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Draft lesson.

Finite projective descent and Morita generators

Written by GPT-6.1 Sol (OpenAI), October 2026. Public domain (CC0).

A finite projective module can be described by a finite dual basis. After a faithfully flat scalar extension, its dual basis need not consist of elements from the original ring. The dual object, evaluation and coevaluation do descend, however. Their two identities recover a finite dual basis over the original ring. We also descend the trace ideal that distinguishes a Morita generator from a finite projective module supported on only part of the ring.

1. Dual bases and arbitrary scalar extension

All rings in the scalar position are commutative and unital. Modules are unital; the zero ring is allowed. A module is projective if maps from it lift through every surjective module map. A finite projective module is a finitely generated projective module. Put \(P^*=\operatorname{Hom}_R(P,R)\).

Lemma 1.1. The following descriptions of an \(R\)-module \(P\) are equivalent: it is finite projective; it is a retract of \(R^n\) for some finite \(n\); there are elements \(p_i\in P\) and functionals \(f_i\in P^*\), for \(1\le i\le n\), with \[ p=\sum_i p_i f_i(p)\quad(p\in P). \tag{1.1} \]

Proof. Finite generation gives a surjection \(R^n\to P\); projectivity lifts \(1_P\) and splits it. From a retraction take the images of the standard basis as \(p_i\) and the coordinates of the section as \(f_i\). Conversely (1.1) supplies the section and retraction. It also proves projectivity directly: for a surjection \(V\to W\) and \(a:P\to W\), lift each \(a(p_i)\) to \(v_i\in V\), and send \(p\) to \(\sum_i v_i f_i(p)\). Its composite is \(a\). The \(p_i\) generate \(P\), completing all implications. \(\square\)

For any module \(Q\), the map \[ Q\otimes_R P^*\longrightarrow\operatorname{Hom}_R(P,Q), \qquad q\otimes f\longmapsto(p\mapsto q f(p)) \tag{1.2} \] is invertible when \(P\) is finite projective. Its inverse sends \(a\) to \(\sum_i a(p_i)\otimes f_i\). One composite is (1.1); the other uses \(f=\sum_i f(p_i)f_i\). These formulas prove naturality in \(Q\) and in maps between finite projective modules: both directions are the inverse of the displayed canonical map, so commute with every precomposition and postcomposition.

For any scalar map \(R\to T\), including a nonflat one, the elements \(1\otimes p_i\) and the functionals \((t\otimes p\mapsto t f_i(p))\) form a dual basis of \(T\otimes_R P\). Formula (1.2) consequently gives the canonical isomorphisms \[ \begin{aligned} T\otimes_R P^*&\simeq(T\otimes_R P)^*,\\ T\otimes_R\operatorname{Hom}_R(P,Q) &\simeq\operatorname{Hom}_T(T\otimes_R P,T\otimes_R Q). \end{aligned} \tag{1.3} \] The second map sends \(t\otimes a\) to \((u\otimes p\mapsto tu\otimes a(p))\). Factoring it through (1.2) proves it is exactly the stated canonical map. For \(Q=P\) it preserves identities and composition, and hence identifies endomorphism algebras, not just modules.

Evaluation \(P^*\otimes P\to R\) and coevaluation \(1\mapsto\sum_i p_i\otimes f_i\) satisfy \[ (1_P\otimes\operatorname{ev})(\operatorname{coev}\otimes1_P)=1_P, \qquad (\operatorname{ev}\otimes1_{P^*})(1_{P^*}\otimes\operatorname{coev})=1_{P^*}. \tag{1.4} \] Indeed the two composites send \(p\) to \(\sum_i p_i f_i(p)\) and \(f\) to \(\sum_i f(p_i)f_i\). The coevaluation is independent of the chosen dual basis: under (1.2) it is \(1_P\). Conversely, suppose modules \(P,L\) have an evaluation \(L\otimes P\to R\) and a coevaluation \(R\to P\otimes L\) satisfying these two identities. Write the image of \(1\) as a finite sum \(\sum_i p_i\otimes l_i\). The first identity supplies (1.1) with \(f_i(p)=\operatorname{ev}(l_i\otimes p)\), so \(P\) is finite projective. The map \(L\to P^*\), \(l\mapsto(p\mapsto\operatorname{ev}(l\otimes p))\), has inverse \(f\mapsto\sum_i f(p_i)l_i\): the two identities prove both inverse composites. Thus the descended dual will be the actual module dual.

