Draft lesson.

Right derived tensor and torsion-free quotients

Written by GPT-6.1 Sol (OpenAI), October 2026. Original text: CC0.

Tensor is usually left derived, because it preserves cokernels. Nevertheless an additive tensor functor can also have a universal right derived functor. Over a principal ideal domain that right derived functor has a particularly concrete description: remove torsion from the fixed module, then tensor by the resulting flat module. Its unit is a quotient map, which can fail to be an isomorphism even in degree zero.

Conventions and prerequisites

Let \(R\) be a nonzero commutative principal ideal domain, and let \(\mathcal A=\operatorname{Mod}(R)\). All modules, character sets and bases lie in a fixed universe. We use choice, including Zorn's lemma and choices of injective embeddings indexed by the integers. Modules need not be finitely generated. A field is allowed as a PID. Put \(S=R\setminus\{0\}\) and let \(K\) be the fraction field.

Complexes are cochain complexes. Translation has \(X[1]^n=X^{n+1}\) and differential \(-d_X\). For a chain map \(f:X\to Y\), the target-first cone is

Write \(*=\varnothing,+,-,b\) for unbounded, bounded-below, bounded-above and bounded complexes, respectively. The localization \(Q:K^*(\mathcal A)\to D^*(\mathcal A)\) inverts quasi-isomorphisms. Every bounded assertion below uses the same construction within the indicated variant.

We use the following complete proof interfaces:

• Hom, tensor and the exactness of module limits, Sections 1, 3 and 4: balanced tensor, its right exactness, and the representative and equality tests for filtered colimits.
• Hereditary categories and cohomology splittings, Section 2: the full submodule theorem for free modules over a PID.
• Countable Mittag–Leffler systems, Section 4: injective characters and the coinduced action, with all size conventions.
• Relative replacements and finite resolution bounds, Lemma 2.2: outgoing replacement density, including the full equality-of-roofs proof.
• Localizing triangulated functors and signed bifunctors, Sections 1–3: absolute right localization and its exact adapted-family construction.
• Universal derived functors and compatible composition, Sections 1–4 and 8: the unit convention, the universal natural-transformation property, tensor rows with a flat quotient, and the left-derived flat-class computation.
• Cohomological bounds and truncation adjunctions, Sections 1–3: the four homotopy localizations, their denominator calculus and the full bounded embeddings.

The module arguments and the finite-width replacement are proved below. In particular, the unbounded assertion does not follow merely from the existence of bounded-below injective resolutions.

1. The torsion-free quotient is flat for all modules

For an \(R\)-module \(M\), define

If \(rx=0\) and \(sy=0\), then \(rs(x+y)=0\). A scalar multiple of a torsion element is torsion. Thus \(T_M\) is a submodule. If \(r(m+T_M)=0\), then \(rm\in T_M\), so \(srm=0\) for some \(s\in S\). The domain hypothesis makes \(sr\ne0\); hence \(m\in T_M\). Therefore \(P_M\) is torsion free.

Every linear map \(\alpha:M\to M'\) sends \(T_M\) into \(T_{M'}\). It induces the uniquely determined map

Uniqueness shows that these maps preserve identities, compositions and sums. No splitting of (1.1) is chosen.

Recall that a module \(P\) is flat when \(P\otimes_R-\) preserves monomorphisms. Together with the right exactness of tensor this means that it preserves every short exact sequence.

Lemma 1.1. A finitely generated torsion-free \(R\)-module is finite free.

Proof. For any module \(N\), the localization \(S^{-1}N\) consists of fractions \(n/s\), with \(n/s=n'/s'\) exactly when some \(u\in S\) kills \(s'n-sn'\). Its map from \(N\) has kernel \(T_N\). Addition and scalar multiplication are the usual common-denominator operations, and they give a \(K\)-vector space when \(N\) is localized at all nonzero scalars. The map \(N\otimes_R K\to S^{-1}N\), \(n\otimes(a/s)\mapsto an/s\), has inverse \(n/s\mapsto n\otimes(1/s)\). The fraction equality makes the latter well defined because \(u\) is invertible in the second tensor factor. The balanced equations and the two pure-tensor formulas prove both inverse identities.

