Module theory

Module theory studies abelian groups with scalar multiplication by a ring. It extends linear algebra: a module over a field is a vector space, while ring scalars let the same language describe integer arithmetic and linear operators. The new phenomena are visible before any abstract classification: some nonzero scalars cannot be divided out, some modules have torsion elements of a module, and some have no basis of a module.

The distinction is useful because an integer relation such as $6m=0$ carries information which disappears if one assumes all nonzero scalars have multiplicative inverses. A finitely generated module over a principal ideal domain can be separated into a finite free module and a torsion module; the structure theorem for finitely generated modules over a principal ideal domain unifies the classification of finitely generated abelian groups with the study of linear operators through polynomial rings. See Keith Conrad's Modules over a PID, Theorem 4.1 and the operator applications.

A historical anchor is Emmy Noether's 1921 paper ^[ref]. Its first section studies rings and ideals under the condition that every ideal is a finitely generated module over its ring, and connects this to the ascending chain condition; later sections develop primary decomposition. This is an early source for the structural viewpoint underlying Noetherian rings. ^[ref] makes those statements accessible in English.

A module is faithful when its annihilator of a module is zero. Equivalently, the action ring homomorphism $R\to {\mathrm{End}}_{\mathbb{Z}}(M)$ to the endomorphism ring of its additive abelian group is injective: every nonzero scalar acts nontrivially somewhere on $M$. A primitive ring has a faithful simple module.

The left regular module of a unital associative algebra $A$ is its underlying additive group with $A$ acting by left multiplication. Its submodules are exactly the left ideals of $A$. The associative algebra is a semisimple algebra precisely when this module is a semisimple module.

If an element $w$ of the module ${\prod }_{j\in J}\mathbb{Z}$ is divisible by every power of two within that module, then every coordinate $w(j)$ is an integer divisible by every power of two. A nonzero integer $a$ cannot be divisible by $2^i>|a|$, so each coordinate is zero and $w=0$. This argument applies to any index set $J$ and to powers of any fixed integer $b$ with $|b|>1$.

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.

The direct sum of a family of modules over the same ring consists of all tuples $(m_j)$ with $m_j\in M_j$ and only finitely many nonzero coordinates, with coordinatewise addition and scalar multiplication. For two submodules $N,P\subseteq M$, an internal direct sum $M=N\oplus P$ means $M=N+P$ and $N\cap P=\{0\}$. Equivalently, every element of $M$ is uniquely $n+p$: equality of two such expressions puts their difference in $N\cap P$. The projection $n+p\mapsto p$ is a module homomorphism with kernel $N$, image $P$, and square equal to itself.

A finite-dimensional algebra has finite representation type when there are only finitely many isomorphism classes of finite-dimensional indecomposable modules. This counts module types, not the dimensions or the number of direct sums formed from them. For a block of a group algebra, finite representation type is equivalent to a cyclic defect group of a block.

For a finite-dimensional algebra, the Auslander–Reiten quiver has one vertex for each isomorphism class of finite-dimensional indecomposable modules, and its arrows record irreducible module homomorphisms. A homomorphism is irreducible when it is neither a split injection nor a split surjection and every factorization forces the first map to split injectively or the second to split surjectively.

For a self-injective finite-dimensional algebra, remove the projective vertices from its Auslander–Reiten quiver. The resulting stable quiver has vertices the nonprojective indecomposable modules. Its translation sends the end term of an almost-split exact sequence to its start term. In particular a quiver of shape $\mathbb{Z}{A}_s/⟨{\tau }^e⟩$ has $es$ vertices.

A topological abelian group $A$ with a continuous action of a topological group $G$ by group automorphisms. Continuity means joint continuity of $G\times A\to A$. For a discrete coefficient group this is equivalent to every element of $A$ having an open stabilizer.