2. Descending the dual and both identities

Fix a faithfully flat scalar map \(R\to S\). This means \(S\otimes_R-\) is exact and faithful. It therefore detects zero modules and isomorphisms: apply it to kernels and cokernels, and use faithfulness on identity maps. The complete effectiveness and all morphism comparisons are proved in Faithfully flat scalar descent, Sections 2–4, specialized to the one-point stack \(\operatorname{Mod}(R)\).

Write \(B=S\otimes_R S\), with maps \(p_0(s)=s\otimes1\) and \(p_1(s)=1\otimes s\). A descent datum on an \(S\)-module \(N\) is an invertible \(B\)-linear map \(\phi:N_0\to N_1\), where \(N_i=B\otimes_{S,p_i}N\), satisfying \(\phi_{12}\phi_{01}=\phi_{02}\) over three scalar factors. The diagonal comparison is the identity. The linked construction gives an \(R\)-module \(M\), its invariant equalizer, and the actual comparison \(S\otimes_R M\simeq N\); maps compatible with \(\phi\) descend uniquely.

Tensor products are compatible with this equivalence. In fact \[ S\otimes_R(M\otimes_R K)\simeq (S\otimes_R M)\otimes_S(S\otimes_R K) \tag{2.1} \] sends \(s\otimes m\otimes k\) to \((s\otimes m)\otimes(1\otimes k)\), with inverse sending \((s\otimes m)\otimes(t\otimes k)\) to \(st\otimes m\otimes k\). These maps respect balancing and the canonical overlap comparisons. On pure tensors they also respect the unit, associator and permutation of factors. Thus the tensor of two descent data descends to the actual tensor of their descended modules.

Theorem 2.1. If \(N\) is finite projective over \(S\), its descended module \(M\) is finite projective over \(R\). Its dual, evaluation and coevaluation descend with it. In particular, an \(R\)-module is finite projective if and only if its extension to \(S\) is finite projective.

Proof. Set \(D=\operatorname{Hom}_S(N,S)\). By (1.3), \(D_i\) is canonically \(\operatorname{Hom}_B(N_i,B)\). Give \(D\) the overlap comparison \[ \phi_D=(\phi^{-1})^*:D_0\longrightarrow D_1. \tag{2.2} \] It satisfies the cocycle: taking inverse duals of a composite preserves its order, since each operation reverses order. Explicitly, \((\phi_{12}^{-1})^*(\phi_{01}^{-1})^*=(\phi_{01}^{-1}\phi_{12}^{-1})^*=(\phi_{02}^{-1})^*\). The identity diagonal remains the identity. Effective descent gives a module \(L\) and \(S\otimes_R L\simeq D\).

Evaluation is compatible with the overlap maps, because \((d\circ\phi^{-1})(\phi n)=d(n)\). Coevaluation is compatible as well: identify \(N_i\otimes_B D_i\) with \(\operatorname{End}_B(N_i)\) by rank-one operators. Transport by \(\phi\otimes\phi_D\) sends an operator \(a\) to \(\phi a\phi^{-1}\); it fixes the identity operator that represents coevaluation. Consequently both maps descend, using (2.1), to \[ L\otimes_R M\longrightarrow R, \qquad R\longrightarrow M\otimes_R L. \tag{2.3} \] After extension to \(S\), their two triangle composites are the identity by (1.4). Faithfulness makes both composites the identity over \(R\). The converse part of Section 1 now proves that \(M\) is finite projective and identifies \(L\) with \(M^*\). Every comparison used here is compatible with the specified descent datum. In the opposite direction a finite dual basis extends along any scalar map, as already proved in Section 1. \(\square\)

This proof descends the coevaluation as a map. It does not require an individually chosen dual-basis vector over \(S\) to be an invariant vector.

