Draft lesson.

Separable algebra objects and absolute balanced tensor

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

A balanced tensor product is usually defined as a quotient. For a separable algebra it can instead be recovered by splitting an explicit projector. This works even when the ambient category has no general coequalizers. The splitting also explains averaging of module maps, coherence of iterated tensor products, and the relation between algebra-derived and base-derived Hom and tensor. Exactness still has its own hypotheses.

1. Algebra objects and split projectors

Let \(\mathcal C\) be an idempotent-complete monoidal category, with tensor \(\otimes\) and unit \(\mathbf1\). Fix universes for its Hom sets and the module constructions used here. Idempotent completeness means that every \(p:X\to X\) with \(p^2=p\) has a splitting \[ X\xrightarrow{q}Y\xrightarrow{i}X,\qquad iq=p,\quad qi=1_Y. \tag{1.1} \] No completeness or additivity is assumed. Associators and unit constraints are inserted in tensor strings. The equations below use their given pentagon and unit axioms.

An algebra object \(A\) has multiplication \(m:A\otimes A\to A\) and unit \(u:\mathbf1\to A\), with associativity and the two unit laws. A left action \(\lambda:A\otimes N\to N\), or right action \(\rho:M\otimes A\to M\), satisfies the corresponding associativity and unit laws. A bimodule has commuting left and right actions.

Tensor actions and absolute algebra presentations, Sections 2–3 constructs these module categories, the free-module adjunction and split presentations. We will construct a different split presentation, for a right module paired with a left module.

For the ordinary algebra and derived examples take \(\mathcal C=\operatorname{Mod}(R)\), where \(R\) is commutative and \(A\) is a unital central \(R\)-algebra. It may be noncommutative. Tensor without a subscript then means tensor over \(R\). Derived Hom and bounded resolution pairs, Sections 6–7 provides enough projectives, the two module sides, bounded-above projective replacements, and bounded derived tensor. Injective objects in locally module stacks, Section 5, specialized to a one-point space, provides enough injectives over every ring.

2. The two equations for separability

The algebra object \(A\) is separable if \(m\) has a section \(\sigma:A\to A\otimes A\) which is an \(A\)-bimodule map. The outer factors carry the left and right regular actions. Thus \[ m\sigma=1_A,\qquad \sigma m=(m\otimes1)(1\otimes\sigma) =(1\otimes m)(\sigma\otimes1). \tag{2.1} \]

Lemma 2.1. Such sections are equivalent to maps \(e:\mathbf1\to A\otimes A\) satisfying \[ me=u,\qquad (m\otimes1)(1\otimes e)=(1\otimes m)(e\otimes1) \quad\text{as maps }A\to A\otimes A. \tag{2.2} \] The correspondences are \(e=\sigma u\) and \(\sigma=(m\otimes1)(1\otimes e)\).

Proof. If \(\sigma\) is a bimodule map, its left-linearity equation in (2.1), precomposed with \(1_A\otimes u\), gives the first formula for \(\sigma\). Its right-linearity equation, precomposed with \(u\otimes1_A\), gives the second formula in (2.2). Also \(me=m\sigma u=u\).

Conversely take (2.2). The first expression for \(\sigma\) is left-linear by associativity of \(m\). The equal second expression is right-linear by that same associativity. These are exactly the two bimodule equations in (2.1). Associativity and \(me=u\) give \(m\sigma=m(1\otimes me)=m(1\otimes u)=1_A\). Precomposing with \(u\) recovers \(e\), by the unit law. The constructions are inverse. \(\square\)

In \(R\)-modules write \(e=\sum_r a_r\otimes b_r\), a finite tensor sum. Equations (2.2) become \[ \sum_r a_rb_r=1,\qquad \sum_r aa_r\otimes b_r=\sum_r a_r\otimes b_ra \quad(a\in A). \tag{2.3} \] The product order in the second equation matters.

