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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.

A module is cyclic if one element $m$ generates it: every element is $rm$ for some scalar $r\in R$. The generating element need not give unique coefficients. The module homomorphism $R\to M$, $r\mapsto rm$, is surjective. Over a commutative ring, its kernel is the annihilator ${\mathrm{Ann}}_R(m)$, givingThus cyclic modules are exactly the quotient modules $R/I$ for ideals $I$ in $R$. For a left module over a noncommutative ring, use a left ideal and interpret the quotient as a module, with no claim that it is a quotient ring. See John A. Beachy's Abstract Algebra Online, section 10.1.

The module $\mathbb{Z}/6\mathbb{Z}$ is generated by $\overline{1}$, and also by $\overline{5}$, whose successive multiples are $\overline{0},\overline{5},\overline{4},\overline{3},\overline{2},\overline{1}$. The element $\overline{2}$ generates only $\{\overline{0},\overline{2},\overline{4}\}$, a submodule with three elements. A cyclic module can be infinite and free, as $\mathbb{Z}=\mathbb{Z}\cdot 1$ shows, or finite and torsion, as $\mathbb{Z}/6\mathbb{Z}$ shows. One generator counts directions of generation, not the number of elements and not freedom from relations.

Evaluation at $e_i$ identifies a module homomorphism $h:kQ{e}_i\to X$ with $h(e_i)\in e_{i}X$. Given $x\in e_{i}X$, the inverse sends $p$ to $px$. The isomorphism is natural and shows that Hom functor evaluation on a vertex projective is exact.

A homomorphism is a map between algebraic structures preserving their specified operations. For a single binary operation, this means $f(a\ast b)=f(a)\star f(b)$, where the source and target may use different operation symbols. A group homomorphism preserves the group operation; a ring homomorphism preserves addition and multiplication, with the identity convention specified; a module homomorphism preserves addition and scalar multiplication.

Reduction modulo $5$ is a ring homomorphism $\mathbb{Z}\to \mathbb{Z}/5\mathbb{Z}$: adding or multiplying first and then reducing gives the same residue as reducing first and then operating. For example, $12\cdot 14=168\equiv 3(\mathrm{mod}5)$, while the reduced operands give $2\cdot 4=8\equiv 3(\mathrm{mod}5)$. This lets a complicated integer calculation be projected into simpler finite arithmetic. It also explains why a polynomial equation over the integers imposes necessary modular arithmetic conditions on any integer solution.

A homomorphism can lose information: $0$ and $5$ have the same image under that reduction. An isomorphism loses none of the structure because it has an operation-preserving inverse. Preservation of the operation, rather than preservation of arbitrary labels or sizes, is the essential condition.

For a module homomorphism $f:M\to N$ over a commutative ring $A$, $f$ is injective if and only if every $f_{\mathfrak{m}}$ is injective, with $\mathfrak{m}$ ranging over maximal ideals. In a localization of a module, $x/s$ maps to zero exactly when some denominator $t$ satisfies $tf(x)=0$. If $f$ is injective this gives $tx=0$, so the original fraction is zero. In general it identifies $\ker (f_{\mathfrak{m}})$ with $(\ker f{)}_{\mathfrak{m}}$. A nonzero element of $\ker f$ has a proper annihilator, contained in some maximal ideal; it remains nonzero in that localization. Thus vanishing of all localized kernels forces $\ker f=0$. No finiteness or Noetherian assumption is required.

A module automorphism is an invertible module endomorphism. Equivalently, it is a bijective module homomorphism from a module to itself; its inverse is automatically a module homomorphism. They form a group under composition.

A module endomorphism is a module homomorphism from a module to itself. These maps form the endomorphism ring ${\mathrm{End}}_R(M)$ under pointwise addition and composition. An invertible module endomorphism is a module automorphism.

A module retraction onto a submodule $U\stackrel{i}{\to }X$ is a module homomorphism $r:X\to U$ with $ri=I_U$. It gives $X=i(U)\oplus \ker r$: write $x=i(r(x))+(x-i(r(x)))$. Thus a nonzero proper submodule admitting a retraction contradicts indecomposability. Extending a projection from a larger submodule can produce such a retraction through the long exact sequence of Ext groups.