3. Other finiteness and exactness properties

Proposition 3.1. For \(R\to S\) faithfully flat and \(N=S\otimes_R M\), finite generation, finite presentation and flatness of \(N\) imply the corresponding properties of \(M\).

Proof. If finitely many elements generate \(N\), express each as a finite tensor sum. Let \(M_0\) be generated by all the resulting elements of \(M\). The map \(S\otimes M_0\to N\) is surjective. Right exactness gives \(S\otimes(M/M_0)=0\), and detection of zero modules gives \(M=M_0\).

For finite presentation, first choose a surjection \(q:R^n\to M\), with kernel \(K\). Flatness of \(S\) identifies \(S\otimes K\) with the kernel of \(S^n\to N\). We justify why that kernel is finitely generated. For any finitely presented module \(V\) over a ring \(A\), take a presentation \(A^a\to A^b\xrightarrow{\pi}V\to0\), and any surjection \(q':A^n\to V\). Lift to maps \(u:A^b\to A^n\), \(v:A^n\to A^b\) satisfying \(q'u=\pi\), \(\pi v=q'\). The kernel of \(q'\) is generated by the images under \(u\) of the finitely many relation generators, together with \((1-uv)e_j\). Indeed for \(k\in\ker q'\), the element \(vk\) is a relation and \(k=u(vk)+(1-uv)k\). Every displayed generator lies in the kernel. Apply this with \(A=S,V=N\), then apply the preceding finite-generation descent to \(K\). It gives a finite presentation of \(M\).

If \(N\) is flat, take any injection \(V\to W\) of \(R\)-modules. After tensoring its tensor-with-\(M\) map with \(S\), the map becomes \[ (S\otimes_R V)\otimes_S N\longrightarrow (S\otimes_R W)\otimes_S N. \tag{3.1} \] It is injective: \(S\) preserves the first injection and \(N\) is flat. Exact faithful scalar extension detects the original kernel, so \(V\otimes_R M\to W\otimes_R M\) is injective. Tensor is right exact, hence \(M\) is flat. \(\square\)

Theorem 2.1 proves projectivity with its own dual construction; it does not infer projectivity from finite generation and flatness alone. Finite projectivity also gives finite presentation directly: a complement in \(R^n\) is generated by the projections of its standard basis, so it supplies finitely many relations.

4. Trace ideals and the generator condition

For a finite projective module \(P\), define its trace ideal \[ J_P=\left\{\sum_{j=1}^m f_j(v_j): f_j\in P^*,\ v_j\in P,\ m\text{ finite}\right\}\subseteq R. \tag{4.1} \] It is an ideal, by additivity and scalar multiplication. It is the image of evaluation \(P^*\otimes_R P\to R\). Here a module is a generator if every module is a quotient of a direct sum of copies of it.

Lemma 4.1. The module \(P\) is a generator if and only if \(J_P=R\). This is equivalent to a finite witness \[ \sum_j g_j(v_j)=1. \tag{4.2} \]

Proof. Such a witness makes \(P^m\to R\), \((x_j)\mapsto\sum_jg_j(x_j)\), surjective. Given a module \(X\), the free module with one generator for every element of \(X\) surjects onto \(X\). Replacing each free summand \(R\) by the above quotient proves the generator property. Conversely, if a direct sum of copies of \(P\) surjects onto \(R\), a preimage of \(1\) has only finitely many nonzero coordinates. Restricting the surjection to these summands gives (4.2). \(\square\)

For any \(R\to T\), formula (1.3) and the evaluation pairing give \[ J_{T\otimes_R P}=J_P T. \tag{4.3} \] To verify both inclusions, every scalar extension of a value \(f(v)\) is an evaluation value after extension; conversely extended vectors and functionals are finite tensor sums, and their evaluations are sums of multiples of these original values. The right side means the extended ideal, the image of \(T\otimes_R J_P\to T\). That map need not be injective for a nonflat scalar map.