Now let \(N\) be torsion free with finitely many generators. It embeds into \(S^{-1}N\), and those generators span a finite-dimensional \(K\)-vector space. Choose a basis of that vector space. In its finitely many coordinates, the given generators have finitely many denominators. Their product \(c\in S\) clears every denominator. Multiplying coordinates by \(c\) embeds \(N\) into \(R^r\), where \(r\) is the dimension.

For completeness, every submodule \(L\subset R^r\) is finite free by induction on \(r\). At \(r=0\) it is zero. Project onto the last coordinate. Its image is an ideal \(aR\). If that image is zero, apply induction inside \(R^{r-1}\). Otherwise choose \(v\in L\) whose last coordinate is the nonzero generator \(a\). Subtracting an appropriate multiple of \(v\) from each element gives

The sum is direct since \(ta=0\) implies \(t=0\). Induction proves the assertion. Applying it to the embedded \(N\) finishes the proof. \(\square\)

Theorem 1.2. Every torsion-free \(R\)-module is flat. Consequently \(P_M\otimes_R-\) is exact for every \(M\), without any finite-generation hypothesis.

Proof. Let \(P\) be torsion free. Its finitely generated submodules form a nonempty directed set: zero belongs to it, and \(P_1+P_2\) contains any two stages. Their union is \(P\), since each element generates a finitely generated submodule. Each stage is finite free by Lemma 1.1.

For every module \(N\), the natural map

is an isomorphism. Indeed a compatible family of balanced maps \(P_i\times N\to V\) gives a balanced map on \(P\times N\): to add two elements of \(P\), put them in a common stage. Conversely restriction gives the compatible family. These operations are inverse, so the tensor and colimit universal properties identify (1.4) with the displayed structural map.

Let \(N'\hookrightarrow N\) be monic. Every stage map \(P_i\otimes N'\to P_i\otimes N\) is monic, because a finite free module tensors it to a finite direct sum of the given injection. The filtered colimit is still monic. More explicitly, represent an element of the first colimit at one stage. If its image in the second colimit is zero, the filtered equality test puts that zero equation at a later stage. At that stage the tensor map is injective, so the representative is zero there; its original colimit class is zero. Using (1.4) proves flatness of \(P\). Right exactness supplies the other two positions of every tensor row. Apply this to \(P_M\). \(\square\)

Conversely a flat module over a domain is torsion free: for each \(r\in S\), tensor the monomorphism \(R\xrightarrow{r}R\) to obtain injective multiplication by \(r\). Thus flatness and torsion freeness coincide here. The theorem does not assert that every torsion-free module is free.

2. Divisibility, injectivity and disappearance of torsion

A module \(J\) is divisible when multiplication \(r:J\to J\) is surjective for every \(r\in S\).

Proposition 2.1. Over \(R\), a module is injective if and only if it is divisible. Quotients and finite direct sums of injective modules are injective.

Proof. Suppose \(J\) is injective and \(j\in J\). For \(r\in S\), define a map \(rR\to J\) by \(ra\mapsto aj\). The domain hypothesis makes this well defined. Extend it over \(rR\subset R\). The image \(y\) of one then satisfies \(ry=j\), proving divisibility.

Conversely, suppose \(J\) is divisible. Given a submodule \(A\subset B\) and a map \(g:A\to J\), order all its partial extensions by inclusion of their domains and agreement of their maps. This is a set. A chain has an upper bound on its union, because the maps agree on intersections. Zorn's lemma gives a maximal extension, with domain \(A'\).

If \(x\in B\setminus A'\), the set \(L=\{a\in R:ax\in A'\}\) is an ideal. Write \(L=rR\). If \(L=0\), choose \(y=0\). Otherwise \(r\ne0\), and divisibility supplies \(y\) with \(ry=g(rx)\). Define on \(A'+Rx\)

If \(a+tx=a'+t'x\), then \((t-t')x\in A'\). In the zero-ideal case \(t=t'\), and then \(a=a'\). In the other case write \(t-t'=ur\). We have \(a-a'=-urx\), and \(g(a)-g(a')+(t-t')y=-u\,g(rx)+ur y=0\). Thus (2.1) is well defined. Its addition and scalar equations follow from the formula. It extends \(g\) to a strictly larger domain, a contradiction. Therefore \(A'=B\), proving injectivity.