A representation of a Banach algebra is a complex-linear multiplicative map $\pi :A\to {\mathrm{End}}_{\mathbb{C}}(X)$. A nonzero irreducible action of a unital algebra automatically has $\pi (1)=I$: the image of the idempotent $\pi (1)$ is a nonzero invariant subspace and hence is all of $X$. An algebraically irreducible representation of a Banach algebra is a nonzero action if it has no nonzero proper invariant submodule, whether closed or not. A normed representation of a Banach algebra has $\pi (a)$ bounded for each $a$ on a preassigned normed vector space $X$; continuity of $\pi$ as an operator-norm-valued map is a further conclusion. This must not be confused with topological irreducibility, which excludes only closed invariant subspaces. Unitizing the algebra extends a nonunital action by $\pi (a+\lambda 1)=\pi (a)+\lambda I$, so it is enough to discuss the unital case.

For $\xi \ne 0$, irreducibility gives $A\xi =X$. The annihilator $L_{\xi }=\{a:\pi (a)\xi =0\}$ is a maximal left ideal: an intermediate left ideal gives an intermediate submodule of $X$. It is closed by the Neumann-series argument for maximal left ideals. Consequently $A/L_{\xi }$ transports a complete quotient norm to $X$, and left multiplication is contractive in this norm. Every algebraically irreducible representation of a Banach algebra can therefore be realized continuously on a Banach space. This alone does not prove continuity for its preassigned norm.

The kernel of $\pi$ is the intersection of the closed left ideals $L_{\xi }$ and is a closed primitive ideal. The Jacobson radical is the intersection of all primitive ideals, equivalently the intersection of maximal left ideals. Thus these quotient representations detect semisimplicity. Equivalent algebra representations are related by a bijective module homomorphism.

The algebra of module endomorphisms is exactly $\mathbb{C}I$. Indeed, if a module endomorphism $D$ sends a fixed cyclic vector $\xi$ to $a\xi$, then $D(b\xi )=ba\xi$. It is induced by bounded right multiplication by $a$ on $A/L_{\xi }$, so is bounded for the quotient norm on an irreducible Banach module. Every nonzero module endomorphism is injective and surjective by irreducibility, and its inverse is bounded by the same argument. This commutant of an operator algebra is a closed division subalgebra of the bounded operators on this Banach module; the Gelfand-Mazur theorem makes it $\mathbb{C}I$.

The Jacobson density theorem now says that for finitely many linearly independent ${\xi }_1,\dots ,{\xi }_n$ and arbitrary ${\eta }_1,\dots ,{\eta }_n$, one element of $A$ sends all ${\xi }_j$ to ${\eta }_j$. Here is the induction behind it. The one-vector assertion is $A{\xi }_1=X$. Assuming the first $n-1$ coordinates can be prescribed, set $L=\{a:a{\xi }_j=0(j<n)\}$. Its image $L{\xi }_n$ is an invariant subspace. If it is $X$, it adjusts the last image without changing the first ones. If it is zero, the rule $(a{\xi }_1,\dots ,a{\xi }_{n-1})\mapsto a{\xi }_n$ is a well-defined module map $X^{n-1}\to X$. Each coordinate map is scalar by the module endomorphism result, forcing ${\xi }_n$ to be a linear combination of the earlier vectors, a contradiction. This proves the assertion.

We can now give Johnson's continuity argument with its crucial construction. If $X$ is finite dimensional, the closed kernel has finite codimension and $A/\ker \pi$ is finite dimensional, so $\pi$ is continuous. Suppose instead that $X$ is infinite dimensional. LetThis is an invariant submodule, because the orbit map of $\pi (b)\xi$ is the orbit map of $\xi$ composed with the bounded map $a\mapsto ab$. Thus either $Y=X$ or $Y=0$.