Theorem 4.2. Finite projective generators descend along faithfully flat scalar maps, and are preserved by every scalar map.

Proof. If \(S\otimes_R M\) is finite projective, Theorem 2.1 first makes \(M\) finite projective. If it is also a generator, (4.3) gives \(J_M S=S\). Right exactness identifies \(S\otimes_R(R/J_M)\) with \(S/J_M S=0\). Faithfulness gives \(R/J_M=0\), so Lemma 4.1 makes \(M\) a generator. In the other direction the finite dual basis and witness (4.2) extend along every scalar map. \(\square\)

5. The actual inverse Morita bimodules

Let \(P\) be a finite projective generator and \(E=\operatorname{End}_R(P)\), with multiplication given by composition. The module \(P\) is an \((E,R)\)-bimodule; \(L=P^*\) is an \((R,E)\)-bimodule with \(f\cdot a=f\circ a\). The scalar actions are central. There are bimodule maps \[ \begin{aligned} P\otimes_R L&\longrightarrow E,&v\otimes f&\longmapsto(x\mapsto v f(x)),\\ L\otimes_E P&\longrightarrow R,&f\otimes v&\longmapsto f(v). \end{aligned} \tag{5.1} \] The first is (1.2), with inverse \(a\mapsto\sum_i a(p_i)\otimes f_i\). Its left and right \(E\)-actions become composition on the corresponding side. The second is well defined because \((f\circ a)(v)=f(av)\).

Proposition 5.1. Both maps (5.1) are isomorphisms. They give inverse equivalences \[ G=P\otimes_R-:\operatorname{Mod}(R)\longrightarrow\operatorname{Mod}(E), \qquad H=L\otimes_E-:\operatorname{Mod}(E)\longrightarrow\operatorname{Mod}(R). \tag{5.2} \]

Proof. It remains to invert the second map. Choose (4.2), and send \(r\in R\) to \(\sum_jg_j\otimes rv_j\). Evaluation gives \(r\). For the other composite, fix \(f\otimes v\), and define \(a_j(x)=v_j f(x)\). Balancing gives \[ \sum_jg_j\otimes v_j f(v) =\sum_j(g_j\circ a_j)\otimes v =\left(\sum_jg_j(v_j)\right)f\otimes v =f\otimes v. \tag{5.3} \] Thus this is an inverse, for every tensor and hence every finite sum. Tensor associativity and (5.1) now give natural isomorphisms \(HG(X)\simeq R\otimes_R X\simeq X\) and \(GH(Y)\simeq E\otimes_E Y\simeq Y\). Their maps are evaluation and the rank-one action, respectively. The compatibility between the two pairings is the elementary identity \((v\otimes f)w=v f(w)\); on the dual side it is \((f\cdot(v\otimes g))(w)=f(v)g(w)\). These also verify the two triangle comparisons on pure tensors. All maps are additive and natural in module maps. This proves the full equivalence and its specified inverses. \(\square\)

The endomorphism comparison (1.3) identifies \(T\otimes_R E\simeq\operatorname{End}_T(T\otimes_R P)\). Under it, both pairings (5.1) extend to the pairings for \(T\otimes_R P\). For the second pairing, use the explicit balanced-tensor comparison \[ T\otimes_R(L\otimes_E P)\simeq (T\otimes_R L)\otimes_{T\otimes_R E}(T\otimes_R P). \tag{5.4} \] It sends \(t\otimes f\otimes v\) to \((t\otimes f)\otimes(1\otimes v)\), and its inverse multiplies the two \(T\)-coefficients. Balancing checks give both inverse composites. Consequently the entire Morita equivalence, including its inverse maps, commutes with arbitrary scalar extension on ordinary modules. Flatness of \(T\) is not needed for this comparison.

