Draft lesson.

Faithfully flat scalar descent and Amitsur resolutions

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

Extending scalars can make a module easier to describe. To recover the original module, one must also remember how its two scalar extensions agree. We express that agreement as a coaction, recover the original object by an equalizer, and check both inverse comparisons. The same argument works in a stack with local noncommutative module descriptions. An explicit contraction then proves exactness of the augmented Amitsur complex.

1. Scalar extension on a locally module stack

Let $X$ be a topological space and $\mathcal R\to\mathcal S$ a map of sheaves of commutative unital rings. Assume that $\mathcal S_x$ is faithfully flat over $\mathcal R_x$ for every point $x$. Here flatness means that tensoring is exact, and faithfulness means that tensoring sends a nonzero module map to a nonzero map. Fix universes in which the covers, Hom sets and diagrams used below are small.

Let $\mathfrak M$ be an $\mathcal R$-linear stack locally equivalent to $\operatorname{Mod}(\mathcal A_i)$ on an open cover, where each unital algebra $\mathcal A_i$ has a central $\mathcal R$-action. The algebras may be noncommutative. Morphisms of the stack need not be invertible.

Write $\mathfrak M_{\mathcal S}$ for objects of $\mathfrak M$ with a unital associative $\mathcal S$-action. Put $\mathcal C=\mathfrak M(X),\qquad \mathcal D=\mathfrak M_{\mathcal S}(X),\qquad F(K)=\mathcal S\otimes_{\mathcal R}K,\qquad U(N)=N\text{ with its scalar action forgotten}. \tag{1.1}$

The global abelian construction proves that these section categories are abelian and that restriction to opens is exact. The tensor/action construction, through its adjunction formulas, applies to the present scalar map: those formulas use centrality, the unit and associativity of the action.

Here are the constructions involved. In an $\mathcal A_i$-description, tensoring an object with an ordinary $\mathcal R$-module sheaf is characterized by tensor–Hom currying. Its universal identification commutes with restriction. Local representability with its specified identification glues these objects and maps. It also glues the tensor unit and associator, since their coherence equations hold locally. Action maps descend with their underlying objects; the action equations can be checked on a cover. Thus $\mathfrak M_{\mathcal S}$ is a stack, with local algebra $\mathcal A_i\otimes_{\mathcal R}\mathcal S$.

The two inverse adjunction rules are $h:FK\to N\longmapsto U(h)\eta_K,\qquad f:K\to UN\longmapsto\bigl(s\otimes k\mapsto s f(k)\bigr), \quad \eta_K(k)=1\otimes k. \tag{1.2}$ The counit $\varepsilon_N:FUN\to N$ is the action, $s\otimes n\mapsto sn$. The rules are inverse by the unit and $\mathcal S$-linearity equations. They are natural and commute with restriction, so give a stack adjunction $F\dashv U$, including both triangle identities.

Lemma 1.1. The functors $F,U$ are exact, and $F$ is faithful. An exact faithful additive functor between abelian categories detects zero objects, isomorphisms and exact sequences.

Proof. In each chart, forgetting the scalar action leaves kernels and cokernels on the same underlying sheaves. Hence $U$ is exact. Tensor sheaves have stalks $(\mathcal S\otimes M)_x=\mathcal S_x\otimes M_x$: a germ tensor has finite representatives on a common neighborhood, and its finite relations hold on a smaller common neighborhood. Flatness and the stalk test for sheaf exactness therefore prove exactness of $F$. If $Ff=0$, faithful flatness gives $f_x=0$ in every chart and at every point. A sheaf map with all zero stalk maps is zero: each image section has zero germs and hence vanishes on a cover. Descent for morphisms gives $f=0$.