A quotient of a divisible module is divisible: choose an \(r\)-preimage in the original module and take its quotient class. Finite direct sums are divisible coordinatewise. The equivalence just proved gives the last assertion. \(\square\)

Theorem 2.2. For every injective module \(J\), the actual quotient map induces a natural isomorphism

Proof. Let \(t\in T_M\) and \(j\in J\). Choose \(r\in S\) with \(rt=0\), and choose \(j'\in J\) with \(rj'=j\), using Proposition 2.1. The balancing relation gives

Pure tensors generate \(T_M\otimes_R J\), so that entire tensor is zero. Right exactness applied to \(T_M\to M\to P_M\to0\) now proves (2.2). The isomorphism is the stated quotient map, rather than one constructed by selecting a section of \(M\to P_M\). Its naturality in both \(M\) and \(J\) follows from the commuting square (1.2) and the tensor functor's commuting squares. \(\square\)

We shall also use, for arbitrary \(N\), the stronger exact row

Here injectivity of the first arrow uses flatness of the quotient \(P_M\). It is the complete flat-quotient tensor-row argument of Universal derived functors, Lemma 8.1. To see its mechanism directly, take a free cover \(U_0\to N\), with kernel \(U_1\). Lift a prospective kernel element from \(T_M\otimes N\) to \(T_M\otimes U_0\). Its image in \(M\otimes U_0\) comes from \(M\otimes U_1\). The image of that lift in \(P_M\otimes U_1\) dies in \(P_M\otimes U_0\), so it is zero by flatness of \(P_M\). Right exactness lifts it from \(T_M\otimes U_1\). Its image in \(T_M\otimes U_0\) equals the original lift, because \(U_0\) is free and hence tensors \(T_M\hookrightarrow M\) injectively. The original element therefore vanishes in \(T_M\otimes N\). This proves the first position; right exactness proves the remainder.

3. Enough injectives with the specified module action

Proposition 3.1. Every \(R\)-module embeds into a small injective \(R\)-module.

Proof. Put \(D=\mathbb Q/\mathbb Z\). It is divisible as an abelian group: divide a rational representative by any positive integer. Proposition 2.1 for the PID \(\mathbb Z\) makes it injective.

Characters into \(D\) detect every nonzero element of an abelian group \(V\). If \(v\) has finite order \(m>1\), map its cyclic subgroup to \(D\) by \(v\mapsto1/m+\mathbb Z\). If it has infinite order, use \(v\mapsto1/2+\mathbb Z\). The injectivity of \(D\) extends this character over \(V\); its value on \(v\) remains nonzero.

Set

This is an \(R\)-module: successive actions evaluate at \(sr_1r_2\), and the unit, addition and scalar distributivity hold pointwise. For every \(R\)-module \(N\), evaluation at one gives inverse natural bijections

The second map is \(R\)-linear by (3.1). For an \(R\)-linear \(g\), \(g(n)(s)=g(sn)(1)\), which proves one inverse identity; evaluation at one proves the other. Extending an additive map into \(D\) across an underlying subgroup, then applying (3.2), extends each \(R\)-linear map from a submodule into \(E\). Thus \(E\) is injective.

For \(M\), let \(A_M=\operatorname{Hom}_{\mathbb Z}(M,D)\). The map

is \(R\)-linear by (3.1). If \(m\ne0\), choose a detecting character and evaluate that coordinate at one; hence the map is monic. A product of injectives is injective: extend each coordinate map from a submodule and assemble the resulting tuple. The set of characters, the Hom module and its indexed product are small in the fixed universe. This proves the assertion. \(\square\)

In particular every module has an exact row

with both \(J^0,J^1\) injective: choose the embedding, and use Proposition 2.1 on its cokernel. This is a finite injective coresolution. The next section turns that quotient property into a replacement for an arbitrary complex.

4. An explicit injective replacement for every complex

Theorem 4.1. Every complex \(X\) admits a monic quasi-isomorphism \(s:X\to I\) with all \(I^n\) injective. The construction works for all four bounded variants. If \(X\) is zero outside \([a,b]\), it can be chosen with \(I\) zero outside \([a-1,b+1]\).

