Draft lesson.

Flat covers and bounded tensor for locally module stacks

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

Flat replacements must respect the algebra acting on an object. We construct them globally from local module descriptions, then construct bounded-above balanced tensor with the dual stack. For external tensor over a commutative base ring, we state a separate algebra-flatness hypothesis. A quotient algebra shows why the two notions of flatness cannot be exchanged.

Let $X$ be a topological space, $\mathcal R$ a sheaf of commutative rings, and $\mathfrak M$ an $\mathcal R$-linear stack locally equivalent to left module sheaves over unital, possibly noncommutative, algebras $\mathcal A_i$ on opens $U_i$. All diagrams and section-indexed sums lie in fixed ambient universes, with choice. Use cochain complexes, with $d$ increasing degree, and $D^-$ for objects whose cohomology vanishes above some integer.

Noncommutative gluing by invertible bimodules supplies the coherent Morita descriptions. Dual and tensor products of locally module stacks constructs the dual $\mathfrak M^\vee$, locally right $\mathcal A_i$-modules, and external stack products. Relative pullback of locally module stacks, Sections 1–2 and 4–6, proves stalk tensor formulas, extension by zero, finite Morita dual bases and pullback coherence. Injective objects in locally module stacks, Sections 1–2, proves the global abelian structure and coproducts. Universal derived functors and compatible composition, Section 8, proves the elementary flat-quotient lemma for ordinary modules. Derived morphisms through quasi-isomorphism roofs, Sections 2–4, supplies the roof equality and localization universal properties.

1. Two flatness conditions and Morita invariance

A left module $P$ over a ring $A$ is flat if $-\otimes_A P$ preserves injections of right modules. Tensor preserves cokernels by its balanced-map universal property, so this is equivalent to exactness. A module sheaf is flat when all its stalk modules are flat. Stalks commute with tensor and detect exactness, so flat stalks make tensor exact on sheaves.

Call $F\in\mathfrak M(X)$ intrinsically flat when its stalks are flat over the local algebra rings in each chart. Call it base-flat when those same local modules are flat over $\mathcal R_x$.

Lemma 1.1. Both conditions are independent of the Morita charts. Intrinsic flatness need not imply base-flatness.

Proof. At an overlap stalk, let $L$ be an invertible $(A,B)$-bimodule, with inverse $K$. If $P$ is left $B$-flat, the functor on right $A$-modules $N\longmapsto N\otimes_A(L\otimes_B P) \simeq(N\otimes_A L)\otimes_B P \tag{1.1}$ is exact: $N\mapsto N\otimes_A L$ is an equivalence to right $B$-modules, with inverse tensor by $K$, and the last tensor is exact by flatness. The inverse equivalence proves the converse.

Also $L$ is finite projective as a right $B$-module. To see the precise reason, write the inverse image of $1_A$ under $L\otimes_B K\simeq A$ as a finite sum $\sum l_s\otimes k_s$. The contraction $K\otimes_A L\simeq B$ gives right $B$-linear functions $\phi_s(l)=k_s l$. The Morita triangle identity says $l=\sum l_s\phi_s(l)$, a finite dual basis. Thus $L$ is a retract of $B^r$, and $L\otimes_B P$ is a retract of $P^r$ as a base-ring module. Finite sums and retracts of base-flat modules are base-flat: tensor a monomorphism into the finite sum, and restrict its injective map to the retract. Applying the inverse Morita bimodule gives the converse. This proves both invariances stalkwise. If $R=\mathbb Z$, $A=\mathbb Z/2$, then $A$ is flat over itself but not over $R$, since tensoring multiplication by two on $\mathbb Z$ gives a zero endomorphism of the nonzero group $A$. $\square$

There is a global balanced pairing $\mathfrak M^\vee(X)\times\mathfrak M(X) \longrightarrow\operatorname{Mod}(\mathcal R),\qquad (N,F)\longmapsto N\otimes_{\mathfrak M}F. \tag{1.2}$ In a chart it is $N_i\otimes_{\mathcal A_i}F_i$. Indeed the right-module transition is $N_i\simeq N_j\otimes_{\mathcal A_j}L_{ji}$, while $F_i\simeq L_{ij}\otimes_{\mathcal A_j}F_j$. Contracting $L_{ji}\otimes_{\mathcal A_i}L_{ij}\simeq\mathcal A_j$ identifies the two local tensors. The coherent contractions and their triangles make these maps satisfy the overlap cocycle, so the sheaves and all their morphisms glue. This also realizes (1.2) as evaluation of the dual functor at $F$.