For the last assertion, if $FZ=0$, then $F(1_Z)=0$, so faithfulness gives $1_Z=0$ and $Z=0$. Exactness identifies images of kernels and cokernels with the corresponding kernels and cokernels of the image map. Consequently it detects isomorphisms. For a complex, exactness identifies $F(H^nK)$ with $H^n(FK)$; detection of zero objects then detects exactness at each degree. $\square$

2. The coaction and its split equalizer

For the moment take any additive adjunction $F:\mathcal C\rightleftarrows\mathcal D:U$ between abelian categories, with $F$ exact and faithful. Its comonad on $\mathcal D$ is $G=FU,\qquad \varepsilon:G\to1,\qquad \delta_N=F(\eta_{UN}):GN\to G^2N. \tag{2.1}$ The triangle identities give $\varepsilon_{GN}\delta_N=1_{GN}$ and $G\varepsilon_N\delta_N=1_{GN}$. Naturality of $\eta$, applied to $\eta_{UN}$, gives $\delta_{GN}\delta_N=G\delta_N\delta_N$. These are the two counit laws and coassociativity.

A $G$-coalgebra is an object $N\in\mathcal D$ and a map $\rho:N\to GN$ satisfying $\varepsilon_N\rho=1_N,\qquad \delta_N\rho=G\rho\,\rho. \tag{2.2}$ A coalgebra map $a:(N,\rho)\to(N',\rho')$ obeys $Ga\,\rho=\rho'a$. Such maps compose and add because $G$ is additive. The canonical coalgebra on $FK$ has coaction $F\eta_K$. Its laws follow from a triangle identity and naturality of $\eta$.

Lemma 2.1. In $\mathcal D$, the map $\rho:N\to GN$ is the equalizer of $\delta_N,G\rho:GN\to G^2N$.

Proof. Equation (2.2) says that $\rho$ equalizes the pair. If $v:Y\to GN$ also equalizes it, then naturality of the counit and its first law give $\rho\varepsilon_Nv =\varepsilon_{GN}G\rho\,v =\varepsilon_{GN}\delta_Nv=v. \tag{2.3}$ Thus the factor is $\varepsilon_Nv$. It is unique because $\varepsilon_N\rho=1_N$. This proves the full equalizer property. $\square$

In scalar notation, $GN=\mathcal S\otimes_{\mathcal R}UN, \quad \delta(s\otimes n)=s\otimes1\otimes n, \quad G\rho=1_{\mathcal S}\otimes\rho. \tag{2.4}$ The $\mathcal S$-action on $GN$ is on its first factor. In particular $\rho$ must be linear for that action.

3. Recovering the invariant object, with both comparisons

Define $M=M(N,\rho)$ in $\mathcal C$ by the equalizer $i:M\longrightarrow UN, \qquad M=\ker\bigl(U\rho-\eta_{UN}:UN\to UGN\bigr). \tag{3.1}$ For a coalgebra map $a$, naturality of $\eta$ and its coaction equation show that $Ua\,i$ factors uniquely through $i':M(N',\rho')\to UN'$. Uniqueness gives identity and composition laws, so $M$ is a functor.

Theorem 3.1. The comparison $\mathcal C\longrightarrow\operatorname{Coalg}(G), \qquad K\longmapsto(FK,F\eta_K) \tag{3.2}$ is an equivalence, with inverse (3.1). Its specified natural comparisons are $\theta_N=\varepsilon_NF i:FM(N,\rho)\longrightarrow N, \qquad h_K:K\longrightarrow M(FK,F\eta_K),\quad i h_K=\eta_K. \tag{3.3}$

Proof. Exact $F$ preserves the equalizer (3.1). Thus $Fi:FM\to GN$ equalizes $G\rho$ and $\delta_N$, and has their equalizer universal property. Lemma 2.1 makes $\theta_N$ an isomorphism with $Fi=\rho\theta_N$. More explicitly, let $v:N\to FM$ be the unique factor of $\rho$ through $Fi$. Then $\theta_Nv=\varepsilon_N\rho=1$, while $Fi\,v\theta_N=\rho\theta_N=Fi$; monicity of $Fi$ gives $v\theta_N=1$.

It is an isomorphism of coalgebras. Indeed naturality of $\eta$ at $i$, followed by the other triangle identity, gives $G\theta_N\,F\eta_M =F\bigl(U\varepsilon_N\,UF i\,\eta_M\bigr) =F\bigl(U\varepsilon_N\,\eta_{UN}i\bigr) =Fi=\rho\theta_N. \tag{3.4}$ The inverse of an underlying isomorphism that respects coactions respects them too: multiply its coaction equation by the inverse and apply $G$ to that inverse.