6. Strict bounded complexes

Suppose a bounded complex \(N^\bullet\) of finite projective \(S\)-modules has descent data in every degree, and its differentials respect those data. Descent of morphisms gives maps \(d_M^i:M^i\to M^{i+1}\). Their consecutive composites vanish, since their extensions are \(d_N^{i+1}d_N^i=0\) and scalar extension is faithful. Theorem 2.1 makes every \(M^i\) finite projective. Zero terms descend to zero terms, so the same finite degree interval bounds the descended complex. The specified comparisons assemble to an isomorphism \(S\otimes_R M^\bullet\simeq N^\bullet\) of complexes.

The same argument descends every compatible chain map and homotopy, and proves uniqueness with the specified comparisons. A map of descended complexes is a quasi-isomorphism if its scalar extension is one: exactness identifies \(S\otimes_R H^i\) with the cohomology after extension, and faithfulness detects its vanishing. These statements concern strict degreewise descent data.

The equivalences (5.2) are exact, because an additive equivalence preserves the universal properties of kernels and cokernels. Applied degreewise, they preserve cohomology, chain homotopies and mapping cones. They therefore give inverse equivalences of bounded derived categories, using the bounded localization construction. This is the derived equivalence supplied by the Morita pairings. Arbitrary scalar extension in (5.4) concerns ordinary tensors; when it is nonflat, it need not preserve quasi-isomorphisms.

7. Local module sheaves

Let \(\mathcal R\to\mathcal S\) be a map of commutative ring sheaves on a topological space, faithfully flat at every stalk. A module sheaf is locally finite projective if near every point it is a retract of a finite free module sheaf. The size of the free module may vary between neighborhoods. The scalar-descent lesson proves effectiveness and faithfulness for these sheaves, including morphisms and restriction.

Suppose a locally finite projective \(\mathcal S\)-module \(\mathcal N\) has scalar descent data. Its sheaf dual \(\mathcal Hom_{\mathcal S}(\mathcal N,\mathcal S)\) satisfies the arbitrary base-change comparison locally by the finite dual bases of Section 1. Thus (2.2), evaluation and coevaluation are maps of sheaves with the same overlap equations. They descend to \(\mathcal L,\mathcal M\) and both triangles hold, since stalkwise faithful extension detects the equalities.

Near a given point, the descended coevaluation is a finite tensor sum \(\sum_i m_i\otimes l_i\). To see this without assuming exact global sections, represent its germ as a finite sum in the tensor stalk, choose representatives of the finitely many entries on a common neighborhood, and shrink until the equality of the two section germs holds. Evaluation gives sheaf maps \(f_i:\mathcal M\to\mathcal R\) there. The first triangle is the sheaf identity \(m=\sum_i m_i f_i(m)\), so these maps exhibit a local finite-free retraction. Hence \(\mathcal M\) is locally finite projective.

Define the trace ideal sheaf as the image of evaluation. Locally, (4.3) identifies its scalar extension with the trace ideal of \(\mathcal N\). If the latter is all of \(\mathcal S\), the quotient of \(\mathcal R\) by the former has zero scalar extension and hence zero stalks. Thus the trace ideal is all of \(\mathcal R\). The witnesses (4.2) then exist locally. The pairings (5.1) are isomorphisms of sheaves, since the explicit inverses can be checked on these neighborhoods. The resulting tensor functors are inverse on every open subset and commute with restriction. They give the local module-stack Morita equivalence with \(\mathcal End_{\mathcal R}(\mathcal M)\). This local statement does not assert that global sections form a projective module or an exact functor.

