The Stacks project

Proof. Assume $N$ is a flat as an $A$-module. Then the functor

is exact. As $B \otimes _ A B \to B$ is flat we conclude that the functor

is exact, hence $N$ is flat over $B$. $\square$

Proof. Let $N$ be a $B$-module. If $d = 0$, then $N$ is flat as an $A$-module, hence flat as a $B$-module by Lemma 15.104.2. Assume $d > 0$. Choose a resolution $F_\bullet \to N$ by free $B$-modules. Our assumption implies that $K = \mathop{\mathrm{Im}}(F_ d \to F_{d - 1})$ is $A$-flat, see Lemma 15.66.2. Hence it is $B$-flat by Lemma 15.104.2. Thus $0 \to K \to F_{d - 1} \to \ldots \to F_0 \to N \to 0$ is a flat resolution of length $d$ and we see that $N$ has tor dimension at most $d$. $\square$

Proof. The equivalence of (1) and (2) is immediate. Assume $A$ is absolutely flat. This implies every ideal of $A$ is pure, see Algebra, Definition 10.108.1. Hence every finitely generated ideal is generated by an idempotent by Algebra, Lemma 10.108.5. If $f \in A$, then $(f) = (e)$ for some idempotent $e \in A$ and $D(f) = D(e)$ is open and closed (Algebra, Lemma 10.21.1). This already implies every ideal of $A$ is maximal for example by Algebra, Lemma 10.26.5. Moreover, if $f$ is nilpotent, then $e = 0$ hence $f = 0$. Thus $A$ is reduced.

Assume $A$ is reduced and every prime of $A$ is maximal. Let $M$ be an $A$-module. Our goal is to show that $M$ is flat. We may write $M$ as a filtered colimit of finite $A$-modules, hence we may assume $M$ is finite (Algebra, Lemma 10.39.3). There is a finite filtration of $M$ by modules of the form $A/I$ (Algebra, Lemma 10.5.4), hence we may assume that $M = A/I$ (Algebra, Lemma 10.39.13). Thus it suffices to show every ideal of $A$ is pure. Since every local ring of $A$ is a field (by Algebra, Lemma 10.25.1 and the fact that every prime of $A$ is minimal), we see that every ideal $I \subset A$ is radical. Note that every closed subset of $\mathop{\mathrm{Spec}}(A)$ is closed under generalization. Thus every (radical) ideal of $A$ is pure by Algebra, Lemma 10.108.4. $\square$

Proof. Let $K_ i$ be a family of fields. If $f = (f_ i) \in \prod K_ i$, then the ideal generated by $f$ is the same as the ideal generated by the idempotent $e = (e_ i)$ with $e_ i = 0, 1$ according to whether $f_ i$ is $0$ or not. Thus $D(f) = D(e)$ is open and closed and we conclude by Lemma 15.104.5 and Algebra, Lemma 10.26.5. $\square$

Proof. Assume $B \otimes _ A B \to B$ is flat. The ring map $B' \otimes _{A'} B' \to B'$ is the base change of $B \otimes _ A B \to B$ by $A \to A'$. Hence it is flat by Algebra, Lemma 10.39.7. This proves (1). Part (2) follows from (1) and the fact (just used) that the base change of a flat ring map is flat. $\square$

Proof. Part (1) follows immediately from Lemma 15.104.2 and the definitions. If $A$ is reduced, then there exists an injection $A \to A' = \prod _{\mathfrak p \subset A\text{ minimal}} A_\mathfrak p$ of $A$ into an absolutely flat ring (Algebra, Lemma 10.25.2 and Lemma 15.104.6). If $A \to B$ is flat, then the induced map $B \to B' = B \otimes _ A A'$ is injective too. By Lemma 15.104.7 the ring map $A' \to B'$ is weakly étale. By part (1) we see that $B'$ is absolutely flat. By Lemma 15.104.5 the ring $B'$ is reduced. Hence $B$ is reduced. $\square$

Proof. Part (1) follows from the factorization

of the multiplication map, the fact that

the fact that a base change of a flat map is flat, and the fact that the composition of flat ring maps is flat. See Algebra, Lemmas 10.39.7 and 10.39.4. Part (2) follows from (1) and the fact (just used) that the composition of flat ring maps is flat. $\square$

Proof. Assume $B \to C$ is faithfully flat and $C \otimes _ A C \to C$ is flat. Consider the commutative diagram

