2 More on local structure
This chapter serves for the second half of the commutative algebra preparation of the remaining part of the paper. The key result is
Theorem 2.46 expressing that every weakly etale map can be covered by ind-etale maps.
The proof goes via the local input .
Let \(R\) be a Henselian local ring and \(S\) a finite \(R\)-algebra. Then \(S\) is a finite product of Henselian local rings, each finite over \(R\).
Omitted, will be in mathlib soon.
A filtered colimit of local rings along local homomorphisms is local.
Let \(R = \operatorname {colim}_i R_i\) with \(R_i\) local rings and \(R_i \to R_j\) a local ring map. Suppose \(x, y \in R_i\) are non-units. Since \(R_i\) is local, their sum is a not a unit. If the image of \(x + y\) in \(\operatorname {colim}_i R_i\) is a unit, their exists a \(j\) such that the image of \(x + y\) is a unit in \(R_j\). But this does not happen because \(R_i \to R_j\) is a local ring map.
A filtered colimit of (strictly) Henselian local rings along local homomorphisms is (strictly) Henselian.
Omitted.
Let \(R\) be a Henselian local ring and \(S\) be an integral \(R\)-algebra and an integral domain. Then \(S\) is a local ring.
Since \(S\) is integral over \(R\), it is the colimit of its finite \(R\)-sub-algebras. Since a subring of an integral domain is still an integral domain, by Lemma 2.2 we may assume that \(S\) is finite over \(R\). By Lemma 2.1, \(S\) is a product of (Henselian) local rings and hence local as an integral domain.
Let \(R \to S\) be an integral ring map of local rings. Then \(R \to S\) is a local ring homomorphism.
This follows from going-up (use Ideal.IsMaximal.under).
Let \(R \to S\) be a local ring homomorphism of local rings. If \(R \to S\) is integral, the extension of residue fields \(\kappa (S) / \kappa (R)\) is algebraic.
It suffices to show that \(\kappa (S)\) is integral over \(R\). Since \(S\) is integral over \(R\) and \(\kappa (S)\) is integral over \(S\) (the map \(S \to \kappa (S)\) is surjective), the composition \(R \to S \to \kappa (S)\) is integral.
Let \(R \to S\) and \(R \to T\) be local ring maps of local rings. Suppose \(R \to S\) is integral and \(\kappa (S) / \kappa (R)\) or \(\kappa (T) / \kappa (S)\) is purely inseparable, then \(T \otimes _{R} S\) is a local ring.
Integral is stable under base change, so \(T \to S' = T \otimes _{R} S\) is integral. Hence by going-up any maximal ideal of \(S'\) lies over \(\mathfrak {m}_T\). We have
Again by going-up and since \(S\) is local, \(\mathrm{Spec}(S / \mathfrak {m}_R S)\) is a single point. Thus \(\mathrm{Spec}(S’ / \mathfrak {m}_T S’) \cong \mathrm{Spec}(\kappa (T) \otimes _{\kappa (R)} \kappa (S))\). The latter is a single point, because purely inseparable extensions induce universal homeomorphisms.
2.1 Weakly étale algebras
Ind-étale is a practical notion, since it usually allows reducing proofs to the étale case. Geometrically, ind-étale is badly behaved though, since it is not local on the (geometric) target. For this reason, when we later define the pro-étale site of a scheme, we use the notion of weakly étale ring maps instead.
An \(R\)-algebra \(S\) is weakly étale if it is flat over \(R\) and flat over \(S \otimes _{R} S\). A ring homorphism \(f \colon R \to S\) is weakly étale if \(S\) is weakly étale as an \(R\)-algebra.
Historically, the property weakly étale is studied under the name of absolutely flat. Lemma 2.15 partially justifies this name. Furthermore, we have an absolute version of this property.
A ring \(A\) is called absolutely flat if every \(A\)-module is flat over \(A\).
Every étale algebra is weakly étale.
Let \(S\) be an étale \(R\)-algebra. Then \(S\) is \(R\)-flat. Also \(S \otimes _{R} S \cong S × T\) for some ring \(T\), in particular \(\mathrm{Spec}(S \otimes _{R} S) \to \mathrm{Spec}(S)\) is an open immersion, hence flat.
A field is absolutely flat.