In the general monoidal category, we use the same indexed tensor notation only as a notation for composites: insert a copy of the morphism \(e\), then apply the indicated multiplications or actions to its ordered factors. Two indices mean two ordered copies of \(e\). No expansion into elements, finite sum of maps, or detection of morphisms by maps from \(\mathbf1\) is being assumed there. The computations below are substitutions of (2.2) into these composites. They require no symmetry of \(\mathcal C\).

3. The actual balancing projector

Let \(M\) be a right \(A\)-module and \(N\) a left \(A\)-module. Put \[ \pi_{M,N}=(\rho\otimes\lambda)(1_M\otimes e\otimes1_N): M\otimes N\longrightarrow M\otimes N. \tag{3.1} \] In the composite notation of Section 2, it sends \(m_0\otimes n_0\) to \(\sum_r m_0a_r\otimes b_rn_0\). Subscripts distinguish module inputs from multiplication.

Write \(\alpha=\rho\otimes1_N\) and \(\beta=1_M\otimes\lambda\), with common source \(M\otimes A\otimes N\).

Lemma 3.1. We have \[ \pi^2=\pi,\qquad \pi\alpha=\pi\beta. \tag{3.2} \] A morphism \(f:M\otimes N\to Z\) is balanced, meaning \(f\alpha=f\beta\), if and only if \(f\pi=f\).

Proof. Action associativity gives for two inserted copies of \(e\) \[ \pi^2(m_0,n_0)= \sum_{r,s}m_0a_ra_s\otimes b_sb_rn_0. \] Apply the central equation (2.2) to move the first factor \(a_r\) past the second inserted pair. The expression becomes \(\sum_{r,s}m_0a_s\otimes b_sa_rb_rn_0\). Contract the first inserted pair by \(me=u\); this leaves \(\sum_s m_0a_s\otimes b_sn_0=\pi(m_0,n_0)\). Formally these steps are, in order, the two action-associativity equations, (2.2) tensored with the untouched module factors, and the multiplication-unit equation. They prove equality of the indicated composites in \(\mathcal C\).

For the second equation, \(\pi\alpha\) acts by \(\sum_r m_0aa_r\otimes b_rn_0\), while \(\pi\beta\) acts by \(\sum_r m_0a_r\otimes b_ra n_0\). These are precisely the two sides of the central equation with the module actions applied.

If \(f\) is balanced, move \(a_r\) from the right module to the left module inside \(f\), then contract \(a_rb_r\). This gives \(f\pi(m_0,n_0)=\sum_r f(m_0,a_rb_rn_0)=f(m_0,n_0)\). In morphism notation this is \(f\alpha=f\beta\) precomposed with the insertion of \(e\), followed by \(me=u\). Conversely \(f=f\pi\) and \(\pi\alpha=\pi\beta\) imply \(f\alpha=f\beta\). \(\square\)

4. An absolute coequalizer without general colimits

Split \(\pi=iq\) as in (1.1), and call its splitting object \(M\otimes_A N\).

Theorem 4.1. The fork \[ M\otimes A\otimes N \overset{\alpha,\beta}{\rightrightarrows} M\otimes N\xrightarrow{q}M\otimes_A N \tag{4.1} \] is an absolute coequalizer: every functor out of \(\mathcal C\) preserves it, whether or not that functor is monoidal or preserves other colimits.

Proof. Since \(q\pi=q\), Lemma 3.1 makes \(q\) balanced. A balanced \(f\) has the factor \(fi\), because \(fiq=f\pi=f\). If \(gq=f\), then \(g=gqi=fi\), proving uniqueness.

For absoluteness define the actual extra map \[ t=(\rho\otimes1_A\otimes1_N)(1_M\otimes e\otimes1_N): M\otimes N\longrightarrow M\otimes A\otimes N. \tag{4.2} \] It acts by \(\sum_r m_0a_r\otimes b_r\otimes n_0\). Thus \[ \alpha t=1_{M\otimes N},\qquad \beta t=\pi=iq,\qquad qi=1. \tag{4.3} \] The first equation uses \(me=u\) and the right-action unit; the second is (3.1).