For a module $M$ over a commutative ring $A$, its character module is the additive-group dual into $D=\mathbb{Q}/\mathbb{Z}$, with action $(a\varphi )(m)=\varphi (am)$. The group $D$ is an injective cogenerator of abelian groups. The natural adjunction ${\mathrm{Hom}}_A(N,M^{\ast })\cong {\mathrm{Hom}}_{\mathbb{Z}}(N{\otimes }_{A}M,D)$ sends $h$ to the balanced pairing $(n,m)\mapsto h(n)(m)$.

An algebra representation is a unital algebra homomorphism into the endomorphisms of a vector space. It is equivalent to making that vector space a left module over the algebra, with $av=\rho (a)v$.

A representation of a complex Banach algebra is a complex-linear multiplicative action on a vector space. For a unital action require $\pi (1)=I$; a nonunital algebra can be handled by extending the action to its unitization. A nonzero action is algebraically irreducible when it has no nonzero proper invariant vector subspace, equivalently when it defines a simple module.

A representation of a C-star algebra on a Hilbert space is a star-homomorphism $\pi :A\to \mathcal{B}(H)$. It is cyclic if some vector $\xi$ has dense orbit span as displayed; the orbit $\pi (A)\xi$ is itself a linear subspace. For a unital representation, a unit cyclic vector defines a state on a C-star algebra by $a\mapsto ⟨\pi (a)\xi ,\xi ⟩$. Conversely, the Gelfand–Naimark–Segal construction realizes every state this way. Cyclicity means density under the action of the whole algebra, not necessarily under powers of a single chosen element.

Algebraic irreducibility excludes every proper nonzero invariant vector subspace, including those that are not closed. It is stronger than topological irreducibility, which excludes only closed invariant subspaces. For a nonzero irreducible action, every nonzero vector is cyclic in the algebraic sense: $A\xi =X$.

For countably many linearly independent vectors in an infinite-dimensional irreducible complex Banach module, there are elements $a_n$ satisfying the displayed annihilation conditions while the tail images $\{\pi (a_n\cdots a_1){\xi }_j:j\ge n\}$ remain linearly independent. Use the quotient norm on an irreducible Banach module to make each finite annihilator a closed Banach subspace. The Jacobson density theorem makes each condition of independence of finitely many tail images open and dense there; the Baire category theorem intersects these countably many conditions. This is stronger than prescribing only finitely many images once.

For a unital irreducible action and $\xi \ne 0$, the annihilator $L_{\xi }=\{a:\pi (a)\xi =0\}$ is a maximal left ideal and is closed. Thus $A/L_{\xi }$ supplies the displayed complete norm on the module, with contractive action. Every module endomorphism is bounded for this quotient norm: its value at the cyclic vector is $a\xi$, so it is induced by bounded right multiplication by $a$ on the quotient. The Gelfand-Mazur theorem then makes the module's commutant consist of scalar operators. This constructs one compatible norm; it does not by itself establish continuity in a different preassigned norm.

A normed representation acts by individually bounded operators on a normed vector space. This condition alone does not assume that $a\mapsto \pi (a)$ is continuous in the operator norm, nor that the representation space is complete. A continuous representation has a common bound $\Vert \pi (a)\Vert \le C\Vert a\Vert$.

Every algebraically irreducible representation of a Banach algebra that is normed is continuous. The continuous orbit vectors form an invariant subspace. If all nonzero orbit maps were discontinuous, apply the countable annihilation sequence in an irreducible Banach module and the gliding-hump continuity principle to right multiplication on the algebra and evaluation of the operator-valued representation. Off-diagonal maps vanish, whereas diagonal maps are discontinuous, a contradiction. Once every orbit map is continuous, the Uniform boundedness principle on the Banach algebra gives the common operator-norm bound. Completing the original normed representation space supplies the Banach operator space needed in the argument.

A finite-dimensional module $X$ degenerates to $Y$ when $Y$ has a representative in the closure of $X$'s change-of-basis orbit in the representation variety of an associative algebra. The full orbit of $Y$ then lies in that closure. The relation is transitive because orbit closures are closed and invariant under change of basis. A split extension as a degeneration shows that a module degenerates to the direct sum of its composition factors.