2. Enough flat objects from section representers

Proposition 2.1. Every object of $\mathfrak M(X)$ is an epimorphic image of an intrinsically flat object. If the chosen algebras $\mathcal A_i$ are flat over $\mathcal R|_{U_i}$, the covering objects can also be chosen base-flat.

Proof. For $V\subseteq U_i$ open, extension by zero $\mathcal A_i[V]$ represents sections: $\operatorname{Hom}_{\mathcal A_i}(\mathcal A_i[V],F_i)=F_i(V). \tag{2.1}$ Its stalk is $(\mathcal A_i)_x$ for $x\in V$, and zero outside $V$. Hence it is intrinsically flat. It is also base-flat under the stated algebra hypothesis. Transfer it to $\mathfrak M(U_i)$, and extend by zero to $X$, obtaining $G_{i,V}$. In any other chart its stalk is either zero or a Morita transform of the old stalk. Lemma 1.1 proves the corresponding flatness of $G_{i,V}$.

For each section $s\in F_i(V)$, (2.1) and the extension-by-zero adjunction give a map $G_{i,V}\to F$. Collect these into $P_F=\bigoplus_{i,V,s\in F_i(V)}G_{i,V}\longrightarrow F. \tag{2.2}$ Every germ of $F_i$ is represented by some such section on a $V$; the regular module stalk of that summand maps its unit to the germ. Thus (2.2) is surjective on each local stalk and is epic globally. Direct sums of flat modules are flat, since tensor commutes with sums and an element has finite support. Stalks commute with coproducts, proving the asserted flatness of $P_F$. The same argument for opposite algebras supplies enough intrinsic flats in $\mathfrak M^\vee(X)$. $\square$

This constructs covers of objects, without claiming that the covers are projective or that their section maps are surjective on every open set.

3. Global bounded-above replacements

Theorem 3.1. A complex $C$ zero above $b$ has a termwise epic quasi-isomorphism $q:P\to C$, where $P$ is zero above $b$ and intrinsically flat term by term. Under the base-flat algebra hypothesis, its terms can also be base-flat.

Proof. Set $P^{b+1}=P^{b+2}=\cdots=0$. Descend in $n$. Having constructed higher terms and their maps, form the global abelian pullback $T_n=C^n\times_{C^{n+1}}Z^{n+1}P, \tag{3.1}$ where the first map is $d_C^n$, the second is the restriction of $q^{n+1}$, and $Z^{n+1}P=\ker d_P^{n+1}$. Choose a flat cover $P^n\twoheadrightarrow T_n$ by Proposition 2.1. Its first projection is $q^n$; its second projection followed by inclusion is $d_P^n$. The pullback equation gives $q^{n+1}d_P^n=d_C^nq^n$, and the cycle target gives $d_P^{n+1}d_P^n=0$.

The construction can be checked in every stalk of every module chart, because those stalk functors are exact and preserve these finite pullbacks. Suppose the map on cycles in degree $n+1$ is surjective. For $x\in C^n_x$, its differential is a cycle, so it has a lift $z\in Z^{n+1}P_x$; lifting $(x,z)$ through the cover proves surjectivity of $q^n_x$. If $x$ itself is a cycle, lifting $(x,0)$ gives a cycle lift, proving surjectivity on $Z^n$. If a cycle $z\in Z^{n+1}P_x$ maps to a boundary $d_Cx$, lifting $(x,z)$ makes $z$ a boundary in $P_x$. Thus the map on $H^{n+1}$ is injective and, by cycle surjectivity, also surjective.