For the canonical coalgebra on $FK$, naturality of $\eta$ gives $UF\eta_K\,\eta_K=\eta_{UFK}\eta_K$, so $h_K$ in (3.3) exists. Applying $F$ and the counit yields $\theta_{FK}\,Fh_K=\varepsilon_{FK}F\eta_K=1_{FK}. \tag{3.5}$ Since $\theta_{FK}$ is invertible, $Fh_K$ is invertible. Lemma 1.1 makes $h_K$ invertible.

Finally, for a coalgebra map $a$, the defining equation for $M(a)$ and counit naturality give $a\theta_N=\theta_{N'}F M(a)$. For $f:K\to K'$, the two maps $K\to UFK'$ obtained from $h$ have the same value $\eta_{K'}f=UFf\,\eta_K$; the equalizer inclusion makes their factors equal. Thus both comparisons are natural. These are inverse equivalence comparisons, on objects and on all morphisms. $\square$

Applying this on every open subset gives an equivalence of stacks $\mathfrak M\simeq\operatorname{Coalg}(FU). \tag{3.6}$ Coalgebra objects descend with their underlying objects; their coaction maps and equations descend because morphisms form sheaves. Equalizers, the adjunction maps and (3.3) commute with restriction, so the equivalence comparisons do as well. No global regular module was chosen.

In an $\mathcal A_i$-chart the answer is especially concrete: $M=\{n:\rho(n)=1\otimes n\},\qquad \theta(s\otimes m)=sm. \tag{3.7}$ This is a kernel sheaf. Membership means equality locally on the open set in question; it imposes no bound on the number of generators. The $\mathcal A_i$-action preserves this kernel because both maps in (3.1) respect it.

4. From a coaction to agreement on overlaps

We check the usual scalar descent datum at a stalk of a module chart. The maps below are intrinsic composites of tensoring, scalar permutations and actions. Stalks detect their identities for sheaves, and their restriction compatibility glues them in the stack. Temporarily write $R,S,A$ for the stalk rings and $N$ for a left $A\otimes_R S$-module. If $\rho(n)=\sum_i s_i\otimes n_i$, define $\varphi:N\otimes_R S\longrightarrow S\otimes_R N, \quad \varphi(n\otimes s)=\sum_i s_i\otimes s n_i. \tag{4.1}$ The two $S$-factors act respectively on $N$ and on $S$ in the domain, and on $S$ and on $N$ in the target. Linearity of $\rho$ makes $\varphi$ linear for these actions. The $A$-action stays on the $N$-factor.

Proposition 4.1. Coalgebra data are equivalent to invertible overlap maps $\varphi$ satisfying $\varphi_{12}\varphi_{01}=\varphi_{02}$ on three scalar factors. Coalgebra maps are precisely maps compatible with the overlap maps.

Proof. Coassociativity reads $\sum_{i,j}s_i\otimes s_{ij}\otimes n_{ij} =\sum_i s_i\otimes1\otimes n_i, \qquad \rho(n_i)=\sum_j s_{ij}\otimes n_{ij}. \tag{4.2}$ On $n\otimes s\otimes t$, the composite overlap route is the left side with middle factor multiplied by $s$ and final module factor by $t$. The direct route is the same modification of the right side. This proves the cocycle equation.

The inverse map has the explicit formula $\psi:S\otimes_R N\longrightarrow N\otimes_R S, \qquad \psi(s\otimes n)=\sum_i s n_i\otimes s_i. \tag{4.3}$ For $\psi\varphi(n\otimes s)$, apply the map $a\otimes b\otimes y\mapsto ay\otimes sb$ to (4.2). Its two sides give the composite and $\sum_i s_i n_i\otimes s=n\otimes s$. For $\varphi\psi(s\otimes n)$, apply $a\otimes b\otimes y\mapsto sb\otimes ay$; its right side is $s\otimes\sum_i s_i n_i=s\otimes n$. Both final equalities are the counit law. These maps are balanced because the scalar rings are commutative, so the two tensor identities prove both inverse composites.