For any functor \(F\), the images of (4.3) remain identities. If \(hF\alpha=hF\beta\), then \[ h=hF\alpha Ft=hF\beta Ft=hFiFq. \] Its unique factor through \(Fq\) is \(hFi\), uniqueness following from \(FqFi=1\). This proves preservation of the whole coequalizer fork directly. \(\square\)

Maps \(a:M\to M'\), \(b:N\to N'\) of the appropriate modules commute with the projectors. Their induced map on chosen splitting objects is \(q'(a\otimes b)i\), and satisfies \[ (a\otimes_A b)q=q'(a\otimes b). \tag{4.4} \] This equation and the epimorphism \(q\) prove functoriality, including identities and composition.

Changing a splitting, or even changing the separability section, gives a unique isomorphism between the balanced objects that commutes with their quotient maps. Both represent the same balanced maps, so their inverse comparisons and three-choice composition law follow by uniqueness. The embedding \(i\) and the projector can change: they are not claimed to be independent of \(e\).

5. Averaging maps and splitting the free-module counit

For a left module \(N\), define \[ s_N=(1_A\otimes\lambda_N)(e\otimes1_N): N\longrightarrow A\otimes N. \tag{5.1} \] The target is free on the underlying object. In composite notation \(s_N(n_0)=\sum_r a_r\otimes b_rn_0\).

Lemma 5.1. This is an \(A\)-linear natural section of the action \(\lambda_N:A\otimes N\to N\).

Proof. The two possible \(A\)-linearity composites are \(\sum_r a_r\otimes b_ra n_0\) and \(\sum_r aa_r\otimes b_rn_0\); (2.2), with the remaining action applied, equates them. Applying \(\lambda_N\) contracts \(e\), giving \(n_0\). A module map commutes with its action, so commutes with the formula for \(s_N\). Thus the section is natural. \(\square\)

For an arbitrary underlying morphism \(f:M\to N\) between left modules, set \[ \operatorname{Av}(f)= \lambda_N(1_A\otimes f)s_M. \tag{5.2} \] Every factor here is a module map, including the free-module map \(1_A\otimes f\), so the average is \(A\)-linear. If \(f\) is already \(A\)-linear, naturality of \(s\) and \(\lambda_Ns_N=1_N\) give \(\operatorname{Av}(f)=f\). Thus (5.2) retracts the inclusion of module morphisms into underlying morphisms. For \(A\)-linear \(a,b\), its formulas give \[ \operatorname{Av}(bfa)=b\operatorname{Av}(f)a. \tag{5.3} \] This is naturality in module maps. It does not assert \(\operatorname{Av}(fg)=\operatorname{Av}(f)\operatorname{Av}(g)\) for arbitrary underlying maps. Exercise 8.2 disproves that assertion.

In \(\operatorname{Mod}(R)\), (5.2) is the \(R\)-linear rule \(\operatorname{Av}(f)(m_0)=\sum_r a_rf(b_rm_0)\). An \(A\)-linear epimorphism with an underlying \(R\)-linear section has an \(A\)-linear section: average the given section and use (5.3) on its composite with the epimorphism. The analogous averaging of a retraction works for a monomorphism. Consequently every \(R\)-split exact row of \(A\)-modules is \(A\)-split.

If \(R=k\) is a field, every underlying short exact row splits by extending a vector-space basis. Hence every \(A\)-module short exact row splits. In particular every module is projective: pull back an epimorphism along the map to be lifted, split that new row, and project its section to the requested lift. Every module is injective too: push out a monomorphism along the map to be extended, split that row by a retraction, and compose to obtain the extension. These assertions concern a separable algebra over a field, not every separable algebra over an arbitrary ring.

For a finite group \(G\), with \(|G|\) invertible in \(R\), the element \[ e_G=\frac1{|G|}\sum_{g\in G}g\otimes g^{-1} \quad\text{in }R[G]\otimes_R R[G] \tag{5.4} \] satisfies (2.3). Its product is one. For a group element \(h\), replace \(g\) by \(hg\) to identify \(\sum_g hg\otimes g^{-1}\) with \(\sum_g g\otimes g^{-1}h\); linearity treats every algebra element. Averaging (5.2) is therefore the familiar group average. Pointwise abelian structure and natural splittings, Exercise 2 proves the underlying-splitting distinction for group actions; the present section derives it from the multiplication section.