The vertical arrows are flat, the top horizontal arrow is flat. Hence $C$ is flat as a $B \otimes _ A B$-module. The map $B \to C$ is faithfully flat and $C = B \otimes _ B C$. Hence $B$ is flat as a $B \otimes _ A B$-module by Algebra, Lemma 10.39.9. This proves (1). Part (2) follows from (1) and the fact that $A \to B$ is flat if $A \to C$ is flat and $B \to C$ is faithfully flat (Algebra, Lemma 10.39.9). $\square$

Proof. The ring map $B \to C$ is flat by Lemma 15.104.2. The ring map $C \otimes _ A C \to C \otimes _ B C$ is surjective, hence an epimorphism. Thus Lemma 15.104.2 implies, that since $C$ is flat over $C \otimes _ A C$ also $C$ is flat over $C \otimes _ B C$. $\square$

Proof. Let $I \subset B \otimes _ A B$ be the kernel of the flat surjective map $B \otimes _ A B \to B$. Then $I$ is a pure ideal (Algebra, Definition 10.108.1), so $I^2 = I$ (Algebra, Lemma 10.108.2). Since $\Omega _{B/A} = I/I^2$ (Algebra, Lemma 10.131.13) we obtain the vanishing. This means $B$ is formally unramified over $A$ by Algebra, Lemma 10.148.2. $\square$

Proof. Part (1) follows from Lemma 15.104.12 and Algebra, Definition 10.151.1. Part (2) follows from part (1) and Algebra, Lemma 10.151.8. $\square$

Proof. An étale ring map is flat and the map $B \otimes _ A B \to B$ is also étale as a map between étale $A$-algebras (Algebra, Lemma 10.143.8). This proves (2).

Let $B_ i$ be a directed system of weakly étale $A$-algebras. Then $B = \mathop{\mathrm{colim}}\nolimits B_ i$ is flat over $A$ by Algebra, Lemma 10.39.3. Note that the transition maps $B_ i \to B_{i'}$ are flat by Lemma 15.104.11. Hence $B$ is flat over $B_ i$ for each $i$, and we see that $B$ is flat over $B_ i \otimes _ A B_ i$ by Algebra, Lemma 10.39.4. Thus $B$ is flat over $B \otimes _ A B = \mathop{\mathrm{colim}}\nolimits B_ i \otimes _ A B_ i$ by Algebra, Lemma 10.39.6.

Part (1) can be proved directly, but also follows by combining (2) and (3). $\square$

Proof. By Lemma 15.104.10 we see that any subfield $K \subset L' \subset L$ the map $L' \otimes _ K L' \to L'$ is flat. Thus we may assume $L$ is a finitely generated field extension of $K$. In this case the fact that $L/K$ is formally unramified (Lemma 15.104.12) implies that $L/K$ is finite separable, see Algebra, Lemma 10.158.1. $\square$

Proof. Parts (1) and (2) are equivalent because every $K$-algebra is flat over $K$. Part (3) implies (1) and (2) by Lemma 15.104.14

Assume (1) and (2) hold. We will prove (3) and the finite statement of the lemma. A field is absolutely flat ring, hence $B$ is a absolutely flat ring by Lemma 15.104.8. Hence $B$ is reduced and every local ring is a field, see Lemma 15.104.5.

Let $\mathfrak q \subset B$ be a prime. The ring map $B \to B_\mathfrak q$ is weakly étale, hence $B_\mathfrak q$ is weakly étale over $K$ (Lemma 15.104.9). Thus $B_\mathfrak q$ is a separable algebraic extension of $K$ by Lemma 15.104.15.

Let $K \subset A \subset B$ be a finitely generated $K$-sub algebra. We will show that $A$ is étale over $K$ which will finish the proof of the lemma. Then every minimal prime $\mathfrak p \subset A$ is the image of a prime $\mathfrak q$ of $B$, see Algebra, Lemma 10.30.5. Thus $\kappa (\mathfrak p)$ as a subfield of $B_\mathfrak q = \kappa (\mathfrak q)$ is separable algebraic over $K$. Hence every generic point of $\mathop{\mathrm{Spec}}(A)$ is closed (Algebra, Lemma 10.35.9). Thus $\dim (A) = 0$. Then $A$ is the product of its local rings, e.g., by Algebra, Proposition 10.60.7. Moreover, since $A$ is reduced, all local rings are equal to their residue fields wich are finite separable over $K$. This means that $A$ is étale over $K$ by Algebra, Lemma 10.143.4 and finishes the proof. $\square$