The cycle condition starts with $0\to0$ in degree $b+1$. Induction proves all these properties; each cohomology degree becomes correct when its next lower term is added. A fixed term, differential and comparison map is never changed afterwards. The resulting complex is termwise epic and a quasi-isomorphism on all local stalks, hence globally. This is degreewise stabilization, not an assertion about exactness of an infinite inverse limit. $\square$

Every object of $D^-$ has a bounded-above representative: if cohomology vanishes above $b$, replace degree $b$ by $\ker d_C^b$, higher degrees by zero, and leave lower degrees unchanged. Its inclusion in $C$ is a quasi-isomorphism, by the kernel and cohomology definitions. Thus the theorem applies to all of $D^-$.

4. Exact tensor rows and finite total diagonals

Two facts justify every replacement comparison below.

Lemma 4.1. A bounded-above acyclic complex of flat left modules remains acyclic after tensoring with any right module. The analogous right-module assertion holds. These statements apply to module sheaves by taking stalks.

Proof. The flat-quotient lemma in the universal-derived-functors lesson proves that an exact row $0\to K\to P\to Q\to0$, with $Q$ flat, stays exact under every tensor, and that $K$ is flat when $P$ is flat too. For an acyclic flat complex $E$ zero above $b$, its top cycle $Z^bE=E^b$ is flat. Descending through $0\to Z^nE\to E^n\to Z^{n+1}E\to0$ proves that every cycle is flat and every tensor cycle row is exact. The image and kernel identities of these rows give acyclicity of the tensor complex. Apply the same proved lemma over the opposite ring for right modules. Stalks detect the sheaf assertions. $\square$

For bounded-above right and left complexes define $(N\otimes P)^n=\bigoplus_{p+q=n}N^p\otimes P^q,\qquad d(u\otimes v)=d_Nu\otimes v+(-1)^p u\otimes d_Pv. \tag{4.1}$ Each diagonal is finite: if the upper bounds are $a,b$, then $n-b\le p\le a$. The square of the differential is zero, since its two cross terms have opposite signs.

Lemma 4.2. If $P$ is bounded above and intrinsically flat term by term, (1.2) with formula (4.1) sends a quasi-isomorphism of bounded-above right complexes to a quasi-isomorphism. If $P$ is also acyclic, $N\otimes_{\mathfrak M}P$ is acyclic for every bounded-above right complex $N$. The dual statements hold. Replace intrinsic flatness and balanced tensor by base-flatness and tensor over $\mathcal R$ to obtain the same assertions there.

Proof. Check acyclicity on local stalks. If the right complex $E$ is acyclic, each column $E\otimes P^q$ is acyclic by flatness of $P^q$. For an acyclic flat $P$, each row $N^p\otimes P$ is acyclic by Lemma 4.1.

Here is why exact rows or columns give exact totalization. Choose the direction called horizontal so it has an upper bound $h$ and the vertical complexes are exact. A total cycle has finitely many horizontal components. At its smallest occupied horizontal index the vertical differential is zero, because there is no component entering from the preceding index. Choose a vertical preimage, with the sign required by (4.1), and subtract its total differential. This removes that component and can add one only at the next horizontal index. Repeating through $h$ removes the entire cycle, producing a finite sum of preimages. This works on stalk modules and proves exactness of the total sheaf complex.

For a quasi-isomorphism take its acyclic cone. Total tensor carries that cone to the cone of its tensor map, with the usual shifted-summand sign; equivalently its two component rows have exactly the differentials from (4.1) and $C_t^n=B^n\oplus A^{n+1}$. For the cone in the right variable the comparison on the shifted factor is identity. For the cone in the left variable, on the summand with right degree $p$ multiply the shifted factor by $(-1)^p$; this matches both the off-diagonal map and the reversed shifted differential. Hence the preceding total acyclicity makes the tensor map a quasi-isomorphism. The opposite-ring and base-ring proofs are identical applications of the stated exact rows. $\square$

5. Balanced derived tensor with the dual stack

Theorem 5.1. There is a bifunctor $D^-(\mathfrak M^\vee(X))\times D^-(\mathfrak M(X)) \longrightarrow D^-(\operatorname{Mod}(\mathcal R)),\qquad (N,F)\longmapsto N\otimes_{\mathfrak M}^{L}F. \tag{5.1}$ It is computed by two intrinsic flat replacements, or by one such replacement with the other bounded-above input left unchanged. These comparisons are natural for all derived morphisms and compatible with changes of replacements.