6. Exterior actions, associativity and units

Let \(M\) be an \(S\)-\(A\) bimodule and \(N\) an \(A\)-\(B\) bimodule. Their exterior \(S\)- and \(B\)-actions commute with \(\pi_{M,N}\): in (3.1) this is precisely the commutation of the two actions on \(M\) or on \(N\). They induce commuting actions on \(M\otimes_A N\). For example its right \(B\)-action is \[ q(1_M\otimes\rho_N)(i\otimes1_B). \tag{6.1} \] It has the desired compatibility with \(q\). To check its unit and associativity, precompose with \(q\), or with \(q\otimes1_B\otimes1_B\). These are epimorphisms: Theorem 4.1 says tensoring preserves the split coequalizer. The resulting equations are the original action equations, so cancellation proves them on the quotient. The same argument gives the commuting exterior actions and their functoriality.

Suppose \(A,B\) are separable, \(M\) is right \(A\), \(N\) is \(A\)-\(B\), and \(P\) is left \(B\). On \(M\otimes N\otimes P\) their balancing projectors commute. In composite notation both orders give \[ \sum_{r,s}m_0a_r\otimes b_rn_0c_s\otimes d_sp_0, \tag{6.2} \] where \(e_A\) and \(e_B\) occupy the indicated ordered positions. Equality is the commuting-action equation on \(N\).

Both iterated products represent maps out of \(M\otimes N\otimes P\) which balance the \(A\)-action and the \(B\)-action. Indeed first factor an \(A\)-balanced map through \(q_{M,N}\otimes1_P\), which is a coequalizer by absoluteness. Its \(B\)-balance descends after precomposition with the epimorphism \(q_{M,N}\otimes1_B\otimes1_P\). Factor it next through the \(B\)-quotient. The reverse argument starts with \(B\). This proves a unique natural isomorphism \[ (M\otimes_A N)\otimes_B P \simeq M\otimes_A(N\otimes_B P) \tag{6.3} \] commuting with the original tensor quotient maps.

For four factors, each composite of these associators agrees after the common iterated quotient map from the fourfold tensor product: both preserve that quotient map. That map is epic, by repeated preservation of the split forks under tensor. Cancellation proves the pentagon. Similarly the actions identify \(A\otimes_A N\simeq N\) and \(M\otimes_A A\simeq M\). For the first, the inverse to the action-induced quotient map is \(n_0\mapsto q(1\otimes n_0)\). Its composite in one direction is the action unit. In the other direction balance gives \(q(a\otimes n_0)=q(1\otimes an_0)\). These are the unit-insertion and balancing equations as composites. The right unit is identical with the right-action order retained. The triangle with (6.3) follows by cancellation of the same epic tensor quotient. Thus all comparisons are coherent maps, including when splitting objects were chosen differently.

7. Bounded Hom and tensor over a flat separable algebra

We now work with a separable central \(R\)-algebra \(A\) in ordinary modules. First averaging respects bounded complex differentials. For a family \(f^i:X^i\to Y^{i+n}\), every differential of \(X,Y\) is \(A\)-linear. Equation (5.3) therefore gives \[ \operatorname{Av}(d_Yf-(-1)^nfd_X) =d_Y\operatorname{Av}(f)-(-1)^n\operatorname{Av}(f)d_X. \tag{7.1} \] It retracts the inclusion of the \(A\)-linear Hom complex into the \(R\)-linear one. In particular an \(R\)-linear chain homotopy between \(A\)-linear chain maps averages to an \(A\)-linear homotopy. An underlying contraction averages to a module contraction. The forgetful functor is consequently faithful on the bounded homotopy category. These conclusions require no flatness assumption.

For the two derived comparisons below assume, in addition, that \(A\) is flat as an \(R\)-module. The general finite-diagonal flat total argument is proved in Flat covers and bounded tensor for locally module stacks, Sections 4–6. Its one-point specialization applies here.