Proof. Let $\mathfrak p$ be a prime of $A$. Then $\kappa (\mathfrak p) \to B \otimes _ A \kappa (\mathfrak p)$ is weakly étale by Lemma 15.104.7. Hence $B \otimes _ A \kappa (\mathfrak p)$ is a filtered colimit of étale $\kappa (\mathfrak p)$-algebras by Lemma 15.104.16. Hence for $\mathfrak q \subset B$ lying over $\mathfrak p$ the extension $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ is a filtered colimit of finite separable extensions by Algebra, Lemma 10.143.4. $\square$

Proof. If $A$ has weak dimension $\leq 1$, then the resolution $0 \to I \to A \to A/I \to 0$ shows that every ideal $I$ is flat by Lemma 15.66.2. Hence (1) $\Rightarrow $ (2).

Assume (4). Let $M$ be an $A$-module. Choose a surjection $F \to M$ where $F$ is a free $A$-module. Then $\mathop{\mathrm{Ker}}(F \to M)$ is flat by assumption, and we see that $M$ has tor dimension $\leq 1$ by Lemma 15.66.6. Hence (4) $\Rightarrow $ (1).

Every ideal is the union of the finitely generated ideals contained in it. Hence (3) implies (2) by Algebra, Lemma 10.39.3. Thus (3) $\Leftrightarrow $ (2).

Assume (2). Suppose that $N \subset M$ with $M$ a flat $A$-module. We will prove that $N$ is flat. We can write $M = \mathop{\mathrm{colim}}\nolimits M_ i$ with each $M_ i$ finite free, see Algebra, Theorem 10.81.4. Setting $N_ i \subset M_ i$ the inverse image of $N$ we see that $N = \mathop{\mathrm{colim}}\nolimits N_ i$. By Algebra, Lemma 10.39.3. it suffices to prove $N_ i$ is flat and we reduce to the case $M = R^{\oplus n}$. In this case the module $N$ has a finite filtration by the submodules $R^{\oplus j} \cap N$ whose subquotients are ideals. By (2) these ideals are flat and hence $N$ is flat by Algebra, Lemma 10.39.13. Thus (2) $\Rightarrow $ (4).

Assume $A$ satisfies (1) and let $\mathfrak p \subset A$ be a prime ideal. By Lemmas 15.104.14 and 15.104.4 we see that $A_\mathfrak p$ satisfies (1). We will show $A$ is a valuation ring if $A$ is a local ring satisfying (3). Let $f \in \mathfrak m$ be a nonzero element. Then $(f)$ is a flat nonzero module generated by one element. Hence it is a free $A$-module by Algebra, Lemma 10.78.5. It follows that $f$ is a nonzerodivisor and $A$ is a domain. If $I \subset A$ is a finitely generated ideal, then we similarly see that $I$ is a finite free $A$-module, hence (by considering the rank) free of rank $1$ and $I$ is a principal ideal. Thus $A$ is a valuation ring by Algebra, Lemma 10.50.15. Thus (1) $\Rightarrow $ (5).

Assume (5). Let $I \subset A$ be a finitely generated ideal. Then $I_\mathfrak p \subset A_\mathfrak p$ is a finitely generated ideal in a valuation ring, hence principal (Algebra, Lemma 10.50.15), hence flat. Thus $I$ is flat by Algebra, Lemma 10.39.18. Thus (5) $\Rightarrow $ (3). This finishes the proof of the lemma. $\square$

Proof. Let $I \subset A$ be a finitely generated ideal. By Lemma 15.104.18 it suffices to show that $I$ is a flat $A$-module. Let $I_ j \subset A_ j$ be the image of $I$. Observe that $I_ j = I \otimes _ A A_ j$, hence $I \to \prod I_ j$ is surjective by Algebra, Proposition 10.89.2. Thus $I = \prod I_ j$. Since $A_ j$ is a valuation ring, the ideal $I_ j$ is generated by a single element (Algebra, Lemma 10.50.15). Say $I_ j = (f_ j)$. Then $I$ is generated by the element $f = (f_ j)$. Let $e \in A$ be the idempotent which has a $0$ or $1$ in $A_ j$ depending on whether $f_ j$ is $0$ or not. Then $f = g e$ for some nonzerodivisor $g \in A$: take $g = (g_ j)$ with $g_ j = 1$ if $f_ j = 0$ and $g_ j = f_ j$ else. Thus $I \cong (e)$ as a module. We conclude $I$ is flat as $(e)$ is a direct summand of $A$. The final statement is true because $K = S^{-1}A$ where $S = \prod (A_ j \setminus \{ 0\} )$. $\square$