A ring is left hereditary if every submodule of a left projective module is projective. Equivalently, every left module has a projective resolution of length at most one. The standard projective resolution of a quiver representation, used also for arbitrary infinite-dimensional modules, proves that path algebras are left hereditary.

Let $A$ be a finite-dimensional basic algebra over an algebraically closed field and a hereditary ring. The radicals of its indecomposable projectives are projective and satisfy $J{P}_i\cong {⨁}_j{P}_j^{a_{ij}}$, where $a_{ij}$ are the Ext quiver arrow multiplicities. Hence $\dim P_i=1+{\sum }_j{a}_{ij}\dim P_j$, forcing strict dimension decrease along arrows and therefore acyclicity. Lifting an arrow basis in $J/J^2$ gives a surjective path algebra homomorphism $kQ\to A$. The number of paths starting at $i$ satisfies the same dimension recursion. Equality of total dimensions proves the homomorphism is an isomorphism. Sums of vertex idempotents over connected components are precisely the primitive central idempotents, so components give the blocks.

The standard projective resolution of a quiver representation has length one, even for quivers with oriented cycles. Thus every path algebra is a hereditary ring, and higher extension groups vanish. Applying a long exact sequence of Ext groups to $0\to I\to K\to C\to 0$ gives a surjection ${\mathrm{Ext}}^1(K,W)\twoheadrightarrow {\mathrm{Ext}}^1(I,W)$, since the next term ${\mathrm{Ext}}^2(C,W)$ is zero. This is the hereditary step in the Ringel lemma on bricks.

For central elements, each multiplication map on the preceding quotient must be injective; a usual convention also requires the final quotient to be nonzero. The augmented Koszul complex on central ring elements tensored with $M$ is then a quasi-isomorphism to the final quotient in degree zero.

A composition series of a module is a finite chain of submodules whose nonzero successive quotient modules are simple modules. Every finite-dimensional Lie algebra representation has such a chain, viewed as a module over the universal enveloping algebra.

Every two finite composition series of a module have the same multiset of simple composition factors, up to isomorphism and order. One proves this by induction on the length: if the first simple submodules differ, their intersection is zero, and the quotient by their sum identifies the remaining common comparison problem; if they coincide, pass directly to the quotient. The theorem makes the multiplicity of a simple module independent of a chosen composition series. It does not assert that the module is a direct sum of its composition factors.

A composition factor is an irreducible successive quotient in a composition series of a module. For a central Casimir operator, its scalar on a composition factor is an eigenvalue of the operator on the original Lie algebra representation.

A nonzero module is uniform if every pair of nonzero submodules has nonzero intersection. Equivalently, every nonzero submodule is an essential submodule. A right Noetherian domain is uniform as a right module: if $aA\cap bA=0$ for nonzero $a,b$, the right ideals $b^{i}aA$, $i\ge 0$, form an infinite direct sum, contradicting the ascending chain condition.

The uniform dimension of a module is the supremum of the numbers of nonzero submodules that can occur in an internal direct sum. Finite uniform dimension is equivalent to having no infinite internal direct sum of nonzero submodules. A finite direct sum of uniform modules contained as an essential submodule gives that number as the dimension. The finite uniform decomposition of a Noetherian module therefore gives finite uniform dimension.

Every nonzero Noetherian module contains a uniform module: otherwise repeatedly split a nonuniform nonzero submodule into two disjoint nonzero submodules, retaining one and collecting the other, to build an infinite direct sum. Repeatedly adding a uniform submodule disjoint from the current finite sum must likewise stop. The final finite sum is an essential submodule, since any disjoint nonzero submodule would allow another addition.

A nonzero module is indecomposable if it cannot be expressed as a direct sum of two nonzero submodules. The zero module is excluded. A module of finite composition length is indecomposable exactly when its endomorphism ring has no idempotents other than zero and one: an idempotent splits the module into its image and kernel.