Proof. Choose embeddings \(i_n:X^n\hookrightarrow J^n\) with \(J^n\) injective; choose \(J^n=0\) when \(X^n=0\). These are embeddings of individual modules, with no differential prescribed on the family \(J^n\). Form instead the complex

The displayed \(d_C\) squares to zero. We have \(d_Ce_X(x)=(i_{n+1}d_Xx,0)=e_Xd_X(x)\), so \(e_X\) is a chain map. Its first component makes it monic in each degree. Each \(C^n\) is injective by Proposition 2.1. The degree-minus-one map \(h_C^n(a,b)=(0,a)\in J^{n-1}\oplus J^n\) satisfies \(d_Ch_C+h_Cd_C=1_C\); thus \(C\) is contractible.

Let \(D=C/e_X(X)\), with quotient chain map \(q:C\to D\). Every \(D^n\) is a quotient of an injective module and is injective by Proposition 2.1. Define, with exactly the cone convention (0.1),

The cone construction makes \(I\) a complex; one may also check that its first second-differential component is \(d_D^2a+d_Dqc-qd_Cc=0\). The equation \(qe_X=0\), together with the chain-map equation for \(e_X\), proves that \(s\) is a monic chain map. All terms of \(I\) are injective finite direct sums.

The actual quotient \(I/s(X)\) identifies in degree \(n\) with \(D^{n-1}\oplus D^n\), by \((a,c)\mapsto(a,qc)\). The quotient differential and a contraction are

Indeed \(dh(a,b)=(a,-d_Da)\) and \(hd(a,b)=(0,d_Da+b)\); their sum is \((a,b)\). The quotient is therefore contractible. The short exact row of complexes makes \(s\) a quasi-isomorphism.

Only two neighboring \(J\)-modules occur in (4.1), and only \(D^{n-1}\) and \(C^n\) occur in (4.2). For support \([a,b]\), the support of \(C\) and \(D\) is contained in \([a-1,b]\); that of \(I\) is contained in \([a-1,b+1]\). The same term formulas retain a finite lower bound, a finite upper bound, or both when these are present. There is no infinite totalization or degreewise infinite sum, so the argument also applies to an unbounded \(X\). \(\square\)

Choices of embeddings were used only to establish existence of replacements. We do not assert that these choices define an additive or homotopy functor on all complexes. The eventual derived functor has an explicit choice-free formula.

5. The absolute universal right derived tensor

Fix an arbitrary module \(M\). On modules and on homotopy categories set

The natural quotient transformation is \(\theta_M=\pi_M\otimes1:F_M\to G_M\). Exactness of \(G_M\) from Theorem 1.2 implies that it preserves acyclic complexes and quasi-isomorphisms. It therefore descends to a functor

On roofs it tensors both arrows; the denominator remains a quasi-isomorphism. The quotient maps of complexes give a specified unit

Theorem 5.1. The pair \((\mathcal R_M,\eta_M)\) is the absolute universal right derived functor of \(F_M\), in every bounded variant.

Proof. Let \(\mathcal J\) be the full subcategory of injective modules. Its homotopy category \(K^*(\mathcal J)\) is a full triangulated subcategory of \(K^*(\mathcal A)\): shifts and cones have finite direct sums of injective terms, and all chain maps and homotopies between such complexes remain available. Theorem 4.1 supplies outgoing quasi-isomorphisms into it in the same bounded variant.

On each complex \(I\) of injectives, Theorem 2.2 makes

an actual chain isomorphism: every degree map is an isomorphism, and naturality makes it commute with every differential. If \(I\) is acyclic, exactness of \(G_M\) makes the right side acyclic; hence \(F_M(I)\) is acyclic too. Thus the full triangulated injective family is \(F_M\)-injective in the relative sense, with precisely the two required properties: outgoing replacements and vanishing on its ambient acyclic objects.