Proof. In each input let $\mathcal K_{\rm fl}$ be the full homotopy subcategory of bounded-above intrinsic flat complexes, with quasi-isomorphisms $S_{\rm fl}$. Its localization is equivalent to $D^-$. Essential surjectivity is Theorem 3.1 and the preceding truncation. For full faithfulness use the roof formula: a derived roof between two flat complexes can be refined by a flat replacement of its middle complex. If that middle complex is not bounded above, first use its good upper truncation: its cohomology has the same upper bound as the source because the denominator is a quasi-isomorphism. The refined roof has all three objects flat. An equality of two such roofs has a common bounded-above refinement by the same roof calculus; replace that refinement by a flat complex to obtain the equality within the flat subcategory. The calculus can be carried out in bounded-above complexes, since its finite cone and homotopy-pullback constructions remain bounded above. Thus the same roofs and equality tests prove fullness and faithfulness, including composition.

The total pairing on two flat homotopy subcategories inverts quasi-isomorphisms in each variable by Lemma 4.2. It respects homotopies: a homotopy in the right variable tensors as $h\otimes1$, and one in the left variable as $u\otimes v\mapsto(-1)^p u\otimes h(v)$. Substituting in (4.1) cancels the cross terms and gives the required homotopy equations. Consequently localization in the two inputs gives the bifunctor and all its transformations.

Choose $Q\to N$, $P\to F$. The comparison maps $Q\otimes_{\mathfrak M}P\longrightarrow N\otimes_{\mathfrak M}P,\qquad Q\otimes_{\mathfrak M}P\longrightarrow Q\otimes_{\mathfrak M}F \tag{5.2}$ are quasi-isomorphisms, by the first assertion of Lemma 4.2 and its dual. Inverting them gives the single-replacement formulas. Naturality on homotopy maps and on inverse denominators gives naturality on every roof. Different replacements identify under the fully faithful localized flat categories; three choices obey the cocycle law there, so their tensor comparisons do too. Local Morita contractions in (1.2) commute with the differentials and all comparison maps, since they are degree-zero bimodule maps. Thus the construction is global and independent of the charts. The sum has upper bound equal to the sum of the input bounds, proving that its value lies in $D^-$. The cone comparisons in Lemma 4.2 also preserve all three cone-triangle maps. A shift of the right input uses identity on total summands; a shift of the left input uses $(-1)^p$ on the summand with right degree $p$. These satisfy the chain equations of (4.1) and identify the last cone projections. Thus the localized bifunctor preserves distinguished triangles in each input. The two orders of these shift comparisons differ by a minus sign, since shifting the right input replaces $p$ by $p+1$. $\square$

6. External base-ring tensor with its actual hypothesis

Let $\mathfrak N$ have charts $\operatorname{Mod}(\mathcal B_j)$. Assume both chosen algebra families are base-flat. Then there is a bifunctor $D^-(\mathfrak M(X))\times D^-(\mathfrak N(X)) \longrightarrow D^-((\mathfrak M\boxtimes_{\mathcal R}\mathfrak N)(X)), \qquad (F,G)\longmapsto F\otimes_{\mathcal R}^{L}G. \tag{6.1}$

Proof. Proposition 2.1 and Theorem 3.1 supply base-flat replacements in both inputs. The localized base-flat homotopy categories are equivalent to their $D^-$ categories by the identical flat-refinement proof in Theorem 5.1: cones, shifts and finite sums remain base-flat, and every middle complex has a base-flat replacement.