Omitted.
lem:weakly-etale-comp If \(A \to B\) and \(B \to C\) are weakly étale, then \(A \to C\) is weakly étale.
Ommited, see [ Sta18 , Tag 092J ] .
If \(B\) is a weakly-étale \(A\)-algebra and \(A'\) is any \(A\)-algebra, the base change \(A' \otimes _{A} B\) is a weakly-étale \(A'\)-algebra.
Calculation, see [ Sta18 , Tag 092H ] .
Let \(B\) be an \(A\)-algebra such that \(B \otimes _{A} B \to B\) is flat. Let \(N\) be a \(B\)-module. If \(N\) is flat as a \(A\)-module, then \(N\) is flat as a \(B\)-module.
If \(N'\) is a second \(B\)-module, then
Hence this follows from the definitions.
Let \(A \to B\) be a ring map such that \(B \otimes _{A} B \to B\) is flat. If \(A\) is an absolutely flat ring, then so is \(B\).
It follows from Lemma 2.15 immediately.
Let \(X\) be a spectral space in which every quasi-compact open subset is closed. Then \(X\) is profinite.
It suffices to show that \(X\) is Hausdorff and totally disconnected. Let \(x \neq y\) be in \(X\). Since \(X\) is quasi-sober and pre-spectral, we may assume that there exists a quasi-compact open \(U\) with \(x \in U\) and \(y \not\in U\). By assumption, its complement is open and \(X\) is Hausdorff. This argument also shows that \(x\) and \(y\) lie in distinct connected components, so \(X\) is totally disconnected.
If \(A\) is an absolutely flat ring, then \(A\) is reduced and every prime is maximal.
It suffices to show that for every \(f \in A\), there exists an idempotent element \(e \in A\) such that \((f) = (e)\). Indeed, if \(f ^n = 0\) for some \(n\), then \(e = e ^n \in (f ^n) = 0\). Hence \(f = 0\). For the second claim observe that every quasi-compact open \(D(f) = D(e)\) is closed. Hence the claim follows from 2.17 and the fact that every quasi-compact open of \(\mathrm{Spec}(A)\) is a finite union of basic opens.
Now let \(f \in A\) be any element. Since \((f)\) is pure by assumption it is nilpotent by Ideal.isIdempotentElem_of_pure and every idempotent finitely generated ideal is generated by an idempotent element by Ideal.isIdempotentElem_iff_of_fg.
If \(A\) is reduced and every prime is maximal, then every local ring of \(A\) is a field.
Localizations of reduced rings are reduced and reduced, local rings of Krull dimension \(\le 0\) are fields.
2.2 Weak dimension
Currently, mathlib does not have \(\mathrm{Tor}\) so we give a non-standard, but equivalent definition which suffices for our purposes.
Let \(A\) be a ring. We say it is of weak dimension \(\le 1\) if every finitely generated ideal of \(A\) is flat over \(A\).
Let \(M\) be a flat \(R\)-module. Then \(M\) is the directed colimit of free, finite \(R\)-modules.
[ Sta18 , Tag 058G ]
If \(A\) is of weak-dimension \(\le 1\), then every ideal \(I \subseteq A\) is flat.
Use the fact that a filtered colimit of flat modules is flat and that any ideal is the filtered colimit of its finitely generated subideals.
Let \(M\) be an \(R\)-module and \(0 = M_0 \subseteq M_1 \subseteq M_n = M\) a sequence of submodules. If \(M_{j + 1} / M_j\) is flat for all \(j\), then \(M\) is flat.
Inductively apply that extensions of flat modules are flat.
Let \(A\) be a ring of weak dimension \(\le 1\) and \(M\) a flat \(A\)-module. Then any submodule of \(M\) is flat.
By Theorem 2.21 we may write \(M = \operatorname {colim}_i M_i\) for finite and free \(A\)-modules \(M_i\). Denote by \(N_i\) the preimage of \(M_i\) in \(N\). Then \(N = \operatorname {colim}_i N_i\) and since the filtered colimit of flat modules is flat, we reduce to the case where \(M = R^{\oplus n}\). By applying Lemma 2.23 to the sequence \(N_j = N \cap R^{\oplus j}\), it remains to show that \(N_{j+1} / N_j\) is flat. For this consider the projection on the \(j + 1\)-th factor \(N_{j+1} \to R\). Its image is an ideal \(I \subseteq R\) and its kernel is \(N_j\). Hence \(N_{j + 1} / N_j \cong I\) and \(I\) is flat by Lemma 2.22.