For a nonzero finitely generated indecomposable module $P$ that is a projective module over a right Artinian ring $A$, its top $P/PJ(A)$ is a simple module. The Jacobson radical $J(A)$ is a nilpotent ideal, so the top is nonzero; the quotient is a semisimple module over the semisimple ring $A/J(A)$. If it decomposed, a nontrivial idempotent of its endomorphism ring would lift by projectivity to an endomorphism of $P$. The Hopkins-Levitzki theorem gives finite composition length, and the Fitting lemma makes the lift either invertible or nilpotent. Its induced map on the top would have the same property, impossible for that nontrivial idempotent.

For an endomorphism $f$ of a finite-length module $M$, sufficiently large $n$ givesIf $M$ is indecomposable, every endomorphism is therefore either invertible or nilpotent, so its endomorphism ring is local.

Every linear endomorphism of a finite-dimensional vector space admits the displayed direct sum of vector spaces. It separates the part on which repeated application is invertible from the part eventually sent to zero.

Proof. Stabilization of images of powers of a linear map gives $\mathrm{im}{h}^n=\mathrm{im}{h}^{2n}$. The rank-nullity theorem and nested null spaces give $\ker h^n=\ker h^{2n}$. If $v=h^{n}u$ also satisfies $h^{n}v=0$, then $u\in \ker h^{2n}=\ker h^n$, so $v=0$. Thus the displayed summands have zero intersection. Their dimensions sum to $n$ by the rank-nullity theorem, proving the decomposition.

Both summands are preserved by $h$. On $\mathrm{im}{h}^n$, stabilization makes $h$ a surjective linear map, hence invertible in finite dimension. On $\ker h^n$, its $n$th power is zero, so the restriction is a nilpotent linear map.

The Krull–Schmidt theorem says that a finite-length module decomposes as a finite direct sum of indecomposable modules, uniquely up to permutation and isomorphism of the summands.

A finite-dimensional module is ${⨁}_a{M}_a^{\oplus m_a}$ with pairwise nonisomorphic indecomposables and uniquely determined multiplicities. The Fitting lemma makes their endomorphism rings local. The multiplicity spaces provide the general linear factors in the Levi decomposition of a quiver automorphism group.

For an $R$-module $M$, its dual module is $M^{\ast }={\mathrm{Hom}}_R(M,R)$, with pointwise addition and scalar multiplication.

The double dual module is $M^{\ast \ast }={\mathrm{Hom}}_R(M^{\ast },R)$. The evaluation homomorphism sends $m\in M$ to the functional $f\mapsto f(m)$.

An $R$-module is finitely generated when there are finitely many elements $m_1,\dots ,m_r$ such that every element is an $R$-linear combination of them.

If a submodule $N\subseteq M$ and the quotient module $M/N$ are finitely generated modules, then $M$ is a finitely generated module. Choose finitely many generators of $N$ and lift finitely many generators of $M/N$. For any $m\in M$, subtract an $R$-linear combination of the lifts to obtain an element of $N$, then express this using the generators of $N$. The union of the two finite lists generates $M$.

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.

Let a finitely generated module over an integral domain have a finite generating set $S$ and a maximal independent subset $T$. For each $s\notin T$, dependence of $T\cup \{s\}$ gives a nonzero coefficient $a_s$ with $a_{s}s\in ⟨T⟩$. The product of those coefficients is nonzero and carries the whole module into $⟨T⟩$.

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.

An $R$-module is Artinian when every descending chain of submodules stabilizes. Equivalently, every nonempty family of submodules has a minimal member.

For an injective module endomorphism $f:M\to M$ of an Artinian module, the chain $M\supseteq fM\supseteq f^{2}M\supseteq \cdots$ stabilizes. If $f^{n}M=f^{n+1}M$, then every $x$ satisfies $f^{n}x=f^{n+1}y$. Injectivity of $f^n$ gives $x=fy$, proving surjectivity. No finite-length or Noetherian hypothesis is needed.