The full replacement lemma and adapted-family theorem cited above apply. For the bounded-above case the same replacement-density proof uses the bounded-above fraction calculus in the bounds prerequisite; its numerator-target replacement and common equality-witness replacement are exactly the proof of Lemma 2.2. It follows in every case that

where \(S_{\mathcal J}\) consists of the ambient quasi-isomorphisms between class complexes. Equation (5.4) identifies the restriction of (5.2) and (5.3) with the localized \(F_M\) and its computing isomorphism.

Here is also the entire absolute natural-transformation test. Let \(\mathcal B\) be any category in a sufficiently large universe and take arbitrary ordinary functors

We must prove that composition with \(L\eta_M\) is a bijection

Given \(\zeta:LQF_M\to HQ\), define on each injective complex \(I\) \(\mu_{QI}=\zeta_I(L(\eta_M)_I)^{-1}\). These components are natural on \(K^*(\mathcal J)\), by naturality of \(\zeta\) and \(\eta_M\). Naturality extends to inverse denominators, since the two functors in question invert them. It consequently extends to every roof in the left side of (5.5). That equivalence extends this transformation uniquely to \(D^*(\mathcal A)\).

For any replacement \(s:X\to I\), naturality explicitly gives the extended component

This is well typed: its successive objects are \(L\mathcal R_M(QX)\), \(L\mathcal R_M(QI)\), \(LQF_M(I)\), \(H(QI)\), and \(H(QX)\). The two inverses exist because \(Qs\) and \((\eta_M)_I\) are invertible. Equation (5.5), including fullness and faithfulness, proves independence of the choice and naturality on all derived arrows; it is not an assumption of functorial embeddings.

The unit's naturality states \(\mathcal R_M(Qs)(\eta_M)_X=(\eta_M)_I QF_M(s)\). Substitute this in (5.7), then use \(\zeta_I LQF_M(s)=H(Qs)\zeta_X\). The inverse factors cancel and give \(\mu_{QX}L(\eta_M)_X=\zeta_X\).

Conversely this last equation forces \(\mu_{QI}=\zeta_I(L(\eta_M)_I)^{-1}\) on every injective complex. Naturality along \(Qs\) then forces (5.7) for every \(X\). This proves uniqueness and both directions of (5.6). Neither \(L\) nor \(H\) was required to be additive, exact or colimit preserving. The assertion is therefore absolute. \(\square\)

For any chosen replacement \(s:X\to I\), the particular computing isomorphism is

This states which map computes the value, and retains the specified quotient unit. It is stronger than an equality of cohomology groups.

6. Exact structure and naturality in the fixed module

Proposition 6.1. The functor \(\mathcal R_M\) is additive and triangulated. Its translation comparison is the identity under the canonical term identifications, and its unit is translation compatible. All four versions agree under their natural inclusions.

Proof. Termwise tensor by \(P_M\) is additive. It carries \(X[1]\) to \((P_M\otimes X)[1]\) using the same terms and the same negative differentials. Its translation comparison is therefore identity. Distributivity over a finite direct sum gives the actual cone isomorphism

The differential equation follows from (0.1), including the negative source component. This map takes the tensor of the cone inclusion to the usual inclusion and takes the tensor of the positive final projection to the positive final projection. Thus it carries all three arrows of each cone triangle, with the identity translation comparison, to a distinguished cone triangle.

Since \(G_M\) inverts quasi-isomorphisms, the exact functor descends to the derived localization and preserves every distinguished triangle, which is an image cone triangle up to isomorphism. The unit (5.3) commutes with shifts termwise: both sides use \(\pi_M\otimes1\) and the same negative differential. On injective complexes its comparison is (5.4), so the compatible adapted-family theorem and uniqueness of the unit-preserving comparison identify this identity with the canonical derived comparison. Tensor, the unit and (6.1) preserve the four term bounds, and have the same formulas on their overlaps. The full bounded embeddings identify these restrictions. \(\square\)

Every map \(\alpha:M\to M'\) induces the natural transformation

The ordinary tensor transformations and the units form a commuting square, because \(\overline\alpha\pi_M=\pi_{M'}\alpha\). Thus (6.2) is the unit-preserving transformation dictated by universality. It commutes with the identity translations and the actual cone maps (6.1). Identity, composition and addition of \(\alpha\)'s are retained by (1.2). This is naturality for a fixed module varying through module maps; no assertion about simultaneously right deriving a complex in that variable is needed.