Conversely, given an invertible cocycle map, set $\rho(n)=\varphi(n\otimes1)$. Pulling $\varphi$ back along multiplication $S\otimes_R S\to S$ gives an invertible map $b:N\to N$. Pulling its cocycle equation back along multiplication of all three factors gives $b^2=b$, so $b=1$. This is the counit law. Evaluating the cocycle on $n\otimes1\otimes1$ gives coassociativity; the first-factor linearity gives $S$-linearity of $\rho$. These constructions are inverse, since second-factor linearity recovers (4.1) from $\varphi(n\otimes1)$. An arrow's overlap equation evaluated on $n\otimes1$ is its coaction equation, and that coaction equation recovers the overlap equation by (4.1). $\square$

This is the ordinary descent datum of Stacks, Definition 35.3.1. Theorem 3.1 recovers effective descent, including the functor and both natural comparisons, in the locally module setting. For the canonical object $N=S\otimes_R M$, its coaction is $s\otimes m\mapsto s\otimes1\otimes m$, and (4.1) is the direct reassociation $(s_0\otimes m)\otimes s_1\mapsto s_0\otimes(s_1\otimes m)$.

5. The augmented Amitsur complex and its contraction

For $K\in\mathcal C$, put $C^{-1}(K)=K$ and $C^n(K)=\mathcal S^{\otimes_{\mathcal R}(n+1)}\otimes_{\mathcal R}K \quad(n\geq0),\qquad d^{-1}(k)=1\otimes k. \tag{5.1}$ For $n\geq0$, let $d^n$ be the alternating sum of the $n+2$ maps inserting $1$ in scalar positions $0,\ldots,n+1$. Thus $d^0(s\otimes k)=1\otimes s\otimes k-s\otimes1\otimes k. \tag{5.2}$ These are morphisms in $\mathcal C$; the local formulas give the intrinsic maps through the tensor comparisons. Each pair of inserted units appears twice in $d^{n+1}d^n$, once in either order, with opposite signs. For the augmentation, the two copies of $1\otimes1\otimes k$ cancel. Hence this is a complex, augmented by zero in degree $-2$.

Theorem 5.1. The augmented complex $C^\bullet(K)$ is exact in every degree $n\geq-1$, naturally in $K$.

Proof. Apply $F$. Its extra first scalar factor will be written $a$. For $n\geq0$, define a homotopy $h^n:FC^n\to FC^{n-1}$ by $h^n(a\otimes s_0\otimes\cdots\otimes s_n\otimes k) =a s_0\otimes s_1\otimes\cdots\otimes s_n\otimes k. \tag{5.3}$ For $n=0$ the right side is $a s_0\otimes k\in FK$. These maps respect the first-factor $\mathcal S$-action and the local $\mathcal A_i$-action. They are intrinsic multiplication maps, so glue.

The term inserting a unit in position zero of the old factors gives the identity after applying $h^{n+1}$. Inserting in position $i\geq1$ and then multiplying the first two factors gives the same tensor as multiplying first and inserting in position $i-1$. Their signs are opposite. Therefore $h^{n+1}F d^n+F d^{n-1}h^n=1_{FC^n}\quad(n\geq0), \qquad h^0F d^{-1}=1_{FK}. \tag{5.4}$ The augmented image complex is contractible. Exactness of $F$ identifies its cohomology with $F(H^nC)$, including degree $-1$; faithfulness detects that these objects vanish. Naturality is visible in every tensor formula. $\square$

In particular $K$ is the equalizer of $C^0(K)\rightrightarrows C^1(K)$. Formula (5.4) explains the exactness statement of Stacks, Lemma 35.3.6 with every augmentation degree included. This is a resolution by scalar-tensor objects. Its terms have no asserted global-section acyclicity, injectivity or finite homological dimension.

6. Exact descent and bounded complexes

In the scalar setting $U$ is exact, so $G=FU$ is exact as well.

Proposition 6.1. The coalgebra category is abelian; kernels, cokernels and exactness are created by its forgetful functor to $\mathcal D$. Theorem 3.1 is an exact equivalence. Consequently it induces an equivalence of bounded derived categories, with strict coalgebra structures on the bounded complexes on its right side.