To rule out $Y=0$, choose linearly independent unit vectors ${\xi }_0,{\xi }_1,\dots$. There are elements $a_n\in A$ such that, with $c_n=a_n\cdots a_1$,For completeness, construct them in the canonical quotient-norm Banach module. At stage $n$ the current tail vectors are independent. In the closed left annihilator of its first vector, density permits arbitrary images of any finite list of the remaining vectors. The condition that finitely many such images are independent is open in this Banach subspace and dense: perturb any element by $t$ times an element prescribing independent images; a suitable determinant is a nonzero polynomial in $t$, with only finitely many exceptional values. The Baire category theorem intersects these open dense conditions for all initial finite tail lists. The resulting $a_n$ kills the first current vector and retains independence of the entire remaining tail. This is the countable annihilation sequence in an irreducible Banach module.

The gliding-hump continuity principle supplies the other ingredient: if $E,F$ are Banach spaces, $S:E\to F$ is linear, $T_j:E\to E$ and $U_n:F\to F_n$ are bounded, and $U_{n}S{T}_1\cdots T_m$ is continuous whenever $m>n$, then $U_{n}S{T}_1\cdots T_n$ is continuous eventually. The product form of this principle is recorded in Proposition 1.3 of Thomas's paper. Here is the contradiction mechanism. Rescale so $\Vert T_j\Vert ,\Vert U_n\Vert \le 1$ and put $P_m=T_1\cdots T_m$. If infinitely many diagonal maps are discontinuous, choose increasing $n_j$ and successively tiny $x_j$ soAlso choose $\Vert x_j\Vert \le 2^{-j}/(1+{\max }_{i<j}\Vert U_{n_i}S{P}_{n_i+1}\Vert )$. The convergent sum $x={\sum }_j{P}_{n_j}{x}_j$ has, after its $j$th term, a tail factoring through $P_{n_j+1}$. Applying that continuous composite bounds the tail contribution by one. The display forces $\Vert U_{n_j}Sx\Vert >j$, contradicting $\Vert U_{n_j}Sx\Vert \le \Vert Sx\Vert$. Only finite decompositions and the continuous tail composites are used; no unjustified interchange of a discontinuous map with an infinite sum occurs.

Apply this principle with $E=A$, $F=\mathcal{B}(\overline{X})$, where $\overline{X}$ is the completion in the original norm, and $S(a)$ the bounded extension of $\pi (a)$. Set $T_j(a)=a{a}_j$ and $U_n(R)=R{\xi }_n$. ThenFor $m>n$ this is zero, hence continuous. For $m=n$ it is the orbit map of a nonzero vector, hence discontinuous if $Y=0$. This contradicts the principle. Therefore $Y=X$.

Finally, apply the Uniform boundedness principle on the Banach space $A$ to the continuous maps $a\mapsto \pi (a)\xi$ indexed by $\Vert \xi \Vert \le 1$. They are pointwise bounded because each individual $\pi (a)$ is bounded in the original norm. It givesThis proves Johnson's continuity theorem for irreducible normed representations, including when the original normed vector space is incomplete. The quotient norm, countable density construction, and gliding hump explain the substantive steps rather than assuming the desired continuity in advance.

The kernel intersection theorem for Specht modules uses maps that move entries between adjacent rows. For $1\le i<\ell (\mu )$ and $0\le v<{\mu }_{i+1}$, let $\mu (i,v)$ replace the row lengths $({\mu }_i,{\mu }_{i+1})$ by $({\mu }_i+{\mu }_{i+1}-v,v)$, keeping every other length unchanged. This is allowed to be a composition rather than a decreasing partition of an integer, and a zero row can be omitted. For a tabloid with row sets $R_j$, defineby summing over all $v$-subsets $B\subset R_{i+1}$: the new row $i+1$ is $B$, the new row $i$ is $R_i\cup (R_{i+1}\setminus B)$, and all other rows are unchanged. These are module homomorphisms, since relabelling commutes with choosing subsets. The theorem isIt holds over every field. Column cancellation shows the inclusion of $S^{\mu }$ in each kernel; the full theorem identifies the common kernel with the Specht module.

