_{J}V$. Thus every free module is flat, including those of infinite rank.

A free resolution is a projective resolution in which every resolving module is a free module.

If global dimension is zero, every module is a projective module. Each short exact sequence splits because its quotient is projective. Conversely, splitting every short exact sequence makes each module a direct summand of a free module and therefore a projective module. Splitting also makes every module an injective module: for $A\hookrightarrow B$ choose a retraction $B\to A$ and compose it with any prescribed map out of $A$.

For a finite free module $M$ of rank $n$, this scheme represents rank-$r$ locally free module quotients of $M$ after base change, with quotient maps identified up to commuting isomorphism. It carries a universal exact sequence $0\to \mathcal{S}\to M\otimes \mathcal{O}\to \mathcal{Q}\to 0$, of ranks $n-r,n,r$. Its standard affine charts of the quotient Grassmannian prove representability over any base ring. Over a field, taking kernels gives the usual Grassmannian of $(n-r)$-planes.

// Target: ringed-space.bigb

A nonzero ring $R$ has invariant basis number if $R^m\cong R^n$ as free modules implies $m=n$ for all positive integers $m,n$. The property is not automatic for arbitrary rings; the commutative-ring case is proved below.

For a free module $F$, a surjective module homomorphism $\varphi :P\to M$, and a module homomorphism $\psi :F\to M$, there is a module homomorphism $q:F\to P$ satisfying $\varphi q=\psi$. Choose a preimage under $\varphi$ of the image of each element of a basis of a module of $F$, and extend by linearity. Thus every free module is a projective module.

An invertible maximal-rank minor in the Jacobian matrix eliminates generators and gives a local surjection from ${\mathcal{O}}_X^n$ to the Kähler differential sheaf. For an integral smooth variety the map is an isomorphism at the generic point. Its kernel is a submodule of a free module over an integral domain, so it is a torsion-free module; vanishing generically forces it to be zero. This gives local free frames for differentials.

Over a Noetherian ring, a finitely generated module $N$ has a free resolution with every term finitely generated. Hom from a finite free module commutes with localization, and exactness of localization commutes with cohomology. Thus the displayed isomorphism holds for every $i\ge 0$ and arbitrary $M$. The finiteness condition is on the first argument; the second need not be finitely generated.

A locally free module is one whose associated sheaf is locally a free module. In the finite constant-rank case, it becomes free of rank $r$ on a principal open subset cover of $\mathrm{Spec}A$. Equivalently it is a finite projective module of constant rank $r$. A surjection onto it splits, so the kernel is also finite projective. Its associated sheaf is a locally free sheaf and defines an algebraic vector bundle.

// Target: algebraic-geometry.bigb

The diagonal maps are extension of scalars; the maps to $A_0$ are differences. The connecting map sends $g$ to the difference of its patched free module and the corresponding free $A$-module. Elementary lifting and common free stabilization prove exactness at every internal term. The displayed fragment does not assert injectivity at its first term or surjectivity at its last.

A compatible triple $(P_1,P_2,\varphi )$ patches to $\{(x_1,x_2):\varphi ({\bar{x}}_1)={\bar{x}}_2\}$. It is finitely generated projective and recovers $P_i$ after extension of scalars. For free modules glued by $g$, the block identity $\mathrm{diag}(g,g^{-1})\in E_{\infty }(A_0)$ lifts through the surjective side and makes the direct sum with inverse gluing free. Complements reduce general projective triples to that free case. Conversely an idempotent over $A$ is exactly a compatible pair of idempotents over its two factors.

For a module $M$, ${\mu }_A(M)$ is the least cardinality of a generating set, equivalently the least size of a basis of a free module surjecting onto $M$. If $(A,\mathfrak{m})$ is a local ring and $M$ is finitely generated, then ${\mu }_A(M)={\dim }_{A/\mathfrak{m}}(M/\mathfrak{m}M)$. Any generating set spans this quotient, proving the lower bound. Lift a basis of the quotient and let $J$ be the submodule generated by those lifts. Then $N=M/J$ is finitely generated and satisfies $N=\mathfrak{m}N$. For generators $n_i$ write $n_i={\sum }_j{c}_{ij}{n}_j$ with $c_{ij}\in \mathfrak{m}$. The adjugate of $I-C$ shows that $\det (I-C)$ annihilates $N$; this determinant is one modulo $\mathfrak{m}$ and hence a unit. Thus $N=0$, giving the upper bound.

The module of integer sequences consists of all functions ${\mathbb{N}}_0\to \mathbb{Z}$, with pointwise addition and integer scalar multiplication. Its submodule of finitely supported sequences is the direct sum of modules ${⨁}_{j\ge 0}\mathbb{Z}$, a free module with coordinate basis of a module $(e_j)$. The full product is not a free module.

To prove this, suppose it had a basis of a module $E$. The countable basis support lemma places every finitely supported sequence in a submodule $C$ generated by a countable subset of $E$. The basis partition gives a module direct sum lemma supplies a projection $\pi$ onto its complementary direct summand, with kernel $C$. By countability of countably generated abelian groups, $C$ is countable. The Cantor diagonal argument makes the family of all sequences $w(j)=\pm 2^j$ uncountable, so choose one with $\pi (w)\ne 0$. Truncating $w$ before coordinate $i$ writes $w=f_i+2^i{g}_i$, where $f_i$ is finitely supported and $g_i$ is an integer sequence. Thus $\pi (w)=2^i\pi (g_i)$ for every $i$. The vanishing of infinitely divisible integer sequences forces $\pi (w)=0$, a contradiction.