On a common chart, use (4.1) over $\mathcal R$, retaining the commuting $\mathcal A_i,\mathcal B_j$ actions. This is a complex of $\mathcal A_i\otimes_{\mathcal R}\mathcal B_j$-modules. The external tensor comparison of the underived stack-product lesson transports its transition by $L_{ii'}\otimes_{\mathcal R}K_{jj'}$; the balancing maps and the descent equations commute with the degreewise differential. Thus these complexes and their maps descend to the stated product stack. Lemma 4.2 over the base ring proves preservation of quasi-isomorphisms in both inputs. Localization gives (6.1), and the two augmentation comparisons give the single-base-flat-replacement formulas just as in (5.2).

The standard operation comparisons are retained. Associativity rebrackets $u\otimes v\otimes w$; both parenthesizations have differential signs $1,(-1)^p,(-1)^{p+q}$, so this is a chain isomorphism. The graded flip sends $u\otimes v$ to $(-1)^{pq}v\otimes u$; comparing the coefficients of $d u\otimes v$ and $u\otimes d v$ verifies its chain equation. Its square is identity. The associativity pentagon has no multiplier on either route; the symmetry hexagons are the identities $(-1)^{p(q+r)}=(-1)^{pq}(-1)^{pr}$ and its counterpart. The stack unit $\operatorname{Mod}(\mathcal R)$, with regular object $\mathcal R[0]$, is base-flat, and its tensor unit comparisons are $r\otimes u\mapsto ru$, with inverse $u\mapsto1\otimes u$. These maps commute with descent, so they also give the derived unit comparisons. Localization preserves all the equations, proving coherence. $\square$

Intrinsic flats alone do not justify this construction over the base ring. The source-algebra hypothesis is sufficient for actual base-flat replacements that keep the module actions; Solution 8.4 explains what can go wrong when it is absent.

7. Exact inverse image preserves the constructions

Let $f:Y\to X$ be continuous, with base sheaf $f^{-1}\mathcal R$ and the transported stacks. There are natural comparisons $f^{-1}(N\otimes_{\mathfrak M}^{L}F) \simeq f^{-1}N\otimes_{f^\boxtimes\mathfrak M}^{L}f^{-1}F, \tag{7.1}$ and, under Section 6's hypotheses, $f^{-1}(F\otimes_{\mathcal R}^{L}G) \simeq f^{-1}F\otimes_{f^{-1}\mathcal R}^{L}f^{-1}G. \tag{7.2}$

Proof. Inverse image is exact by the stalk formula proved in the relative-pullback and injective-stack lessons. A pulled-back module has at $y$ the same stalk module over the same algebra or base stalk ring at $f(y)$. It therefore preserves intrinsic and base flatness, and preserves quasi-isomorphisms of the replacements. It commutes with tensor and finite direct sums by that same stalk formula and their canonical section comparisons.

Apply it to the flat complexes computing (5.1) or (6.1). The degreewise tensor comparison becomes an isomorphism of total complexes, since every diagonal is finite and the comparison respects the two differential terms and their signs. It commutes with the replacement augmentations, hence induces (7.1) and (7.2) for every derived arrow after localization. Morita contractions pull back to the transported contractions, so the balanced pairing comparisons glue. For composable maps, all comparisons pull back the same germs and the same ordered tensor factors, giving composition coherence; identity and the unit, associativity and symmetry comparisons are preserved for the same reason. $\square$

8. Four graded exercises with complete solutions

Exercise 8.1 — A flat cover of a sheaf arrow (introductory)

Let $X=\{0,1\}$ have opens $\varnothing,\{1\},X$, with constant ring sheaf $A$. Given a sheaf arrow $r:M_0\to M_1$, construct a cover by flat sheaves using only free covers of the two modules. Over a field, compare flatness with injectivity.

Solution. The sheaf category is the arrow category of left $A$-modules, as proved in the injective-stack lesson. Choose free surjections $a_0:F_0\to M_0,\ a_1:F_1\to M_1$. The arrow $P=(F_0\xrightarrow{x\mapsto(x,0)}F_0\oplus F_1) \tag{8.1}$ maps to $M$ by $a_0$ at $0$ and $(x,y)\mapsto r(a_0x)+a_1y$ at $1$. This is a morphism because the restriction square gives $r a_0$ on the first summand. Both components are surjective, hence it is epic. Both stalk modules of $P$ are free, so $P$ is intrinsically flat.