For a short exact sequence $0\to N\to M\to Q\to 0$, the middle module is Artinian exactly when both outer modules are Artinian. Submodules and quotients inherit the descending chain condition. Conversely, for a descending chain $M_i$, both $M_i\cap N$ and its image in $Q$ eventually stabilize. If $x\in M_i$, choose $y\in M_{i+1}$ with the same image in $Q$; then $x-y\in M_i\cap N=M_{i+1}\cap N$, giving $M_i=M_{i+1}$. Induction shows that finite direct sums of Artinian modules are Artinian.

A filtered colimit of modules is a direct limit indexed by a filtered category. Filtered colimits preserve exact sequences, and tensor products commute with them.

A direct system indexed by a directed set $A$ consists of abelian groups $G_a$ and homomorphisms ${\varphi }_{ab}:G_a\to G_b$ for $a\le b$, with ${\varphi }_{aa}=1$ and ${\varphi }_{bc}{\varphi }_{ab}={\varphi }_{ac}$.

The direct limit of a direct system of abelian groups is the quotient of ${⨁}_a{G}_a$ by the relations identifying $g\in G_a$ with ${\varphi }_{ab}(g)\in G_b$ whenever $a\le b$.

For nonzero integers $a_i$ and $P_n=a_0\cdots a_{n-1}$, the compatible embeddings of stage $n$ into $\mathbb{Q}$ send $m$ to $m/P_n$. They identify the direct limit with

The additive group ${\mathbb{Q}}_{\mathrm{sq}}$ consists of the rational numbers whose reduced denominators are square-free integers. If $p_0,p_1,\dots$ lists the primes without repetition, then

An $(R,S)$-bimodule is simultaneously a left $R$-module and a right $S$-module, with commuting scalar actions $(rm)s=r(ms)$.

For a group $G$ acting on a module $M$, the invariant submodule is $M^G=\{m\in M:g\cdot m=m\text{ for all }g\in G\}$.

For a group $G$ acting on a module $M$, the coinvariant module is

Suppose $1\to X\to \mathcal{G}\to \Gamma \to 1$ is an exact sequence of pro-p groups, with $X$ abelian and $\Gamma \cong {\mathbb{Z}}_p$. Conjugation gives $X$ a compact Galois module structure. The closed commutator subgroup is $(\gamma -1)X$: its image is closed by compactness, and after quotienting by it, a lift of $\gamma$ centralizes $X$ and topologically generates the remaining quotient. Hence $0\to X_{\Gamma }\to {\mathcal{G}}^{\mathrm{ab}}\to \Gamma \to 0$ is exact.

The support of an $R$-module $M$ isFor a nonzero finitely generated module, the support is nonempty and contains a maximal ideal.

The tensor product of an $R$-module $M$ and an $R$-module $N$ is an $R$-module $M{\otimes }_{R}N$ generated by pure tensors $m\otimes n$, subject to additivity in both variables and the balancing relation

For a short exact sequence $0\to U\stackrel{i}{\to }V\stackrel{q}{\to }W\to 0$ and any module $E$, the sequence $U{\otimes }_{A}E\to V{\otimes }_{A}E\to W{\otimes }_{A}E\to 0$ is exact. The last map is surjective since every pure tensor $w\otimes e$ can be lifted using a preimage of $w$. Put $T=(V{\otimes }_{A}E)/\mathrm{im}(i\otimes 1_E)$. There is a well-defined balanced map $W\times E\to T$, sending $(q(v),e)$ to the class of $v\otimes e$: replacing $v$ by $v+i(u)$ changes it by an element of the discarded image. This induces $W{\otimes }_{A}E\to T$, inverse to the map induced by $q\otimes 1_E$. Thus the kernel of $q\otimes 1_E$ is exactly the image of $i\otimes 1_E$. Consequently a module is flat exactly when tensoring with it also preserves injections.