7. Cohomology, computing objects and the boundary of the PID hypothesis

Proposition 7.1. For every complex \(X\), the cycle and boundary maps give natural isomorphisms

In particular, for a module \(N\) in degree zero,

The degree-zero unit is precisely the quotient map in (2.4). An ordinary module \(N\) computes this right derived value if and only if \(T_M\otimes_R N=0\).

Proof. Apply the exact functor \(G_M\) to the short exact cycle and boundary rows of \(X\). It takes the kernel of \(d_X^n\) to the kernel of \(G_M(d_X^n)\), and its image to the image. Applied to \(0\to\operatorname{im}d_X^{n-1}\to\ker d_X^n\to H^nX\to0\), it identifies the quotient with \(P_M\otimes H^nX\). This is the specified cycle-class map, so it commutes with chain maps. It remains natural after inverse quasi-isomorphisms and hence after roofs. No bound on \(X\) was used.

For \(N[0]\), (5.2) is the stalk \((P_M\otimes N)[0]\), giving (7.2). Its unit is (5.3). By (2.4), that unit is an isomorphism exactly when its kernel \(T_M\otimes N\) vanishes. \(\square\)

If \(M\) is torsion, the entire right derived functor is zero, including its degree-zero value. If \(M\) is torsion free, it is flat and the unit is identity, so ordinary tensor already computes the functor on every complex. If \(M\) has both parts, (7.2) keeps its flat quotient and discards its torsion. Although all positive groups in (7.2) vanish, this does not make every \(N\) a computing object: the degree-zero unit still matters. The equality \(R^0F_M=F_M\) would require left exactness, which an arbitrary tensor functor does not have.

The PID assumption on flatness cannot be replaced simply by the domain assumption. Here is an explicit test. Let \(k\) be a nonzero field, \(A=k[x,y]\), \(\mathfrak m=(x,y)\), and \(U=(x,y)\subset A\). This is a torsion-free module, since \(A\) is a domain. In \(U\otimes_A U\), the element

is nonzero. To verify this, map the tensor to \((U/\mathfrak mU)\otimes_k(U/\mathfrak mU)\). The classes of \(x,y\) form a basis of \(U/\mathfrak mU\): their linear terms are independent, and all other monomials of \(U\) have degree at least two and lie in \(\mathfrak mU\). The image of \(z\) has coefficients one and minus one on two different basis tensors, so it is nonzero in every characteristic.

Tensor the monomorphism \(U\hookrightarrow A\) with \(U\). After the identification \(A\otimes_A U\simeq U\), the image of (7.3) is \(xy-yx=0\). Thus \(U\) is not flat. The argument proving Theorem 1.2 used principality to make every finitely generated torsion-free submodule free; this example shows why that step is substantive.

8. Four graded exercises with complete solutions

Exercise 1 — A mixed module and its degree-zero unit (basic)

Work over \(\mathbb Z\). Let \(M=\mathbb Z\oplus\mathbb Z/12\oplus\mathbb Q/\mathbb Z\). Determine \(T_M,P_M\), the right derived functor in all four bounds and its unit. Compute the ordinary and right derived tensors at \(N=\mathbb Z/6\). Then compute the ordinary tensor with an arbitrary injective abelian group \(J\).

Solution. The first summand has no nonzero torsion, whereas every element of each of the last two summands has finite order. Hence \(T_M=0\oplus\mathbb Z/12\oplus\mathbb Q/\mathbb Z\) and \(P_M\simeq\mathbb Z\), with \(\pi_M\) the first projection. Theorems 5.1 and 6.1 give \(\mathcal R_M\simeq1_{D^*(\mathbb Z)}\) in every bound. Its unit on any complex is the first projection of the termwise tensor decomposition.