Proof. For every $x \in K$, $x \not\in A$ pick $V_ x \subset K$ as in Algebra, Lemma 10.50.11. Set $V = \prod _{x \in K \setminus A} V_ x$ and $L = \prod _{x \in K \setminus A} K$. The ring $V$ has weak dimension at most $1$ by Lemma 15.104.19 which also shows that $V \to L$ is a localization. A localization is flat and an epimorphism, see Algebra, Lemmas 10.39.18 and 10.107.5. $\square$

Proof. Let $x \in B$ be integral over $A$. Let $A' = A[x] \subset B$. Then $A'$ is a finite ring extension of $A$ by Algebra, Lemma 10.36.5. To show $A = A'$ it suffices to show $A \to A'$ is an epimorphism by Algebra, Lemma 10.107.6. Note that $A'$ is flat over $A$ by assumption on $A$ and the fact that $B$ is flat over $A$ (Lemma 15.104.18). Hence the composition

is injective, i.e., $A' \otimes _ A A' \cong A'$ and the lemma is proved. $\square$

Proof. Choose a diagram as in Lemma 15.104.20. As $A \to B$ is flat, the base change gives a cartesian diagram

of rings. Note that $V \to B \otimes _ A V$ is weakly étale (Lemma 15.104.7), hence $B \otimes _ A V$ has weak dimension at most $1$ by Lemma 15.104.4. Note that $B \otimes _ A V \to B \otimes _ A L$ is a flat, injective, epimorphism of rings as a flat base change of such (Algebra, Lemmas 10.39.7 and 10.107.3). By Lemma 15.104.21 we see that $B \otimes _ A V$ is integrally closed in $B \otimes _ A L$. It follows from the cartesian property of the diagram that $B$ is integrally closed in $B \otimes _ A K$. $\square$

Proof. Write $B$ as a filtered colimit $B = \mathop{\mathrm{colim}}\nolimits B_ i$ of finite $A$-sub algebras. If we prove the results for each $B_ i$, then the result follows for $B$. See Algebra, Lemma 10.154.8. If $A \to B$ is finite, then $B$ is a product of local henselian rings by Algebra, Lemma 10.153.4. Since $B$ is a domain we see that $B$ is a local ring. The maximal ideal of $B$ lies over the maximal ideal of $A$ by going up for $A \to B$ (Algebra, Lemma 10.36.22). If $A$ is strictly henselian, then the field extension $\kappa (\mathfrak m_ B)/\kappa (\mathfrak m_ A)$ being algebraic, has to be purely inseparable. Of course, then $\kappa (\mathfrak m_ B)$ is separably algebraically closed and $B$ is strictly henselian. $\square$

Proof. We will show that for all $\mathfrak p \subset A$ there is a unique prime $\mathfrak q \subset B$ lying over $\mathfrak p$ and $\kappa (\mathfrak p) = \kappa (\mathfrak q)$. This implies that $B \otimes _ A B \to B$ is bijective on spectra as well as surjective and flat. Hence it is an isomorphism for example by the description of pure ideals in Algebra, Lemma 10.108.4. Hence $A \to B$ is a faithfully flat epimorphism of rings. We get $A = B$ by Algebra, Lemma 10.107.7.

Note that the fibre ring $B \otimes _ A \kappa (\mathfrak p)$ is a colimit of étale extensions of $\kappa (\mathfrak p)$ by Lemmas 15.104.7 and 15.104.16. Hence, if there exists more than one prime lying over $\mathfrak p$ or if $\kappa (\mathfrak p) \not= \kappa (\mathfrak q)$ for some $\mathfrak q$, then $B \otimes _ A L$ has a nontrivial idempotent for some (separable) algebraic field extension $L/\kappa (\mathfrak p)$.

Let $L/\kappa (\mathfrak p)$ be an algebraic field extension. Let $A' \subset L$ be the integral closure of $A/\mathfrak p$ in $L$. By Lemma 15.104.23 we see that $A'$ is a strictly henselian local ring whose residue field is a purely inseparable extension of the residue field of $A$. Thus $B \otimes _ A A'$ is a local ring by Algebra, Lemma 10.156.5. On the other hand, $B \otimes _ A A'$ is integrally closed in $B \otimes _ A L$ by Lemma 15.104.22. Since $B \otimes _ A A'$ is local, it follows that the ring $B \otimes _ A L$ does not have nontrivial idempotents which is what we wanted to prove. $\square$