For completeness, the semistandard homomorphism theorem concerns a partition of an integer $\lambda$, a composition $\alpha$, and fillings $T$ of shape $\lambda$ with ${\alpha }_j$ occurrences of $j$. Such a filling gives ${\Theta }_T:M^{\lambda }\to M^{\alpha }$: replace its labels by the numbered entries of an input tableau, and sum the output tabloids over distinct permutations of the labels within each input row. Label $j$ assigns an entry to output row $j$. The use of distinct assignments avoids dividing by any factorial. Let ${\hat{\Theta }}_T$ be its restriction to $S^{\lambda }$. The restrictions indexed by Semistandard Young tableaux are linearly independent. They form a basis of ${\mathrm{Hom}}_{F{S}_n}(S^{\lambda },M^{\alpha })$ when $p\ne 2$, or when $\lambda$ is two-regular. In characteristic two for a two-singular shape they need not span; omitting this exception would make the statement incorrect.

We now prove the invariant vector criterion for a Specht module directly from the first theorem. The group acts transitively on the tabloid basis of $M^{\mu }$, so any invariant vector has all coefficients equal. Thus $(M^{\mu }{)}^{S_n}=F{z}_{\mu }$, where $z_{\mu }$ is the sum of all its tabloids and is nonzero over every field. Fix $i,v$, write $a={\mu }_i$, $b={\mu }_{i+1}$, and count preimages of one output tabloid. The $v$ retained entries are prescribed; among the $a+b-v$ entries of the enlarged row, exactly $a$ must have come from its old row. HenceThe common-kernel criterion becomesfor every adjacent pair, where $j=b-v$.

Suppose $p>0$, and let $t$ be least with $b<p^t$. If $a\equiv -1(\mathrm{mod}{p}^t)$, its lowest $t$ base-$p$ digits are all $p-1$. For any $0<j<p^t$, adding $j$ to $a$ creates a carry at the lowest nonzero digit of $j$. At that position the digit of $a+j$ is smaller than the corresponding digit of $j$. Lucas's theorem therefore gives $\left(\genfrac{}{}{0ex}{}{a+j}{j}\right)\equiv 0$. Conversely, if one of the first $t$ digits $a_d$ is not $p-1$, take $j=p^d$. Minimality of $t$ gives $p^d\le b$, while addition at that digit creates no carry. Lucas then yieldsThis proves the claimed equivalence, and thereforeThe invariant space, when present, is one-dimensional. At the last row put ${\mu }_{i+1}=0$; then $t_i=0$ and the condition is vacuous. This congruence formulation is for positive characteristic. In characteristic zero the positive binomial coefficients cannot vanish, so an invariant vector exists precisely for the one-row shape $(n)$.

Finally suppose $\mu$ is regular. It has a unique simple head $D^{\mu }$, by Question 1. The given sign-twisted duality of Specht modules impliesThus $D^{\mu }$ is sign exactly when the conjugate shape ${\mu }^{\prime }$ satisfies the invariant congruences.

Here is an explicit solution of those congruences under regularity. Write the positive column heights of $\mu$ as $a_1\ge \cdots \ge a_l>0$, so ${\mu }^{\prime }=(a_1,\dots ,a_l)$. Regularity says $a_i-a_{i+1}<p$, including $a_l-a_{l+1}=a_l<p$, because this difference counts rows of length $i$. Starting with $0<a_l<p$, the congruence for its preceding row forces $a_{l-1}\equiv -1(\mathrm{mod}p)$. Since $a_l\le a_{l-1}<a_l+p\le 2p-1$, the only possibility is $a_{l-1}=p-1$. Repeating upwards forces every earlier height to be $p-1$. Consequentlywhere a zero final part is omitted. Conversely this shape satisfies every invariant congruence, so its conjugate is the desired regular label of the modular sign representation:Zero parts are omitted. For $p=2$ this is $(n)$, as it should be because sign is trivial. In characteristic zero the answer is $(1^n)$.

Suppose $f$ is injective. If $f_{\mathfrak{m}}(x/s)=0$, some $t\notin \mathfrak{m}$ satisfies $tf(x)=f(tx)=0$. Injectivity gives $tx=0$, and the vanishing criterion in a module localization gives $x/s=0$. Hence every localized module homomorphism is injective.