The summand $(F_0\xrightarrow1F_0)$ is a sum of the regular representer on $X$; $(0\to F_1)$ is a sum of regular representers extended by zero from $\{1\}$. This identifies (8.1) with the local-representer construction. If $A$ is flat over the base ring, the free modules and their sums are base-flat too.

Over a field every sheaf arrow is flat, since both stalks are vector spaces. The injective-stack lesson proves that an arrow is injective exactly when its restriction is surjective. Thus $(0\to k)$ is flat but not injective, and (8.1) need not be injective when $F_1\ne0$. The two replacement constructions solve different extension and exactness problems. $\square$

Exercise 8.2 — A single replacement and its double comparison (intermediate)

On the two-point space of Exercise 8.1 with ring $\mathbb Z$, let $N=F=(\mathbb Z/2\xrightarrow1\mathbb Z/2)$, regarding $N$ as a right-module sheaf. Compute $N\otimes^L_{\mathbb Z}F$ using one flat replacement and specify the map from the double replacement, including its signs.

Solution. Write $\Delta B=(B\xrightarrow1B)$. The complex $P=(\Delta\mathbb Z\xrightarrow2\Delta\mathbb Z)$ in degrees $-1,0$, augmented by reduction modulo two, is a flat replacement of $F$. Its only cohomology is $F$ in degree zero; its terms have flat integer stalks. The single-replacement complex $N\otimes P$ is $\Delta(\mathbb Z/2)\xrightarrow0\Delta(\mathbb Z/2) \tag{8.2}$ in degrees $-1,0$. Hence those two degrees of cohomology are both $\Delta(\mathbb Z/2)$, and every other degree is zero.

Take the same complex $Q$ as a right flat replacement of $N$. Order the degree-minus-one summands of $Q\otimes P$ as $Q^{-1}\otimes P^0,\ Q^0\otimes P^{-1}$. Formula (4.1) gives the total complex $\Delta\mathbb Z\xrightarrow{t\mapsto(-2t,2t)} \Delta(\mathbb Z^2)\xrightarrow{(u,v)\mapsto2u+2v} \Delta\mathbb Z \tag{8.3}$ in degrees $-2,-1,0$. Its first differential is injective. The middle kernel is the set $(-s,s)$, and its quotient by $(-2t,2t)$ is $\mathbb Z/2$; the final cokernel is also $\mathbb Z/2$.

The comparison induced by $Q\to N$ is zero in degree $-2$, $(u,v)\mapsto v\bmod2$ in degree $-1$, and reduction modulo two in degree zero. The chain equations hold because $2t$ reduces to zero and $2u+2v$ does too. On the middle kernel it sends $(-s,s)$ to $\bar s$, an isomorphism after the indicated quotient; it is the evident isomorphism on the final cokernel. Thus it is the precise quasi-isomorphism giving the single-replacement comparison, not merely an equality of cohomology counts. Restriction to either point gives these same formulas. $\square$

Exercise 8.3 — Graded symmetry and a contractible total (advanced)

Over the constant integer ring sheaf on any space, let $P=(\mathbb Z_X\xrightarrow2\mathbb Z_X)$ and $Q=(\mathbb Z_X\xrightarrow3\mathbb Z_X)$, in degrees $-1,0$. Compute their total tensor, its graded flip, and a contracting homotopy.

Solution. The terms are base-flat, since their stalks are free integers. With minus-one summands ordered $P^{-1}\otimes Q^0,\ P^0\otimes Q^{-1}$, the complex $T=P\otimes Q$ is $\mathbb Z_X\xrightarrow{t\mapsto(-3t,2t)} \mathbb Z_X^2\xrightarrow{(u,v)\mapsto2u+3v} \mathbb Z_X. \tag{8.4}$ The composite is $-6t+6t=0$. The reversed total $T'=Q\otimes P$ has differentials $(-2t,3t)$ and $3u+2v$. The graded flip has components $t\mapsto-t,\qquad (u,v)\mapsto(v,u),\qquad z\mapsto z \tag{8.5}$ in these three degrees: only the product of the two odd degrees contributes a minus sign. At the first differential, the two routes both send $t$ to $(2t,-3t)$; at the second they both give $2u+3v$. Thus (8.5) is a chain isomorphism with inverse the same graded flip.