8. Four graded exercises with complete solutions

Exercise 8.1 — A matrix and its descended dual (4 points)

Take \(R=\mathbb Z\), \(S=R\times R\) with diagonal scalar map, and an \(S\)-module whose two components are \(R^2\). Give the overlap from component zero to component one by \[ T=\begin{pmatrix}1&1\\0&1\end{pmatrix}, \tag{8.1} \] the reverse overlap by \(T^{-1}\), and the diagonal overlaps by identities. Determine the invariant module, its dual invariant module and their evaluation. Verify the cocycle and exhibit a descended dual basis.

Solution. The two-component overlaps \(\phi_{ji}\) satisfy \(\phi_{kj}\phi_{ji}=\phi_{ki}\). If the indices agree the comparison is identity; if a route traverses the two components and returns, its product is \(TT^{-1}\) or \(T^{-1}T\), also identity. These exhaust the triples, proving the cocycle. The invariant module is \(M=\{(v,Tv):v\in R^2\}\), isomorphic to \(R^2\) by the first projection. Tensoring with \(S\) gives the original two components, with the specified overlap.

Use row vectors for functionals. The dual comparison sends \(g\) to \(gT^{-1}\), so the dual invariant module is \(L=\{(g,gT^{-1}):g\in(R^2)^*\}\). Evaluation on the two components gives \(g v\) and \((gT^{-1})(Tv)=gv\); it therefore descends to that integer. Let \(e_1,e_2\) be the standard columns and \(e_1^*,e_2^*\) the standard rows. The elements \((e_i,Te_i)\) and functionals represented by \((e_i^*,e_i^*T^{-1})\) are a descended dual basis. Their evaluations are \(\delta_{ij}\), and their rank-one sum is the identity on both components. This gives both triangle identities directly. The opposite transpose appears if functionals are written as columns; with the row convention the formula is exactly \(gT^{-1}\).

Exercise 8.2 — Two boundaries for the Hom comparison (6 points)

(a) For \(R=\mathbb Z\), \(P=\mathbb Z/2\), \(Q=\mathbb Z\), \(T=\mathbb Z/2\), compute the two sides of the second map (1.3). (b) Take \(P=\bigoplus_{n\ge1}\mathbb Z\), \(Q=\mathbb Z\), \(T=\mathbb Q\). Describe the image of the same map and show it is not surjective. State precisely which hypotheses these examples test.

Solution. (a) A homomorphism \(\mathbb Z/2\to\mathbb Z\) is zero because its value on the generator is killed by two. Thus the left side is zero. On the right, both scalar-extended modules are \(\mathbb F_2\), so their Hom module is \(\mathbb F_2\). The comparison is the zero map into a nonzero module. The first argument is finitely presented but not projective, and the scalar map is nonflat. Finite presentation alone does not give the arbitrary scalar comparison of Section 1.

(b) A map from a direct sum is specified by an unrestricted sequence of integer images of its basis vectors. Hence \(\operatorname{Hom}_{\mathbb Z}(P,Q)=\prod_{n\ge1}\mathbb Z\), whereas the Hom module after extension is \(\prod_{n\ge1}\mathbb Q\). Tensoring with \(\mathbb Q\) localizes the former: each element has one common nonzero integer denominator and an integer sequence as numerator. Its comparison is coordinatewise division. It is injective, since a sequence whose every coordinate becomes zero already has every coordinate zero. Its image consists exactly of rational sequences with a common integer denominator. For distinct primes \(p_n\), the sequence \((1/p_n)_n\) has none: an integer clearing every denominator would be divisible by every \(p_n\). Infinitely many distinct primes exist, since a prime divisor of one plus a finite product belongs to none of the factors. Thus the map is not surjective. Here the scalar map is flat and \(P\) is free, but not finitely generated. The example tests finiteness, not projectivity or effectiveness of faithful-flat descent.

Exercise 8.3 — Matrix corners and an invisible component (8 points)

For \(P=R^n\), \(n\ge1\), compute the pairings (5.1) with matrix units and write an inverse to evaluation \(P^*\otimes_{\operatorname{Mat}_n(R)}P\to R\). Then take \(R=k\times k\) for a nonzero field and \(P=k\times0\). Compute the endomorphism ring and trace ideal, and explain why this finite projective module does not generate \(\operatorname{Mod}(R)\).

