Lemma (Properties of faithfully flat maps). Let $A \to B$ be a faithfully flat ring map. Then:

(a) A sequence $M' \to M \to M''$ of $A$-modules is exact if and only if $M' \otimes_A B \to M \otimes_A B \to M'' \otimes_A B$ is exact.

(b) An $A$-module $M$ is zero if and only if $M \otimes_A B = 0$.

(c) A map $f: M \to N$ of $A$-modules is injective (resp. surjective, bijective) if and only if $f \otimes \operatorname{id}_B: M \otimes_A B \to N \otimes_A B$ is.

Proof of Lemma. Since $B$ is a flat $A$-module, $- \otimes_A B$ is exact, so the "only if" direction in (a) is immediate. For the "if" direction: let $H = \ker(M \to M'') / \operatorname{im}(M' \to M)$ be the homology. Then $H \otimes_A B = 0$ by hypothesis. Since $A \to B$ is faithfully flat, $H = 0$, giving exactness. Parts (b) and (c) are special cases of (a).

Step 1a: The Amitsur complex.

For an $A$-module $N$, consider the augmented cosimplicial complex (the Amitsur complex):

$0 \to N \xrightarrow{d^0} N \otimes_A B \xrightarrow{d^0 - d^1} N \otimes_A B \otimes_A B \xrightarrow{d^0 - d^1 + d^2} N \otimes_A B^{\otimes 3} \to \cdots$

where $d^i: N \otimes_A B^{\otimes n} \to N \otimes_A B^{\otimes (n+1)}$ inserts $1_B$ in the $(i+1)$-th tensor position. Explicitly:

$d^0(n \otimes b_1 \otimes \cdots \otimes b_n) = n \otimes 1 \otimes b_1 \otimes \cdots \otimes b_n$ $d^i(n \otimes b_1 \otimes \cdots \otimes b_n) = n \otimes b_1 \otimes \cdots \otimes b_i \otimes 1 \otimes b_{i+1} \otimes \cdots \otimes b_n$

Step 1b: Reduction to the split case.

After tensoring the Amitsur complex with $B$ over $A$ (applying $- \otimes_A B$), we get the same complex for the ring map $B \to B \otimes_A B$, $b \mapsto b \otimes 1$.

This map has a section: $s: B \otimes_A B \to B$ defined by $s(b_1 \otimes b_2) = b_1 b_2$ (the multiplication map). The existence of a section means we are in the "split" case.

Step 1c: Exactness in the split case.

When $A \to B$ has a section $s: B \to A$ (i.e., $s \circ \varphi = \operatorname{id}_A$), the Amitsur complex is exact. We construct a contracting homotopy.

Define $h^n: N \otimes_A B^{\otimes (n+1)} \to N \otimes_A B^{\otimes n}$ by

$h^n(n \otimes b_0 \otimes b_1 \otimes \cdots \otimes b_n) = s(b_0) \cdot n \otimes b_1 \otimes \cdots \otimes b_n$

We claim this is a contracting homotopy: $h^{n+1} \circ \partial^{n+1} + \partial^n \circ h^n = \operatorname{id}$ where $\partial^n = \sum_{i=0}^n (-1)^i d^i$.

To verify: $h^{n+1} \circ d^0 (n \otimes b_1 \otimes \cdots \otimes b_n) = h^{n+1}(n \otimes 1 \otimes b_1 \otimes \cdots \otimes b_n) = s(1) \cdot n \otimes b_1 \otimes \cdots \otimes b_n = n \otimes b_1 \otimes \cdots \otimes b_n$. And $h^{n+1} \circ d^i = d^{i-1} \circ h^n$ for $i \geq 1$, from which the homotopy identity follows.

Step 1d: Conclusion of exactness.

Since $A \to B$ is faithfully flat, a complex of $A$-modules is exact if and only if it becomes exact after tensoring with $B$. After tensoring the Amitsur complex with $B$, we are in the split case (Step 1c), so the tensored complex is exact. By faithful flatness, the original complex is exact.

Therefore, for any $A$-module $N$:

$0 \to N \to N \otimes_A B \rightrightarrows N \otimes_A B \otimes_A B$

is exact. In particular, $N = \ker(N \otimes_A B \rightrightarrows N \otimes_A B^{\otimes 2})$.

Step 2a: Setup.

We show that for any $A$-modules $N_1, N_2$, the natural map

$\operatorname{Hom}_A(N_1, N_2) \to \operatorname{Hom}_{B}^{\mathrm{desc}}(N_1 \otimes_A B, N_2 \otimes_A B)$

is a bijection, where $\operatorname{Hom}^{\mathrm{desc}}$ denotes $B$-module homomorphisms compatible with the canonical descent data.