Conversely, suppose all $f_{\mathfrak{m}}$ are injective. For $x\in \ker f$, its image $x/1$ in $M_{\mathfrak{m}}$ maps to zero and therefore is zero for every maximal ideal $\mathfrak{m}$. The annihilator argument in part (b) implies $x=0$. Thus $\ker f=0$ and $f$ is injective. We have proved that injectivity of a module map is detected at maximal localizations:Equivalently, exactness of localization gives $(\ker f{)}_{\mathfrak{m}}\cong \ker (f_{\mathfrak{m}})$, and part (b) detects whether this kernel is zero.

Use the defining criterion that a flat module $E$ makes $u\otimes 1_E$ injective for every injective module homomorphism $u$. By right exactness of the tensor product, preserving injections is equivalent to preserving every short exact sequence.

First assume $E$ is flat over $A$, and fix a maximal ideal $\mathfrak{m}$. Let $u:U\hookrightarrow V$ be an injection of $A_{\mathfrak{m}}$-modules. Regard it as an injection of $A$-modules. There are natural module isomorphismsFor example, $u_0\otimes (e/s)$ maps to $s^{-1}{u}_0\otimes e$, with inverse $u_0\otimes e\mapsto u_0\otimes (e/1)$. These are well-defined because every $s\notin \mathfrak{m}$ already acts invertibly on $U$. Under these isomorphisms the tensor map is $u{\otimes }_A{1}_E$, which is injective by flatness. Thus $E_{\mathfrak{m}}$ is flat over $A_{\mathfrak{m}}$.

For the converse, assume every $E_{\mathfrak{m}}$ is flat over $A_{\mathfrak{m}}$ and take any injection $u:U\hookrightarrow V$ of $A$-modules. By part (c), each $u_{\mathfrak{m}}$ is injective. Since localization commutes with tensor products, the localization of $u{\otimes }_A{1}_E$ identifies withIt is injective by the assumed local flatness. Apply part (c) again, now to $u{\otimes }_A{1}_E$, to conclude that it is injective globally. Since $u$ was arbitrary, $E$ is flat. ThereforeThis proves flatness is local without any finite-generation hypothesis.

Take $B=\mathbb{Z}$, $N=\mathbb{Z}/2\mathbb{Z}$, and the short exact sequenceMultiplication by two is injective, and the reduction map has image all of $\mathbb{Z}/2\mathbb{Z}$ and kernel $2\mathbb{Z}$, proving exactness. Any module homomorphism from $\mathbb{Z}/2\mathbb{Z}$ to $\mathbb{Z}$ is zero: its generator must map to an integer killed by two. On the other hand, a homomorphism from $\mathbb{Z}/2\mathbb{Z}$ to itself is determined arbitrarily by the image of its generator, so its Hom group is $\mathbb{Z}/2\mathbb{Z}$. Applying the Hom functor therefore givesThis sequence is not exact at $\mathbb{Z}/2\mathbb{Z}$: the incoming image is zero while the outgoing kernel is the whole nonzero group. In particular, the identity of $N$ cannot lift through the reduction map to a map $N\to \mathbb{Z}$.

Here the Hom groups must be taken over $A$; the printed subscript $B$ in this part has no defined coefficient ring. For an arbitrary module $P$, form the free moduleand define a module homomorphism $\pi :F\to P$ by $e_p\mapsto p$. Each element of $F$ has finite support, so this prescription extends uniquely and is well-defined. It is surjective because every $p\in P$ is the image of $e_p$. Thus every module is a quotient of a free module.

Let $K=\ker \pi$. The short exact sequence $0\to K\to F\stackrel{\pi }{\to }P\to 0$ and the defining exactness condition for a projective module makesurjective. Lift $1_P$ to a map $s:P\to F$. Then $\pi s=1_P$, so $s$ is injective. Every $f\in F$ has the decompositionwhose first term lies in $K$ and second lies in $s(P)$. If $s(p)\in K$, applying $\pi$ gives $p=0$, so the intersection is zero. ConsequentlyThis is a split short exact sequence and proves that projective modules are direct summands of free modules, including modules that are not finitely generated.