Let \(A\) be of weak dimension \(\le 1\) and \(B\) a weakly étale \(A\)-algebra. Then \(B\) is of weak dimension \(\le 1\).
Let \(M\) be a flat \(B\)-module and \(N \subseteq M\) a submodule. By Lemma 2.15 it suffices to show that \(N\) is \(A\)-flat. Since \(B\) is \(A\)-flat, \(M\) is also \(A\)-flat and hence its submodule \(N\) is \(A\)-flat by Lemma 2.24.
Let \((A_i)_{i \in I}\) be a family of rings and for every \(i\) an \(A_i\)-module \(M_i\). Let \(N\) be a \(\prod _{i \in I} A_i\)-submodule of \(\prod _{i \in I} M_i\). Then \(N \subseteq \prod _{i \in I} N_i\). If \(N\) is finitely generated, equality holds.
TBA.
Let \((A_i)_{i \in I}\) be a family of valuation rings. Then \(\prod _{i \in I} A_i\) is of weak dimension \(\le 1\).
Let \(I \subseteq \prod _{i \in I} A_i\) be a finitely generated ideal. Let \(I_i \subseteq A_i\) be the image of \(I\) in \(A_i\). By Lemma 2.26, \(I\) is the product of the \(I_i\). Since each \(A_i\) is a valuation ring, there exists \(f_i \in A_i\) such that \(I_i = (f_i)\). Hence \(I = (f)\) where \(f = (f_i)\). Let \(e \in A\) be the idempotent element defined by
Let \(A\) be of weak dimension \(\le 1\). Let \(B\) be a flat \(A\)-algebra such that \(A \to B\) is injective and an epimorphism of rings. Then \(A\) is integrally closed in \(B\).
Let \(x \in B\) be integral over \(A\) and \(A' = A[x]\). The map \(A \to A'\) is injective and finite, so to show \(A = A'\) it suffices to check that \(A \to A'\) is an epimorphism. By Lemma 2.24, \(A'\) is flat over \(A\). Hence the composition
is injective.
Let \(A\) be a domain and \(L\) an algebraic extension of \(\mathrm{Frac}(A)\). If \(A\) is integrally closed in \(L\), then there exists a cartesian diagram of rings
with \(S\) of weak dimension \(\le 1\) and \(S \to T\) a flat, injective epimorphism of rings.
TBA.
Let \(A\) be a domain and \(B\) a weakly étale \(A\)-algebra. Let \(L\) be an algebraic extension of \(\mathrm{Frac}(A)\) and assume that \(A\) is integrally closed in \(L\). Then \(B\) is integrally closed in \(B \otimes _{A} L\).
2.3 Fields
Let \(L/K\) be an extension of fields. If \(L \otimes _K L \to L\)is flat, then \(L\) is an algebraic separable extension of \(K\).
TBA.
Let \(K\) be a field. Suppose that \(K \to B\) is weakly étale, then
every finitely generated \(K\)-subalgebra of \(B\) is étale over K,
\(B\) is a filtered colimit of étale \(K\)-algebras, i.e. \(B\) is ind-étale.
It suffices to show the first statement. A field is absolutely flat ring by Lemma 2.11, hence \(B\) is a absolutely flat ring by Lemma 2.16. And \(B\) is reduced and every local ring is a field, by Lemma 2.18 and Lemma 2.19.
Let \(q \subset B\) be a prime. The ring map \(B \to B_q\) is weakly étale by Lemma 2.10, hence \(B_q\) is weakly étale over \(K\) (??). Thus \(B_q\) is a separable algebraic extension of \(K\) by Lemma 2.31.
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.
TBA.
Then every minimal prime \(p \subset A\) is the image of a prime \(q\) of \(B\), see Algebra, Lemma 00FK. Thus κ(p) as a subfield of Bq=κ(q) is separable algebraic over K. Hence every generic point of Spec(A) is closed (Algebra, Lemma 00GA). Thus dim(A)=0. Then A is the product of its local rings, e.g., by Algebra, Proposition 00KJ. Moreover, since A is reduced, all local rings are equal to their residue fields which are finite separable over K. This means that A is étale over K by Algebra, Lemma 00U3 and finishes the proof.