Step 2b: Injectivity.

Let $f: N_1 \to N_2$ be an $A$-module map such that $f \otimes \operatorname{id}_B = 0$. By faithful flatness (Step 1, property (c)), $f = 0$. So the map is injective.

Step 2c: Surjectivity.

Let $g: N_1 \otimes_A B \to N_2 \otimes_A B$ be a $B$-module map compatible with the canonical descent data. "Compatible" means the diagram

$N_1 \otimes_A B \otimes_A B \xrightarrow{g \otimes \operatorname{id}} N_2 \otimes_A B \otimes_A B$

commutes with the two natural maps from $N_i \otimes_A B$ to $N_i \otimes_A B \otimes_A B$ (inserting $1$ on the left vs. on the right).

Consider the two maps $\alpha, \beta: N_1 \otimes_A B \to N_2 \otimes_A B \otimes_A B$ defined by:

• $\alpha = (g \otimes \operatorname{id}_B) \circ d^1$ (first map $N_1 \otimes B \to N_1 \otimes B \otimes B$, then apply $g$ on the first two factors)
• $\beta = d^1 \circ g$ (apply $g$ first, then map to the triple tensor)

Wait, let us be more precise. The compatibility condition states that

$(\operatorname{id}_{N_2} \otimes d^0_B) \circ g = (g \otimes \operatorname{id}_B) \circ (\operatorname{id}_{N_1} \otimes d^0_B)$

where $d^0_B: B \to B \otimes_A B$ sends $b \mapsto 1 \otimes b$ and $d^1_B$ sends $b \mapsto b \otimes 1$.

By the exactness of the Amitsur complex (Step 1), $g$ maps the submodule $N_1 \hookrightarrow N_1 \otimes_A B$ to $N_2 \hookrightarrow N_2 \otimes_A B$. Here $N_i \hookrightarrow N_i \otimes_A B$ is the map $n \mapsto n \otimes 1$, and the image is precisely the kernel of $N_i \otimes B \rightrightarrows N_i \otimes B \otimes B$.

Indeed, the compatibility condition says $g$ preserves the equalizer of the two maps to the double tensor, and this equalizer is $N_i$ (by Step 1). So $g$ restricts to an $A$-module map $f: N_1 \to N_2$, and $g = f \otimes \operatorname{id}_B$.

Step 3a: Setup.

Let $(M, \sigma)$ be a descent datum: $M$ is a $B$-module and $\sigma: M \otimes_A B \xrightarrow{\sim} B \otimes_A M$ is a $B \otimes_A B$-module isomorphism satisfying the cocycle condition on $B^{\otimes 3}$.

We need to construct an $A$-module $N$ with $N \otimes_A B \cong M$ (compatibly with $\sigma$).

Step 3b: Construction of $N$.

Define $N$ as the equalizer:

$N = \ker\left(M \xrightarrow{d^0} B \otimes_A M \xleftarrow{\sigma} M \otimes_A B \xleftarrow{d^1} M\right)$

More precisely, we have two maps $M \rightrightarrows B \otimes_A M$:

• $\alpha: M \to B \otimes_A M$ defined by $m \mapsto 1 \otimes m$ (the map $d^0$).
• $\beta: M \to M \otimes_A B \xrightarrow{\sigma} B \otimes_A M$ defined by $\beta(m) = \sigma(m \otimes 1)$.

Set $N = \{m \in M \mid 1 \otimes m = \sigma(m \otimes 1) \text{ in } B \otimes_A M\}$.

This is an $A$-submodule of $M$ (the map $\alpha - \beta$ is $A$-linear, and $N$ is its kernel).

Step 3c: The map $N \otimes_A B \to M$.

There is a natural $B$-module map $\mu: N \otimes_A B \to M$ defined by $n \otimes b \mapsto bn$ (using the $B$-module structure on $M$). We claim this is an isomorphism.

Step 3d: Proof that $\mu$ is an isomorphism (after base change).

We use the faithful flatness strategy: show that $\mu \otimes_A B$ is an isomorphism, then conclude by faithful flatness.

After base changing along $A \to B$, we get:

• The module $M$ becomes $M \otimes_A B$ (a $B \otimes_A B$-module).
• The descent datum $\sigma: M \otimes_A B \to B \otimes_A M$ becomes $\sigma': M \otimes_A B \otimes_A B \to B \otimes_A M \otimes_A B$.