For a homomorphism of commutative rings $R\to A$ and two $A$-modules, the tensor product of modules over $A$ is the quotient of the tensor product of modules over $R$ imposing all relations $am\otimes n=m\otimes an$. The quotient map is $A$-linear when $A$ acts through the first factor on the source. This expresses the extra balancing imposed by change of rings.

For commutative $R$-algebras, their module tensor product becomes a commutative algebra by $(a\otimes b)(a^{\prime }\otimes b^{\prime })=a{a}^{\prime }\otimes b{b}^{\prime }$. Its universal property is that pairs of compatible $R$-algebra maps from $A$ and $B$ into a commutative algebra $C$ correspond to maps $A{\otimes }_{R}B\to C$. This coproduct of algebras gives, contravariantly, the fibre product of schemes on affine charts.

If $L,L^{\prime }$ are field extensions of $K$, choose a $K$-basis of $L$ containing $1$. After tensoring with $L^{\prime }$, that basis expresses $L{\otimes }_K{L}^{\prime }$ as a nonzero direct sum of copies of $L^{\prime }$, and $1\otimes 1$ is nonzero. A prime ideal of this nonzero ring supplies a point used in the point-lifting property of a scheme fibre product.

A pure tensor is an element of a tensor product that can be written as $m\otimes n$. General tensors are finite sums of pure tensors and need not themselves be pure.

If $M{\otimes }_{R}N\cong R^n$ for a positive integer $n$, then both $M$ and $N$ are projective modules. Choose a finite expression for the inverse image of one basis vector. It produces a split surjection $N^r\to R$; tensoring the splitting with $M$ exhibits $M$ as a direct summand of the finite free module $(M{\otimes }_{R}N{)}^r$. Symmetry gives the result for $N$.

An $R$-module $M$ is tensor-nilpotent when $M^{\otimes k}=0$ for some positive integer $k$. A nonzero finitely generated module cannot be tensor-nilpotent: localize at a maximal ideal in its support and reduce modulo that maximal ideal, obtaining a nonzero tensor power of a nonzero vector space.

A map $b:M\times N\to P$ is balanced when it is additive in each variable and satisfies $b(rm,n)=b(m,rn)$.

Every balanced map $b:M\times N\to P$ factors uniquely through an $R$-module homomorphism $\tilde{b}:M{\otimes }_{R}N\to P$ satisfying $\tilde{b}(m\otimes n)=b(m,n)$.

For modules over a commutative ring, there is a natural isomorphismgiven by $f\mapsto (m\mapsto (n\mapsto f(m\otimes n)))$.

Tensor product need not commute with an infinite direct product. For example,The element $(1\mathrm{mod}n{)}_n$ has infinite order in the product, so it survives rational localization.

An $R$-module $M$ is flat when tensoring with it preserves injections, equivalently when $-{\otimes }_{R}M$ is an exact functor.

For $D=k[\epsilon ]/({\epsilon }^2)$, a module $M$ is a flat module exactly when $\ker (\epsilon :M\to M)=\epsilon M$. Flatness implies this by tensoring the exact sequence $0\to (\epsilon )\to D\to k\to 0$. Conversely lift a $k$-basis of $M/\epsilon M$ to $M$. Every element is a finite combination of the lifts plus $\epsilon$ times another such combination, proving generation over $D$. A relation first reduces to zero coefficients modulo $\epsilon$; the kernel condition then reduces its remaining coefficients to zero as well. The lifts form a free $D$-basis, proving flatness.

For a flat module $M$ over a commutative ring, every finite relation among elements of $M$ is a finite combination of relations among elements of the ring itself, in the displayed sense. To obtain this, tensor the kernel of the linear map $A^r\to A$, $(u_i)\mapsto {\sum }_i{a}_i{u}_i$, with $M$. Flatness says that this tensor maps onto the kernel in $M^r$. Writing an element of that tensor as a finite sum gives the columns $b_{ij}$ and the elements $n_j$. The converse follows by applying this condition to relations in a tensor product to prove that tensoring preserves injections.