Theorem 7.1. Forgetful maps on bounded derived Hom groups have natural \(R\)-linear retractions. For \(X\) bounded above and \(Y\) bounded below, this includes \[ \operatorname{Hom}_{D(A)}(X,Y[n]) \longrightarrow \operatorname{Hom}_{D(R)}(UX,UY[n]) \tag{7.2} \] for every integer \(n\). In particular \(U:D^b(A)\to D^b(R)\) is faithful.

Proof. The left adjoint of \(U\) is \(A\otimes_R-\), by the action/underlying-map adjunction, and it is exact by the base-flat hypothesis. Hence \(U\) preserves injectives: transpose an extension problem through that exact left adjoint, extend into the \(A\)-injective object, and transpose back. Choose a bounded-below \(A\)-injective replacement \(Y\to I\). Since \(U\) is exact, \(UY\to UI\) is an \(R\)-injective replacement.

Both derived Hom values are computed by their respective Hom complexes into \(I\) and \(UI\), by the bounded Hom prerequisite. Averaging (7.1) gives an actual chain retraction of their inclusion. Its cohomology in degree \(n\) is the retraction in (7.2). Comparison maps between \(A\)-injective resolutions are \(A\)-linear and unique up to \(A\)-linear homotopy. Equation (5.3) makes the averaging maps commute with these comparisons and with precomposition by module chain maps. Source denominators and resolution comparisons are inverted on both sides, so the retractions are natural for all derived arrows and independent of models. \(\square\)

Theorem 7.2. For \(M\in D^-(A^{\mathrm{op}})\) and \(N\in D^-(A)\), there are natural maps \[ M\otimes_R^L N\xrightarrow{q^L}M\otimes_A^L N \xrightarrow{i^L}M\otimes_R^L N, \qquad q^Li^L=1. \tag{7.3} \] Their composite \(i^Lq^L\) is represented by the separability projector. Consequently \(\operatorname{Tor}_j^A(M,N)\), for modules in degree zero, is naturally a direct summand of \(\operatorname{Tor}_j^R(UM,UN)\).

Proof. Choose bounded-above right and left \(A\)-projective models \(P\to M,Q\to N\). A projective \(A\)-module is a retract of a free one. The underlying free module is a direct sum of copies of the base-flat \(A\), and is \(R\)-flat; its retract is \(R\)-flat too. Thus \(T=\operatorname{Tot}(P\otimes_R Q)\) computes the base-derived tensor. The total has finite diagonals.

In each summand \(P^r\otimes_R Q^s\), use (3.1). It commutes with both differentials since they are \(A\)-linear. With the total sign \((-1)^r\) on the second differential it is still a chain projector \(\pi:T\to T\). In \(R\)-modules its image and kernel give a direct-sum decomposition of complexes: for any element \(z\), the decomposition is \(\pi z+(z-\pi z)\), and both summands are preserved by the differential. Its image complex is canonically \(\operatorname{Tot}(P\otimes_A Q)\), by the balanced quotient and splitting from Sections 3–4. That complex computes the \(A\)-derived tensor by the two-projective replacement theorem.

The resulting chain quotient and inclusion give (7.3). All maps commute with right and left module chain maps. Common projective refinements compute every derived roof and its equality tests; the corresponding base-flat comparisons are quasi-isomorphisms. Inverting those comparisons therefore gives naturality and model independence of these maps. Taking cohomology preserves their retraction equation. This proves the Tor assertion, with \(\operatorname{Tor}_j=H^{-j}\). \(\square\)

The base-flat hypothesis justifies these particular derived models; separability alone supplies the ordinary absolute fork. Even when that hypothesis holds, (7.3) is a retraction, not a general vanishing theorem for Tor.

8. Four graded exercises with complete solutions

Exercise 8.1 — Integral matrix units and two different projectors (intermediate)

For any commutative ring \(R\), integer \(d\ge1\), and \(A=\operatorname{Mat}_d(R)\), prove separability without assuming \(d\) invertible. Compute the balanced tensor of the row module \(R^{1\times d}\) and column module \(R^{d\times1}\). Compare the projectors obtained from two chosen matrix corners.