2.4 Local rings
Let \(A\) be a ring, let \(f: R \to R',\, g : S \to S'\) be two ring maps of \(A\)-algebras. Then the kernel of \(R \otimes _A S \to R' \otimes _A S'\) is generated by \(\ker f \otimes _A S\) and \(R \otimes _A \ker g\).
Omitted.
Let \(k\) be a field, \(A\) be a local \(k\)-algebra of dimension zero, i.e. there is a unique prime ideal in \(A\). Suppose that the residue field of \(A\) is identified with \(k\). Then \(A \otimes _k A\) is a local \(k\) algebra of dimension zero.
The unique prime ideal of \(A\) is the nilradical \(N\) of \(A\). By Lemma 2.33, the kernel of \(A \otimes _k A \to k = A/N\) is generated by \(N \otimes _k A\) and \(A \otimes _k N\), which are all consisiting of nilpotent elements. Thus the nilpotent ideal of \(A \otimes _k A\) is also the maximal ideal. This finishes the proof.
Let \(A \to B\) be a ring map. For all \(\mathfrak {p} \subset A\), there is a unique prime \(\mathfrak {q} \subset B\) lying over \(\mathfrak {p}\) and \(\kappa (\mathfrak {p}) = \kappa (\mathfrak {q})\). Then \(B \otimes _A B \to B\) is bijective on spectra.
It suffices to show that, for every prime \(p\) of \(A\), there is exactly one prime ideal \(q'\) of \(B \otimes _A B\) lying over \(p\). To detect this, we may assume \(A\) is a field by replacing \(A\) with \(\kappa (\mathfrak {p})\), \(B\) with \(\kappa (\mathfrak {p}) \otimes _A B\) which is a local ring. Now the goal follows from Lemma 2.34.
Let \(A \to B\) be a ring map that is bijective on spectra, as well as surjective and flat, then it is an isomorphism.
Recall that a pure ideal \(I\) in \(A\) is an ideal such that \(A/I\) is flat. (Use Ideal.Pure.) In our case, let \(I\) be the kernel of the map \(A \to B\). Then \(I\) is pure and have empty vanishing locus. But we know that pure ideals are determined by their vanishing locus (Ideal.zeroLocus_inj_of_pure) and zero ideal is also a pure ideal having the same vanishing locus with \(I\), thus \(I = 0\).
Let \(A \to B\) be a ring map. The following are equivalent:
\(A \to B\) is an epimorphism;
the two ring maps \(B \to B \otimes _A B\) are equal;
either of the ring maps \(B \to B \otimes _A B\) is an isomorphism, and
the ring map \(B \otimes _A B \to B\) is an isomorphism.
Omitted. Use Algebra.IsEpi.
Let \(A \to B\) be a faithfully flat ring epimorphism. Then \(A \to B\) is an isomrophism.
By ??, the map \(B \to B \otimes _A B\), which is a base change of \(A \to B\), is isomorphism. So is \(A \to B\) by faithfully flatness.
Let \(A \to B\) be a local homomorphism of local rings. For all \(\mathfrak {p} \subset A\), there is a unique prime \(\mathfrak {q} \subset B\) lying over \(\mathfrak {p}\) and \(\kappa (\mathfrak {p}) = \kappa (\mathfrak {q})\). Then \(A \to B\) is an isomorphism.
To see this, suppose that we are under the above hypothesis. This implies that \(\mu : B \otimes _A B \to B\) is bijective on spectra by Lemma 2.35, as well as surjective and flat (immediately by Definition 2.8). Hence it is an isomorphism by Lemma 2.36. Together with the fact that \(A \to B\) is a faithfully (by surjectivity on spectrum) flat, we can apply Lemma 2.38 to conclude \(A = B\).
Let \(I\) be a directed set. Let \(T_i, \, i \in I\) be a series of totally disconnected topological spaces. Then
is also totally disconnected.
Suppose \(x, y\) are two distinct points that lives in the same connected component. Then the projection of \(x, y\) will be always fall in the same connected components, thus equal. This is a contradiction.
Let \(B\) be a ind-étale algebra over some field \(K\). If there are two different prime ideals \(q_1\) and \(q_2\) in \(B\), then \(B\) has a nontrivial idempotent element \(e\) (i.i. \(e^2 = e\)).