A degree-minus-one homotopy $h$ from identity to zero on $T$ is $h^0(z)=(-z,z),\qquad h^{-1}(u,v)=-(u+v), \tag{8.6}$ with all other components zero. In degree zero, $d h^0(z)=-2z+3z=z$. In degree minus one, $d h^{-1}(u,v)+h^0d(u,v) =(3u+3v,-2u-2v)+(-2u-3v,2u+3v)=(u,v).$ In degree minus two, $h^{-1}d(t)=-(-3t+2t)=t$. These are all the homotopy equations, valid as sheaf maps on every open. The total is contractible. Consequently the derived tensor of $\mathbb Z_X/2$ and $\mathbb Z_X/3$, computed by these replacements, is zero. The displayed flip and contraction specify the signs and maps behind that vanishing. $\square$

Exercise 8.4 — A nonflat algebra obstructs base-flat replacements (advanced)

Let $X$ be a point, $R=\mathbb Z$, $A=R/2R$, and $\mathfrak M=\operatorname{Mod}(A)$. Show that nonzero objects have no base-flat covering object. Compare intrinsic balanced tensor, the derived restricted external functor, and ambient derived tensor over $R$.

Solution. Every $A$-module is killed by two. If it is flat as a $\mathbb Z$-module, tensoring the injection $\mathbb Z\xrightarrow2\mathbb Z$ makes multiplication by two injective on that module. Since this map is zero, the module must be zero. Thus zero is the only base-flat object of $\mathfrak M$, and it cannot cover $A$. In contrast, all $A$-modules are vector spaces over the field $\mathbb F_2$ and are intrinsically flat.

For the dual regular object and regular object, intrinsic balanced derived tensor is $A\otimes_A^L A=A[0]$. Also $A\otimes_R A=A$, and for arbitrary $A$-modules the restricted external functor $M\otimes_R N$ is naturally $M\otimes_A N$: an $R$-balanced map is $A$-balanced because the actions factor through $R\to A$, and the converse is immediate. This functor is exact in each input, so its derived functor within these vector-space categories is the ordinary tensor, again $A[0]$ on $(A,A)$. This follows directly by using the intrinsically projective vector spaces in degree zero.

The ambient derived tensor in $D^-(\mathbb Z)$ has a different value. Resolve the first underlying group by $\mathbb Z\xrightarrow2\mathbb Z$ in degrees $-1,0$ and tensor with the second $A$. The result is $A\xrightarrow0A$, giving $H^{-1}(A\otimes_{\mathbb Z}^{L}A)=A,\qquad H^0(A\otimes_{\mathbb Z}^{L}A)=A. \tag{8.7}$ The resolution's free integer terms cannot be made $A$-modules with their usual integer action: an $A$-module is killed by two. Thus this ambient resolution is not a base-flat replacement inside $\mathfrak M$. The calculation compares underlying ambient integer complexes; it does not construct a general tensor-valued functor on arbitrary nonflat algebra charts.

Even finite weak global dimension of the base alone cannot give such covers. Here that dimension is one: every abelian group has a free surjection, whose kernel is torsion free and hence flat by Right derived tensor and torsion-free quotients, Theorem 1.2. This gives flat dimension at most one for every group. Calculation (8.7), or tensoring the injection by $A$, shows that $A$ is not flat, so the bound is attained. The obstruction is the algebra action, and Section 6's algebra-flatness hypothesis addresses it. $\square$

9. Further reading

The Stacks Project, extension by zero of a regular module is flat and bounded-above flat complexes, gives the ringed-space counterparts of the flat generators and tensor exactness used here.

Andrea D'Agnolo and Pietro Polesello, Stacks of twisted modules and integral transforms, Section 2.6, discusses derived operations for locally module stacks. The statements above distinguish balanced flatness from base-ring flatness and give explicit hypotheses for replacements retaining the algebra actions.