For any group \(V\), the actual balanced maps give \((\mathbb Z/a)\otimes V\simeq V/aV\): send \(\overline r\otimes v\) to \(rv+aV\), and send \(v+aV\) back to \(\overline1\otimes v\). The tensor relation and \(\overline a=0\) prove well-definedness and the two inverse equations. With \(V=\mathbb Z/6\), this gives a second copy of \(\mathbb Z/6\) for the \(\mathbb Z/12\) summand. Symmetry of tensor and the same quotient formula give \((\mathbb Q/\mathbb Z)\otimes(\mathbb Z/6)\simeq(\mathbb Q/\mathbb Z)/6(\mathbb Q/\mathbb Z)=0\), by divisibility. Consequently

The unit is \((u,v)\mapsto u\); its kernel is the second copy. Every nonzero-degree right derived group is zero. Thus \(\mathbb Z/6\) does not compute the value despite positive-degree vanishing.

For injective \(J\), Proposition 2.1 makes it divisible. The \(\mathbb Z/12\) tensor is \(J/12J=0\). Each pure tensor of \(\mathbb Q/\mathbb Z\) with \(J\) vanishes by (2.3), because its first factor is torsion. Thus \(M\otimes J\simeq J\), and this isomorphism is the first projection prescribed by (2.2). It is natural in \(J\).

Exercise 2 — Infinite generation does not prevent exactness (intermediate)

For a PID \(R\) that is not a field, show that its fraction field \(K\) is torsion free, flat and divisible, but is not a free \(R\)-module. Identify the right derived functors for the fixed modules \(K\) and \(K/R\), including their units and their values on \(R[0]\).

Solution. Multiplication by each \(r\in S\) is invertible in \(K\). This proves both torsion freeness and divisibility. Theorem 1.2 makes \(K\) flat, and Proposition 2.1 makes it injective. It is not free: choose a nonzero nonunit \(a\in R\), which exists because \(R\) is not a field. A nonzero free module is not \(a\)-divisible, since any basis vector would require its coefficient one to lie in \(aR\). A hypothetical empty basis would make \(K=0\), contrary to the nonzero field. Thus no free basis exists. This explains why finite freeness of the stages in Theorem 1.2 does not imply freeness of their union.

For \(M=K\), torsion is zero. Hence \(\mathcal R_K(QX)=Q(K\otimes_R X)\), with identity unit. The fraction maps in Lemma 1.1 identify it termwise with \(S^{-1}X\); this identification commutes with every differential and chain map, so it also identifies the localized derived functor. In particular \(\mathcal R_K(R[0])=K[0]\).

For \(M=K/R\), every class \(a/b+R\) is killed by the nonzero denominator \(b\). Thus it is a torsion module and \(P_M=0\). Its right derived functor is zero in all four bounds, with the zero unit. At \(R[0]\), the ordinary tensor is the nonzero module \(K/R\): the class \(1/a+R\) is nonzero because \(a\) is a nonunit. The degree-zero unit \((K/R)[0]\to0\) is therefore not an isomorphism. Each injective computing tensor is zero by Theorem 2.2.

Exercise 3 — The unit on an unbounded complex (advanced)

Let \(M=\mathbb Z\oplus\mathbb Z/3\). Consider the doubly infinite complex \(X^n=\mathbb Z\) for all integers \(n\), with \(d_X^n=3\) for even \(n\) and \(d_X^n=0\) for odd \(n\). Compute \(H^nX\), the cohomology of the ordinary tensor complex \(M\otimes X\), and the map on cohomology induced by its right-derived unit. Explain why the unit is not a quasi-isomorphism, and why the unbounded replacement theorem still applies.

Solution. Each consecutive differential composite is zero. At an even degree the outgoing multiplication by three is injective, so the cycle group is zero and \(H^{2r}X=0\). At an odd degree all integers are cycles, and the incoming image is \(3\mathbb Z\), so \(H^{2r+1}X=\mathbb Z/3\), for every integer \(r\).

Here \(T_M=\mathbb Z/3\) and \(P_M=\mathbb Z\). Termwise distributivity gives \(M\otimes X=X\oplus(X/3X)\). The latter summand has \(\mathbb Z/3\) in every degree and zero differential: both three and zero become zero modulo three. Consequently

The unit is the projection \(X\oplus(X/3X)\to X\). In even degrees its cohomology map is \(\mathbb Z/3\to0\); in odd degrees it is the first projection onto \(\mathbb Z/3\). It fails to be injective in every degree, so it is not a quasi-isomorphism. Theorem 7.1 agrees with the derived cohomology just computed.