Proof. For a coalgebra map $a:N\to N'$, take its underlying kernel $k:L\to N$. Since $G$ preserves kernels, the equation $Ga\,\rho_Nk=\rho_{N'}ak=0$ uniquely gives $\rho_L:L\to GL$ with $Gk\,\rho_L=\rho_Nk$. Naturality of $\varepsilon$ and monicity of $k$ give $\varepsilon_L\rho_L=1$. Naturality of $\delta$, the coalgebra law on $N$, and monicity of $G^2k$ give $\delta_L\rho_L=G\rho_L\rho_L$. Thus this is a coalgebra kernel. Any underlying factor through $k$ respects coactions after composing with the monic map $Gk$, proving its universal property among coalgebra arrows.

For the underlying cokernel $q:N'\to Q$, preservation of cokernels gives the unique $\rho_Q$ such that $\rho_Qq=Gq\,\rho_{N'}$. Its counit and coassociativity equations hold after precomposing with the epic $q$, and hence hold. The same epimorphism test proves the coalgebra cokernel universal property. Zero objects and biproducts have the component coactions. The underlying coimage-to-image map is an isomorphism in the abelian category $\mathcal D$; its inverse respects coactions by the inverse-isomorphism observation in Theorem 3.1. This proves the abelian axiom and the assertions about exactness.

The comparison sends an exact row to an underlying exact row by exactness of $F$, and thus to an exact coalgebra row. Its inverse is exact because an equivalence preserves kernel and cokernel universal properties. Apply these two additive inverse functors term by term to bounded complexes. They preserve chain-map equations, homotopy equations and the natural inverse comparisons. Exactness identifies their cohomology objects and therefore preserves quasi-isomorphisms. Their homotopy-category equivalence descends to the localization at quasi-isomorphisms, with the same inverse comparisons. Shift and the signed biproduct mapping-cone formulas are preserved, so it is a triangulated equivalence of bounded derived categories. $\square$

Here a complex of coalgebras means that every degree has (2.2) and every differential is a coalgebra arrow. An arbitrary collection of derived coaction maps is a different datum; the theorem uses the actual degreewise structures and equations.

7. Scalar base change includes the invariant comparison

In a module chart let $R\to R'$ be any commutative scalar map. Put $S'=R'\otimes_R S$, $A'=R'\otimes_R A$, and $N'=R'\otimes_R N$, with the base-changed coaction. Then $S'$ is faithfully flat over $R'$. Indeed on any $R'$-module map tensoring with $S'$ identifies with tensoring the underlying $R$-map with $S$. The latter functor is exact and faithful, and the forgetful functor for $R'$-modules is exact and faithful.

Proposition 7.1. If $M$ is the invariant module of the descent datum on $N$, then the canonical map $R'\otimes_R M\longrightarrow M(N',\rho') \tag{7.1}$ is an isomorphism, even when $R'$ is not flat over $R$.

Proof. Theorem 3.1 supplies the coaction-compatible isomorphism $S\otimes_R M\to N$. Tensoring an isomorphism with $R'$ preserves it, and reassociation yields $S'\otimes_{R'}(R'\otimes_R M)\xrightarrow{\sim}N'. \tag{7.2}$ It respects the canonical coaction because the original comparison did and all the scalar insertion maps commute with base change. Applying the invariant functor to (7.2) and then its specified comparison $h$ from (3.3) gives (7.1). On $r'\otimes m$ this comparison is the inclusion of $r'\otimes m$ as an invariant element, so it is the stated canonical map. This argument establishes the particular kernel comparison using effective descent; it assumes no preservation of arbitrary kernels by $R'\otimes_R-$. $\square$

For sheaves these formulas apply at every stalk, where they detect the invariant comparison isomorphism. For the stack, tensor, scalar actions and the comparison maps are the local universal constructions of Section 1, so their compatibility on overlaps gives the corresponding intrinsic scalar base change. The same argument is valid on every open set. No arbitrary derived Hom base-change assertion is needed.