Prove the submodule theorem for free modules over a principal ideal domain by induction on the rank of $R^n$. For a submodule $N$, project onto its first coordinate. The image is an ideal $aR$. If $a=0$, then $N$ is a submodule of $R^{n-1}$ and induction applies. Otherwise choose $v\in N$ projecting to $a$. Every $w\in N$ has first coordinate $ra$ and hence $w-rv$ lies in the projection kernel. The domain property shows $Rv$ meets that kernel only at zero. Thusand the kernel is a free module by induction. This proves that $N$ is free, of rank at most $n$.

To prove free modules are projective, choose a basis $(e_{\lambda })$ of a free module $P$. For the given surjection $f$, choose $m_{\lambda }\in M$ with $f(m_{\lambda })=g(e_{\lambda })$ and extend $e_{\lambda }\mapsto m_{\lambda }$ linearly. Finite support of basis expansions makes the resulting module homomorphism well defined, and $f\circ h=g$.

Finally, take a finite generating set of a projective module $P$ to obtain a surjection $\pi :R^n\to P$. Projectivity applied to ${\mathrm{id}}_P$ gives a section $s:P\to R^n$ with $\pi s={\mathrm{id}}_P$. Thus $s$ is injective and identifies $P$ with a submodule of $R^n$. The proved submodule theorem for free modules over a principal ideal domain shows that every finitely generated projective module over a principal ideal domain is free.

Here the usual homological algebra convention is that the right-exact functor $F$ is additive. Module categories have enough projective modules, because any module is a quotient of a free module. Choose a projective resolutionApply $F$ to the unaugmented chain complex and define the left derived functors byAdditivity ensures that $F$ preserves zero maps, sums of maps and homotopy identities, so the transformed differentials still compose to zero.

For a module homomorphism $u:M\to N$, lift it to a chain map between chosen projective resolutions. In degree zero, lift through $Q_0\twoheadrightarrow N$ using projectivity of $P_0$. In each higher degree, the already constructed map lands in the next kernel; exactness of $Q_{\bullet }$ and projectivity of $P_i$ supply the next lift. Any two lifts are chain homotopic: their degree-zero difference lands in $\ker (Q_0\to N)=\mathrm{im}{d}_1^Q$, so it lifts to $h_0:P_0\to Q_1$. Inductively lift the remaining difference after subtracting the preceding homotopy term. This givesApplying $F$ preserves this identity, so the lifts induce the same homology map. Identity lifts and composable lifts show that $L_{i}F$ is a functor. Comparison lifts between two resolutions over the identity of $M$ are inverse up to chain homotopy; hence the construction is independent of the chosen resolution up to its canonical natural isomorphism.

Right exactness identifies $L_{0}F(M)=\mathrm{coker}(F(P_1)\to F(P_0))$ naturally with $F(M)$. For a projective module $P$, the length-zero resolution gives $L_{i}F(P)=0$ for $i>0$. For a short exact sequence $0\to M^{\prime }\to M\to M^{\prime \prime }\to 0$, the horseshoe lemma builds a degreewise split short exact sequence of projective resolutions. Concretely, lift the projective generators covering $M^{\prime \prime }$ into $M$ and combine them with generators covering $M^{\prime }$; repeat on the kernels. The middle projective term in each degree is the direct sum of the two outer ones. An additive functor preserves these split sequences, and lifting cycles, taking their differentials and returning to the subcomplex produces the connecting maps inThis is the natural long exact derived-functor sequence. In particular, if $0\to K\to P\to M\to 0$ has $P$ projective, then $L_{i}F(M)\cong L_{i-1}F(K)$ for $i\ge 2$, and $L_{1}F(M)=\ker (F(K)\to F(P))$. These are dimension-shifting formulas. Positive degrees can be killed by the surjection $P\to M$. If $F$ is exact, it preserves every resolution's exactness and all its positive left derived functors vanish. Conversely, if $L_{1}F$ vanishes on all modules, the displayed long exact sequence makes $F$ exact. A natural transformation between additive right-exact functors induces transformations on the left derived functors by applying it termwise.