For this unbounded \(X\), choose an injective embedding of every \(\mathbb Z=X^n\), for instance into \(\mathbb Q\). The maps in (4.1) form \(C\) without assuming that the chosen \(\mathbb Q\)'s were already a complex. The quotient \(D\) has injective terms; the signed two-column complex \(I\) of (4.2) is injective termwise; and its quotient by \(X\) has the explicit contraction (4.3). Thus \(X\to I\) is a monic quasi-isomorphism even though the degrees run in both directions. The computing isomorphism is (5.8). No truncation or convergence of an infinite totalization is used.

Exercise 4 — Left and right derivations of cyclic tensor (advanced)

Let \(m,n\ge2\) be integers and \(d=\gcd(m,n)\). For the fixed module \(M=\mathbb Z/m\), compare its universal right derived tensor at \((\mathbb Z/n)[0]\) with the usual left derived tensor there. Compute the two possible cohomology groups of the left derived object with explicit cyclic generators. Which of these objects computes the ordinary tensor at this stalk?

Solution. The fixed module \(M\) is torsion, so Theorem 5.1 gives the zero right derived object. Its unit has source \((\mathbb Z/m)\otimes(\mathbb Z/n)\simeq\mathbb Z/d\). Indeed the cyclic quotient formula gives \((\mathbb Z/m)/n(\mathbb Z/m)=\mathbb Z/(m\mathbb Z+n\mathbb Z)\), and Bézout's identity identifies this ideal with \(d\mathbb Z\). Bézout's identity follows from the Euclidean algorithm: each division leaves the same ideal generated by the two successive remainders, and the final nonzero remainder is their gcd. Since \(d\) can be greater than one, this ordinary tensor need not vanish.

For left derivation, take the free resolution \(P=(\mathbb Z\xrightarrow{n}\mathbb Z)\) in degrees \(-1,0\), augmented by reduction modulo \(n\). Multiplication by \(n\) is injective, and its cokernel is \(\mathbb Z/n\), so this is a quasi-isomorphic projective replacement. The complete flat-class computation in Universal derived functors, Section 8, shows that its tensor computes the left derived functor, with the augmentation as its counit. The resulting complex is

Its \(H^0\) is the just computed \(\mathbb Z/d\), with generator the class of one modulo the image of multiplication by \(n\). Its \(H^{-1}\) is the kernel of multiplication by \(n\). Write \(m=dm'\) and \(n=dn'\), with \(\gcd(m',n')=1\). For an integer \(a\), the condition \(m\mid na\) is equivalent to \(m'\mid n'a\), hence to \(m'\mid a\), by Bézout for \(m',n'\). The kernel consists of the classes of multiples of \(m'\); its generator is \(\overline{m'}=\overline{m/d}\) in \(\mathbb Z/m\). The map \(\mathbb Z/d\to H^{-1}\), \(\overline t\mapsto\overline{m't}\), is surjective and has zero kernel, since \(m\mid m't\) is equivalent to \(d\mid t\). Thus both \(H^{-1}\) and \(H^0\) are \(\mathbb Z/d\), and there are no other cohomology groups.

If \(d>1\), the stalk does not compute either derivation: its right unit has nonzero source and zero target, and the left counit from (8.3) to the ordinary tensor stalk has a nonzero kernel cohomology in degree \(-1\). If \(d=1\), the ordinary tensor is zero and (8.3) is acyclic, because multiplication by \(n\) is invertible modulo \(m\) by Bézout. Both the right unit and left counit are then isomorphisms in the derived category. In either case the left and right universal properties have different directions, and the two explicit computations preserve those directions.

9. Freely accessible references

• The Stacks Project, Torsion free modules, tag 0549, especially the PID flatness assertion in Lemma 15.22.11. Section 1 above gives a direct finite-stage proof for arbitrary modules.
• The Stacks Project, Injective abelian groups, tag 01D6. Sections 2–3 give the ideal extension proof over a PID and specify the coinduced module action.
• The Stacks Project, Derived functors on derived categories, tag 05T3. Sections 4–6 provide the explicit replacement, absolute natural-transformation proof, quotient unit and exact structural maps used here.