Solution. Write \(e_i\) for columns and \(e_j^*\) for rows. The first pairing sends \(e_i\otimes e_j^*\) to the matrix unit \(E_{ij}\), giving its basis inverse. The second sends \(e_i^*\otimes e_j\) to \(\delta_{ij}\). An inverse is \(r\mapsto e_1^*\otimes r e_1\). For every basis tensor, \[ e_i^*\otimes e_j =(e_1^*E_{1i})\otimes e_j =e_1^*\otimes E_{1i}e_j =\delta_{ij}e_1^*\otimes e_1. \tag{8.2} \] This proves the inverse on all tensors, including their scalar multiples. Consequently both pairings are the asserted bimodule isomorphisms. If \(n=0\) and \(R\ne0\), the trace ideal is zero and there is no generator; when \(R=0\), both module categories contain only the zero module.

For \(R=k\times k\), the idempotent \(e=(1,0)\) gives \(P=eR\), a retract of \(R\). Its endomorphism ring is \(eR\simeq k\), and its dual is also \(eR\). Evaluation has image \(eR\), so \(J_P=k\times0\), a proper ideal. Every direct sum of copies of \(P\), and every quotient of that sum, is annihilated by \(1-e\). The module \(0\times k\) is not annihilated by \(1-e\), so cannot be such a quotient. The Morita equivalence applies to the supported component \(eR\), not to all of \(R\).

Exercise 8.4 — Losing flatness or faithfulness (10 points)

(a) Let \(R=\mathbb Z\), \(S=\mathbb Q\times\mathbb F_2\), with its diagonal scalar map, and \(M=\mathbb Z/2\). Show that the scalar map is injective and \(S\) is a faithful \(R\)-module in the annihilator sense, but nonflat. Show that \(S\otimes_R M\) is finite projective although \(M\) is not. (b) For a prime \(p\), take the flat map \(\mathbb Z\to\mathbb Z[1/p]\). Give failures of projectivity descent and finite-generation descent when faithfulness is omitted.

Solution. (a) The map is injective because of its \(\mathbb Q\)-coordinate. If an integer annihilates all of \(S\), it annihilates \(1\in\mathbb Q\), so is zero. Nevertheless tensoring is not exact: the injection given by multiplication by two on \(\mathbb Z\) becomes multiplication by two on \(S\), whose kernel is \(0\times\mathbb F_2\ne0\). Further, \(S\otimes_R M\simeq S/2S=0\times\mathbb F_2\), the idempotent summand \((0,1)S\), so it is finite projective over \(S\). The module \(M\) is not projective over \(\mathbb Z\): a section of \(\mathbb Z\to\mathbb Z/2\) would send the generator to an integer killed by two, necessarily zero, contradicting the section identity. Annihilator faithfulness is not the exact faithful tensor condition used in Theorem 2.1; this example violates flatness as well.

(b) Localization is flat: for an injection of abelian groups, if a fraction maps to zero, some denominator kills its numerator in the target and hence in the source, so that fraction was zero; localization is also right exact. It is not faithful, because it kills \(\mathbb Z/p\). That nonprojective module becomes the zero module, which is finite projective. For finite generation take \(V=\bigoplus_{n\ge1}\mathbb Z/p\). It too becomes zero. Any finite family in \(V\) has support in finitely many coordinates, so cannot generate an element in a new coordinate. Thus \(V\) is not finitely generated although its extension is. The projectivity argument for \(\mathbb Z/p\) is the same section obstruction as in part (a), with two replaced by \(p\). These examples demonstrate the hypotheses of the descent theorems without contradicting the arbitrary scalar preservation of a module that was already finite projective.

Freely accessible references

Original lesson text: CC0 1.0.