Solution. For a fixed corner \(j\), put \[ e_j=\sum_{r=1}^d E_{rj}\otimes E_{jr}. \tag{8.1} \] Its product is \(\sum_rE_{rr}=1\). For a matrix unit \(E_{ab}\), multiplying the first factor on the left or the second on the right gives in both cases \(E_{aj}\otimes E_{jb}\). Every matrix is a linear combination of matrix units, so (2.3) holds. No division by \(d\) occurs.

For a row \(v\) and column \(w\), the projector is \[ \pi_j(v\otimes w)= \left(\sum_r v_rw_r\right)\mathbf r_j\otimes\mathbf c_j. \tag{8.2} \] Indeed \(vE_{rj}=v_r\mathbf r_j\) and \(E_{jr}w=w_r\mathbf c_j\). Dot product is balanced: \((vT)w=v(Tw)\) by associative matrix multiplication. It has section \(z\mapsto z\mathbf r_j\otimes\mathbf c_j\), and their composite is (8.2). Thus the balanced tensor is \(R\). The extra map \(t(v,w)=\sum_r vE_{rj}\otimes E_{jr}\otimes w\) satisfies \(\alpha t(v,w)=v\otimes w\) and \(\beta t=\pi_j\), as in (4.3).

For \(d\ge2\) and a nonzero \(R\), \(\pi_1\ne\pi_2\): on \(\mathbf r_1\otimes\mathbf c_1\) their values are the two distinct corner basis tensors. In fact \(\pi_1\pi_2=\pi_1\) and \(\pi_2\pi_1=\pi_2\), since either projector retains the same dot-product scalar. The two splitting images are nevertheless canonically identified by that scalar, and both quotient maps are the same balanced dot product. Equality of balanced objects does not require equality of their projectors.

Exercise 8.2 — Averaging is not a functor on arbitrary maps (intermediate)

Let \(k\) have characteristic different from two. On \(V=k^2\), the generator \(g\) of \(C_2\) acts by \(\operatorname{diag}(1,-1)\). Compute the average of the coordinate swap \(f\), and compare averaging \(f^2\) with squaring its average.

Solution. The separability element is \(\tfrac12(1\otimes1+g\otimes g)\), since \(g^{-1}=g\). Consequently \(\operatorname{Av}(h)=\tfrac12(h+ghg)\). For \(f=\begin{pmatrix}0&1\\1&0\end{pmatrix}\), we have \(gfg=-f\). Thus \(\operatorname{Av}(f)=0\), whereas \(f^2=1_V\) and \(\operatorname{Av}(f^2)=1_V\). The two results differ. Equivariant endomorphisms are exactly diagonal matrices: commuting with \(\operatorname{diag}(1,-1)\) kills both off-diagonal entries because two is invertible. Averaging is the projection to those diagonal entries. Its naturality with multiplication by an equivariant map is precisely (5.3); the swap is not such a map.

Exercise 8.3 — A field can be semisimple without being separable (advanced)

Let \(p\) be prime, \(L=\mathbb F_p(u)\), and \(k=\mathbb F_p(u^p)\). Show that \(L\) has dimension \(p\) over \(k\), that every short exact sequence of \(L\)-modules splits, and that \(L\) is not a separable \(k\)-algebra.

Solution. The elements \(1,u,\ldots,u^{p-1}\) span \(L\) over \(k\). Polynomial powers reduce by their exponent modulo \(p\). For a nonzero polynomial \(h(u)\), its \(p\)-th power is a polynomial in \(u^p\), so \(h(u)^{-1}=h(u)^{p-1}/h(u)^p\) also lies in this span. This handles every rational function. They are independent: clear denominators in a relation with coefficients in \(k\). Its polynomial summands have disjoint exponent classes modulo \(p\), forcing each coefficient polynomial to vanish. Thus the dimension is \(p\).

Every \(L\)-module is a vector space over the field \(L\). Extending a basis of a subspace gives an \(L\)-linear complement, so all short exact rows split.