8. Four graded exercises with complete solutions

Exercise 8.1 — Two copies and an actual gluing map (intermediate)

Let $S=R\times R$, with the diagonal scalar map. A left $S$-module is $N=P\oplus Q$. Classify its coactions, calculate its invariant module and reconstruction map, and classify coalgebra arrows.

Solution. Tensoring with $S$ sends an $R$-module to two copies, so it is exact and faithful. Identify $S\otimes_R N=N\oplus N$, with the two outer idempotents selecting these summands. Linearity and the counit force every coaction to have the form $\rho(p,q)=\bigl((p,\alpha p),(\beta q,q)\bigr), \qquad \alpha:P\to Q,\quad\beta:Q\to P. \tag{8.1}$ Indeed an input in $P$ can land only in the first outer summand, whose projection to $P$ is the identity by the counit. The input in $Q$ gives the second formula. Write $S\otimes S\otimes N$ in its four outer-idempotent components. For an input $p$, the mixed component of $(1\otimes\rho)\rho=\delta\rho$ requires $\beta\alpha p=p$; for an input $q$ it requires $\alpha\beta q=q$. The other two component equations are the identities already in (8.1). Hence coactions correspond exactly to isomorphisms $\alpha:P\simeq Q$, with $\beta=\alpha^{-1}$.

The invariant equation compares (8.1) with $((p,q),(p,q))$. It gives $q=\alpha p$, so $M$ is the graph of $\alpha$, naturally isomorphic to $P$. Reconstruction is $(R\times R)\otimes_R P\longrightarrow P\oplus Q, \qquad (s_1,s_2)\otimes p\longmapsto(s_1p,s_2\alpha p). \tag{8.2}$ Its inverse is the decomposition into the two idempotent summands, with $\alpha^{-1}$ on the second. An $S$-linear map is a pair $(f_P,f_Q)$; its coaction equation is $f_Q\alpha=\alpha'f_P$. Thus it is determined by the invariant map $f_P$, and every such map yields a coalgebra arrow. This checks the equivalence on morphisms, not just its object classification.

Exercise 8.2 — A flat extension that loses a component (intermediate)

Take $R=k\times k$ and let $S=k$ be first-component projection. Compute the coalgebra comparison and show where recovery and the augmented complex fail.

Solution. An $R$-module is a pair $(P,Q)$. Scalar extension gives $F(P,Q)=P$, while $UN=(N,0)$. Both functors are exact, but $F(0,k)=0$. The comonad $FU$ on $k$-modules is the identity, with identity counit. The counit law therefore forces the coaction to be the identity, and all $k$-linear maps are coalgebra maps. Invariants recover $(N,0)$. The comparison $(P,Q)\to(P,0)$ is invertible exactly when $Q=0$. In particular both zero and $(0,k)$ have the same image, so there is no equivalence with all $R$-modules.

For $K=(0,k)$, every term $C^n(K)$ for $n\geq0$ is zero, while $C^{-1}(K)=K\neq0$. Its degree-minus-one cohomology is $K$, so the augmented complex is not exact. After applying $F$ it is the zero complex and has the contraction of Section 5. Faithfulness is precisely the missing inference back to the original complex.

Exercise 8.3 — Faithfulness without pointwise kernels (advanced)

Let $S=\mathbb Z\times\mathbb F_p$ as a $\mathbb Z$-algebra. Show that scalar extension is faithful but not flat. For the canonical coalgebra on $S=F\mathbb Z$, examine the kernel of multiplication by $p$ in $S$-modules and decide whether its inclusion can be a coalgebra arrow.

Solution. For a group $M$, the underlying group of $FM$ is $M\oplus M/pM$. On a map its first component is the original map, so the functor is faithful. Applying it to the injection $p:\mathbb Z\to\mathbb Z$ gives multiplication by $p$ on both summands, and is zero on the second summand $\mathbb F_p$. Thus it does not preserve this monomorphism and is not flat.