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.

If $F=P\oplus Q$ is a flat module, then $P$ and $Q$ are flat. For an injection $u:U\hookrightarrow V$, distributivity of the tensor product of modules identifies $u\otimes F$ with $(u\otimes P)\oplus (u\otimes Q)$. This map is injective by flatness of $F$. Each component is therefore injective, proving flatness of both summands.

The arrow is the canonical multiplication map, not an arbitrary abstract module isomorphism. Its inclusion into $M$ is ${\mu }_I:I{\otimes }_{A}M\to M$. Dualizing identifies ${\mu }_I^{\ast }$ with the restriction ${\mathrm{Hom}}_A(A,M^{\ast })\to {\mathrm{Hom}}_A(I,M^{\ast })$. All these maps are surjective precisely when $M^{\ast }$ is injective, by the Baer criterion. The injective cogenerator of abelian groups detects injectivity of ${\mu }_I$, and the Lambek theorem gives flatness. No Noetherian or finite-generation hypothesis is needed.

A module over a commutative ring is flat exactly when its character module is injective. For a submodule inclusion $N^{\prime }\hookrightarrow N$, the character adjunction identifies restriction ${\mathrm{Hom}}_A(N,M^{\ast })\to {\mathrm{Hom}}_A(N^{\prime },M^{\ast })$ with the dual of $N^{\prime }{\otimes }_{A}M\to N{\otimes }_{A}M$. The injective cogenerator of abelian groups detects whether the latter map is injective. Thus the two extension/exactness conditions are equivalent.

A flat module $M$ is faithfully flat when $N{\otimes }_{R}M=0$ implies $N=0$. Every field extension is faithfully flat over its base field.

An $R$-module $M$ is flat if and only if $M_{\mathfrak{p}}$ is flat over $R_{\mathfrak{p}}$ for every prime ideal $\mathfrak{p}$, equivalently for every maximal ideal.

Over a principal ideal domain, a module is flat exactly when it is torsion-free. Hence the tensor product of two torsion-free modules over a principal ideal domain is torsion-free: both tensor functors are exact, so their composite is exact.

If an $R$-algebra $A$ is flat as an $R$-module, then for ideals $I,J\subseteq R$,Tensor the exact sequence $0\to I\cap J\to I\oplus J\to I+J\to 0$ with $A$.

A nonzero commutative ring $R$ has the property that $M{\otimes }_{R}N=0$ forces $M=0$ or $N=0$ exactly when $R$ is local with maximal ideal $\mathfrak{m}$ and $M=\mathfrak{m}M$ forces $M=0$ for every $R$-module. Reduction modulo $\mathfrak{m}$ turns a tensor product into a tensor product of vector spaces.

If a finitely generated faithful module $M$ over a commutative ring $R$ satisfies $xM\subseteq IM$, the determinant of a matrix presenting these relations gives a monic equationIn particular, an element of an extension field that preserves a nonzero finitely generated ideal is integral over $R$.

A short exact sequence is an exact sequenceThus $j$ is injective, $q$ is surjective, and $\mathrm{im}j=\ker q$.

A short exact sequence with specified kernel module $A$ and quotient module $C$. Its equivalence class is determined by the degree-one Ext functor; equivalence must preserve both end modules.

With trivial action on both end modules, an extension has underlying abelian group ${\mathbb{Z}}^r\oplus {\mathbb{Z}}^s$ and generator action $\left(\begin{array}{cc}I& T\\ 0& I\end{array}\right)$. The integer matrix $T$ is invariant under equivalences fixing the ends. The one-element projective resolution with differential $z-1$ computes the same classification over the group ring.

Pushing an inclusion $i:A\to B$ along $f:A\to A\prime$ gives a new module extension with kernel $A\prime$ and the same quotient. This pushout realizes the covariance of the degree-one Ext functor in its second argument.