For extension of scalars $F(M)=B{\otimes }_{A}M$, these derived functors are the Tor functors ${\mathrm{Tor}}_i^A(B,M)$. When $B=A/(a)$ and $a$ is a non-zero-divisor, there is the length-one free resolutionAlthough the construction above resolves $M$, one may instead use this resolution of $B$. To justify that calculation, tensor a free resolution of $M$ with the two-term resolution of $B$ to form a double complex. Taking homology first along the resolution of $B$ leaves $B{\otimes }_A{P}_{\bullet }$; taking it first along the resolution of $M$ leaves $M\stackrel{a}{\to }M$. The free terms preserve exactness under tensoring. The double complex is first-quadrant with finitely many terms on each total diagonal, so both filtrations compute the same total homology. This proves the required balancing directly.

The kernel and cokernel of multiplication by $a$ now give Tor for a principal non-zero-divisor quotient:These are naturally $B$-modules. For $M=A$, the degree-one group is zero; for $M=B$, multiplication by $a$ is zero and both degrees zero and one are $B$. Thus, when $B\ne 0$, degree one explicitly detects the failure of this base-change functor to be exact.

Use the local-presentation definition of a quasi-coherent sheaf: locally it admits a presentation ${\mathcal{O}}^{(I)}\to {\mathcal{O}}^{(J)}\to \mathcal{F}\to 0$ by direct sums of copies of the structure sheaf, with no finiteness requirement on the index sets. If $M$ is an $A$-module, choose a free module presentation $A^{(I)}\to A^{(J)}\to M\to 0$. Exactness of localization of a module gives the corresponding local presentation of $\tilde{M}$. Hence a sheaf which is associated with a module on every affine open subset is a quasi-coherent sheaf.

For the converse, begin with a quasi-coherent sheaf and a local presentation on an affine open subset $V=\mathrm{Spec}R$. The free sheaves have global sections $R^{(I)}$ and $R^{(J)}$: a section has locally finite support, and quasi-compactness of $V$ makes its support finite globally. Part (i) identifies the presentation map with an $R$-module homomorphism. Exactness of localization then identifies its cokernel sheaf with ${\tilde{M}}_V$ for the module cokernel $M_V$. Thus there is an affine open cover on which $\mathcal{F}$ is associated with modules.

It remains to show this on an arbitrary affine open subset $U=\mathrm{Spec}A$, rather than only the selected cover. Around each point of $U$, choose an affine neighborhood contained in $U$ with such a local presentation. Choose a finite cover $U={\bigcup }_{i=1}^{r}D(s_i)$ refining these neighborhoods. On each $D(s_i)$ the restriction is associated with a module. Its restriction to $D(s_i{s}_j)$ is therefore the corresponding localization, and the same is true after intersecting with $D(a)$. Put $M=\Gamma (U,\mathcal{F})$. The sheaf gluing axiom giveswhere the arrow is the difference of the two restriction maps. Localization of a module commutes with kernels and finite products. Localizing this formula at $a$ therefore gives exactly the gluing formula for $D(a)$, covered by the $D(a{s}_i)$. Consequentlycompatibly with restriction. These isomorphisms on the principal open subsets identify $\tilde{M}$ with $\mathcal{F}{|}_U$. We have proved the global-section localization for quasi-coherent sheaves and, in particular,

An $A$-module homomorphism $h:M\to N$ induces, by localization of a module, compatible mapson all principal open subsets $D(a)$. These glue to an ${\mathcal{O}}_X$-linear morphism of sheaves $\tilde{h}:\tilde{M}\to \tilde{N}$.

Conversely, taking global sections of an ${\mathcal{O}}_X$-linear morphism of sheaves gives an $A$-module homomorphism $h:M\to N$, since $\Gamma (X,\tilde{M})=M$ and $\Gamma (X,\tilde{N})=N$. On $D(a)$, compatibility with restriction determines the image of $m/1$, and ${\mathcal{O}}_X$-linearity determines the image of $a^{-r}m$. It must be $h(m)/a^r$. Thus the morphism of sheaves is uniquely recovered from $h$. Both constructions preserve addition and composition, proving the affine module-sheaf equivalence on morphisms:No finite-generation assumption is used.