In \(L\otimes_kL\), put \(z=1\otimes u-u\otimes1\). In characteristic \(p\), \(z^p=1\otimes u^p-u^p\otimes1=0\). The map \[ L[z]/(z^p)\longrightarrow L\otimes_kL \tag{8.3} \] is an isomorphism of left \(L\)-algebras: the powers \(z^i\), for \(0\le i<p\), form a triangular change of basis with diagonal one from \(1\otimes u^i\). Multiplication to \(L\) sends \(z\) to zero.

If a separability element \(e\) existed, its central equation for \(u\) would give \(ze=0\). In the truncated polynomial algebra the annihilator of \(z\) is \(Lz^{p-1}\): compare the coefficients of \(z,z^2,\ldots,z^{p-1}\) after multiplying by \(z\). Since \(p-1\ge1\), multiplication to \(L\) kills that annihilator. Thus \(me=0\), contradicting \(me=1\). The failure concerns the relative \(k\)-algebra pairing, despite the complete splitting of rows of \(L\)-vector spaces.

Exercise 8.4 — An absolute tensor still has nonzero derived Tor (advanced)

Let \(A=\operatorname{Mat}_2(\mathbb Z)\), \(e=E_{11}\), and \(n\ge2\). Put \(M=eA/neA\), a right module, and \(N=Ae/nAe\), a left module. Compute their ordinary and derived balanced tensor. Explain why this does not contradict separability.

Solution. Exercise 8.1 makes \(A\) separable over \(\mathbb Z\), and \(A\) is base-free of rank four. The underlying modules of \(M,N\) are respectively the row and column over \(\mathbb Z/n\). Dot product and the first-corner section show \(M\otimes_A N\simeq\mathbb Z/n\), with projector the reduction of (8.2). Their ordinary base tensor has four copies of \(\mathbb Z/n\), and this projector selects the diagonal sum into its first corner. Its split coequalizer is absolute.

As a left \(A\)-module \(Ae\) is projective, since \(A=Ae\oplus A(1-e)\). The sequence \[ 0\longrightarrow Ae\xrightarrow{\,n\,}Ae\longrightarrow N\longrightarrow0 \tag{8.4} \] is exact: \(Ae\) is a free abelian column module, so multiplication by \(n\) is injective. It is a projective resolution of length one.

For any right module \(T\), multiplication gives \(T\otimes_A Ae\simeq Te\), with inverse \(te\mapsto te\otimes e\). To check the second composite, write \(ae\) in a pure tensor and balance \(t\otimes ae=tae\otimes e\). For \(T=M\), the group \(Me\) is \(\mathbb Z/n\). Tensoring (8.4) therefore gives the two-term complex \(\mathbb Z/n\xrightarrow{0}\mathbb Z/n\) in degrees minus one and zero. Hence \[ \operatorname{Tor}_0^A(M,N)=\mathbb Z/n,\qquad \operatorname{Tor}_1^A(M,N)=\mathbb Z/n,\qquad \operatorname{Tor}_j^A(M,N)=0\quad(j\ge2). \tag{8.5} \] The base Tor groups in degrees zero and one are \((\mathbb Z/n)^4\): use the analogous multiplication-by-\(n\) resolution of the abelian row or column, and its tensor differential is zero. The same first-corner projector retracts each to the displayed \(\mathbb Z/n\), illustrating (7.3).

Neither \(M\) nor \(N\) is base-flat. The row (8.4) does not even split over \(\mathbb Z\): any map from its torsion quotient to the free abelian \(Ae\) is zero and cannot be a section. The relative splitting theorem in Section 5 applies to base-split rows, so does not apply to this one. Preservation of the particular balanced coequalizer by every functor is entirely compatible with this nonzero Tor.

Freely accessible references

• Paul Balmer, Separability and triangulated categories, author manuscript, Section 3, for separable ring objects and natural retractions on morphisms.
• Emily Riehl, Category Theory in Context, author-hosted edition, Sections 5.4–5.6, for split coequalizers and monadic presentations.