The base-changed descent datum is "split": via the multiplication map $B \otimes_A B \to B$ (sending $b_1 \otimes b_2 \mapsto b_1 b_2$), we have a section of $B \to B \otimes_A B$.

In the split case, the descent datum comes from a $B$-module: explicitly, $M$ itself viewed as a $B$-module (with the section providing the splitting). The equalizer $N \otimes_A B$ in the split case is isomorphic to $M$ (this is the content of the contracting homotopy from Step 1).

Therefore $\mu \otimes_A B$ is an isomorphism. Since $A \to B$ is faithfully flat, $\mu$ is an isomorphism.

Step 3e: Compatibility with descent data.

We must verify that the isomorphism $\mu: N \otimes_A B \xrightarrow{\sim} M$ is compatible with the descent data, i.e., the canonical descent datum on $N \otimes_A B$ corresponds to $\sigma$ under $\mu$.

The canonical descent datum on $N \otimes_A B$ is the isomorphism $(N \otimes_A B) \otimes_A B \xrightarrow{\sim} B \otimes_A (N \otimes_A B)$ given by $n \otimes b_1 \otimes b_2 \mapsto b_1 \otimes n \otimes b_2$.

Under $\mu$: the left side maps to $M \otimes_A B$ (via $n \otimes b_1 \otimes b_2 \mapsto b_1 n \otimes b_2$), and the right side maps to $B \otimes_A M$ (via $b_1 \otimes n \otimes b_2 \mapsto b_1 \otimes b_2 n$). The composite isomorphism is $\sigma$, which holds because $n \in N$ satisfies $1 \otimes n = \sigma(n \otimes 1)$, and extending by $B$-linearity gives the full compatibility.

Step 4a: $N$ is functorial.

The construction $N = \ker(M \rightrightarrows B \otimes_A M)$ is functorial in $(M, \sigma)$: a morphism of descent data $(M, \sigma) \to (M', \sigma')$ (a $B$-module map $f: M \to M'$ compatible with $\sigma, \sigma'$) restricts to a map $f|_N: N \to N'$.

Step 4b: The unit.

For an $A$-module $N_0$, setting $M = N_0 \otimes_A B$ with canonical descent datum, the equalizer recovers $N_0$: this is the content of Step 1 (exactness of the Amitsur complex), which says $N_0 = \ker(N_0 \otimes_A B \rightrightarrows N_0 \otimes_A B^{\otimes 2})$.

So the composite $N_0 \mapsto (N_0 \otimes_A B, \text{can}) \mapsto \ker(\cdots) = N_0$ is the identity (up to canonical isomorphism).

Step 4c: The counit.

For a descent datum $(M, \sigma)$, we constructed $N = \ker(M \rightrightarrows B \otimes_A M)$ and showed $N \otimes_A B \cong M$ in Step 3. So the composite $(M, \sigma) \mapsto N \mapsto (N \otimes_A B, \text{can}) \cong (M, \sigma)$ is an isomorphism.

This completes the proof that $\Phi$ and its quasi-inverse form an equivalence of categories.

Step 5: Properties preserved by descent.

The equivalence $\Phi$ preserves many properties of modules:

(a) $N$ is finitely generated over $A$ $\iff$ $N \otimes_A B$ is finitely generated over $B$.

Proof. The "only if" direction is clear. For "if": suppose $M = N \otimes_A B$ is generated by $m_1, \ldots, m_r$. Write $m_j = \sum_i n_{ij} \otimes b_{ij}$ for some $n_{ij} \in N$ and $b_{ij} \in B$. Let $N_0 = \sum_{i,j} A \cdot n_{ij} \subseteq N$. Then $N_0 \otimes_A B \to N \otimes_A B = M$ is surjective (since the $m_j$ are in the image). By faithful flatness, $N_0 \to N$ is surjective, so $N$ is finitely generated.

(b) $N$ is finitely presented over $A$ $\iff$ $N \otimes_A B$ is finitely presented over $B$. (Similar argument using that faithful flatness reflects finite presentation.)

(c) $N$ is projective over $A$ $\iff$ $N \otimes_A B$ is projective over $B$. (Uses that faithful flatness reflects projectivity.)

(d) $N$ is flat over $A$ $\iff$ $N \otimes_A B$ is flat over $B$. (Because flatness is detected by the vanishing of $\operatorname{Tor}_1$, and $\operatorname{Tor}$ commutes with flat base change.)

These results are essential for applications: they show that descent preserves the "type" of module (finitely generated, projective, etc.).