The underlying kernel of $p:S\to S$ is $L=0\times\mathbb F_p$. The canonical coaction is $\rho_S(s)=s\otimes1$. Decompose $S\otimes_{\mathbb Z}S \simeq\mathbb Z\oplus\mathbb F_p\oplus\mathbb F_p\oplus\mathbb F_p, \tag{8.3}$ with factors in the order $(1,1),(1,2),(2,1),(2,2)$. The image of $S\otimes L\to S\otimes S$ occupies only the second and fourth factors. For $x=(0,1)\in L$, the element $\rho_S(x)=x\otimes1$ has components $(0,0,1,1)$. It does not belong to that image. Consequently no coaction on $L$ can make its given inclusion a coalgebra arrow: the required equation would factor this element through $S\otimes L$. This is a concrete failure of the kernel construction in Proposition 6.1. It proves that faithfulness alone does not supply its exactness hypothesis; it makes no claim that every other descent argument for this scalar map fails.

Exercise 8.4 — Real forms of a complex vector space (advanced)

For $\mathbb R\to\mathbb C$, classify coactions on a complex vector space $V$ as semilinear involutions. Compute the invariant real form and its reconstruction. For $V=\mathbb C$, classify $\sigma_u(v)=u\overline v$ and their coalgebra morphisms.

Solution. Complexification is exact and faithful because $\mathbb C$ is free of rank two over $\mathbb R$. The real-linear isomorphism $\Phi:\mathbb C\otimes_{\mathbb R}V\longrightarrow V\oplus V, \qquad z\otimes v\longmapsto(zv,\overline z v) \tag{8.4}$ has inverse $(v,w)\mapsto1\otimes(v+w)/2+i\otimes(v-w)/(2i)$. The outer complex action is standard in the first component and conjugate in the second. The counit is first projection. Thus a linear coaction has $\Phi\rho(v)=(v,\sigma v)$ with $\sigma(zv)=\overline z\sigma(v)$.

Put $a=(v+\sigma v)/2$ and $b=(v-\sigma v)/(2i)$, so $\rho(v)=1\otimes a+i\otimes b$. Coassociativity is $1\otimes\rho(a)+i\otimes\rho(b) =1\otimes1\otimes a+i\otimes1\otimes b. \tag{8.5}$ The real basis $1,i$ of the outer factor makes this equivalent to $\rho(a)=1\otimes a$ and $\rho(b)=1\otimes b$, hence to $\sigma a=a$ and $\sigma b=b$. Those equalities imply $\sigma^2v=\sigma(a-ib)=a+ib=v$. Conversely a semilinear involution makes both $a,b$ fixed and therefore proves (8.5).

The invariants are $W=V^\sigma$. Every $v$ has the decomposition $a+ib$ with $a,b\in W$. It is unique: applying $\sigma$ to $a+ib=0$ gives $a-ib=0$, and subtraction gives $b=0$, then $a=0$. Thus $\mathbb C\otimes_{\mathbb R}W\to V$, $z\otimes w\mapsto zw$, is the reconstruction isomorphism, for any vector-space dimension.

For a line, $\sigma_u^2=1$ exactly when $u\overline u=1$. Choose $t_u=1+u$ if $u\neq-1$, and $t_{-1}=i$. In both cases $t_u/\overline{t_u}=u$ and the fixed line is $t_u\mathbb R$: dividing its fixed-point equation by $t_u$ gives a real number. A complex-linear map from $(\mathbb C,\sigma_u)$ to $(\mathbb C,\sigma_v)$ is multiplication by $c$, with coaction condition $v\overline c=cu$. In the fixed bases it is multiplication by $c t_u/t_v$, which is real by that equation. Conversely every real scalar gives such a $c$. This gives the full morphism comparison between the descended real lines.

9. Freely accessible sources

The Stacks Project, Descent for modules, Section 35.3, defines overlap descent data and treats their augmented complexes. Proposition 35.3.9 gives effective faithfully flat module descent. Sections 2–4 here supply the exact-adjunction argument, all inverse comparisons and the locally module stack specialization; Section 5 gives the explicit contraction.

Andrea D'Agnolo and Pietro Polesello, Stacks of twisted modules and integral transforms, freely accessible version, motivates stacks with local noncommutative module descriptions and their tensor operations. The needed local representability, scalar action and global abelian arguments are proved in the linked course prerequisites.