Writing \(B\) as a filtered colimit \(\operatorname {colim}B_i\) of étale algebra over \(K\). Each \(B_i\) is a product of separable field extensions of \(K\). In particular, the spectrum of \(B_i\)’s are totally disconnected. By Lemma 2.40, the spectrum of \(B\) is again totally disconnected. Thus different prime ideals \(q_1\) and \(q_2\) livs in different connected components. This giving the desire nontrivial idempotent element.
Let \(B\) be a ind-étale algebra over some field \(K\). If there exists a prime ideal \(q\) of \(B\), such that the residue field of \(q\) is not \(K\), then \(\kappa (q) \otimes _A B\) has a nontrivial element.
Writing \(B\) as a filtered colimit \(\operatorname {colim}B_i\) of étale algebra over \(K\). (Each \(B_i\) is a product of separable field extensions of \(K\).) Let \(q_i\) be the inverse image of \(q\) in \(B_i\), \(\kappa (q_i)\) be the residue field of \(q_i\), then \(K_i\) is a separable field extensions of \(K\). Now \(\kappa (q) = \operatorname {colim}\kappa (q_i)\) is again a separable extension of \(K\). Then \(\kappa (q) \otimes _A B = \kappa (q) \otimes _K \kappa (q)\) splits. The conclusion follows.
Let \(R\) be a Henselian local ring with separably closed residue field and \(S\) a weakly étale local \(R\)-algebra with \(R \to S\) a local homomorphism. Let \(\mathfrak {p}\) be a prime ideal of \(R\) and \(L\) an algebraic field extension of \(\kappa (\mathfrak {p})\). Then \(L \otimes _{R} S\) has no non-trivial idempotents.
Let \(R'\) be the integral closure of \(R / \mathfrak {p}\) in \(L\). Since \(R \to R / \mathfrak {p} \to R'\) is integral, \(R'\) is local by Lemma 2.4. Moreover, the residue field \(\kappa (R')\) is an algebraic extension of \(\kappa (R)\) by Lemma 2.6. Since \(\kappa (R)\) is separably closed, the extension is purely inseparable. Hence \(R' \otimes _{R} S\) is local by Lemma 2.7. By Lemma 2.30 and since \(R'\) is integrally closed in \(L\), the tensor product \(R' \otimes _{R} S\) is integrally closed in \(L \otimes _{R} S\). Since \(R' \otimes _{R} S\) is local and any idempotent of \(L \otimes _{R} S\) is integral over \(R' \otimes _{R} S\), the latter can not have any non-trivial idemponent elements.
Let \(A \to B\) be a local homomorphism of local rings. If \(A\) is henselian, the residue field of \(A\) is separably closed, and \(A \to B\) is weakly étale, then \(A = B\).
It suffices to show that for all \(\mathfrak {p} \subset A\) there is a unique prime \(\mathfrak {q} \subset B\) lying over \(\mathfrak {p}\) and \(\kappa (\mathfrak {p}) = \kappa (\mathfrak {q})\) by Lemma 2.39
Note that the fibre ring \(\kappa (\mathfrak {p}) \otimes _A B\) is weakly etale over \(\kappa (\mathfrak {p})\) by Lemma 2.14, thus it is a colimit of étale extensions of \(\kappa (\mathfrak {p})\) by Theorem 2.32. If the conclusion does not hold, at least one of the following cases is true:
there exists more than one prime \(\mathfrak {q}\_ 1, \mathfrak {q}\_ 2\) lying over \(\mathfrak {p}\);
if \(\kappa (\mathfrak {p}) \neq \kappa (\mathfrak {q})\) for some \(\mathfrak {q}\).
In the first case, by Lemma 2.41; in the second case, by Lemma 2.42, we can always find some \(L\) then \(L \otimes _A B\) has a nontrivial idempotent for some (separable) algebraic field extension \(L/\kappa (\mathfrak {p})\).
Now the statement follows from Lemma 2.43.
Let \(A \to B\) be a local homomorphism of local rings. If \(A\) is strictly henselian, and \(A \to B\) is weakly étale, then \(A = B\).
This is simply Theorem 2.44 plus Lemma 1.47.
Let \(R \to S\) be weakly étale. Then there exists a faithfully flat ind-étale morphism \(S \to T\) such that the composition \(R \to T\) is ind-étale.
TBA.