Étale Cohomology

Report on Meeting 3

The previous report lied; it included descriptions of descent theory that we hadn’t actually finished at that time. We started this week by going over that material. We began by spending more time than I thought humanly possible on the following facts.

Lemma 1. Let A → B be a ring homomorphism. Let N and M be arbitrary B-modules. Then N ⊗_{A} M is a B ⊗_{A} B module in a natural way.

Proof: The natural action is given by the formula

(b ⊗ c) · (n ⊗ m) = bn ⊗ cm.

Lemma 2. Let A → B be a ring homomorphism. Let p_{i} : B → B ⊗ B denote the map that puts a 1 in the i^{th} slot. Let N be any B module. Then there are isomorphisms of B ⊗ B-modules

N ⊗_{p_{1}} B ⊗_{A} B⟶^{q_{1}}B ⊗_{A} N

and

N ⊗_{p_{2}} B ⊗_{A} B⟶^{q_{2}}N ⊗_{A} B.

Proof: The modules on the right hand sides have the actions coming from the previous lemma. In order to define the first map, observe that the identification of tensor elements using the map p_{1}(b) = 1 ⊗ b allows us to normalize general tensors by

n ⊗ (b ⊗ c) = n ⊗ (1 ⊗ c)(b ⊗ 1) = cn ⊗ (b ⊗ 1).

Now the B ⊗_{A} B structure comes from

(d ⊗ e) · (n ⊗ (b ⊗ c)) = n ⊗ (db ⊗ ec) = ecn ⊗ (db ⊗ 1).

So, we should clearly define

q_{1}(n ⊗ b ⊗ c) = b ⊗ cn.

We check that the B ⊗_{A} B structure is preserved by computing

q_{1}((d ⊗ e) · (n ⊗ b ⊗ c)) = q_{1}(n ⊗ db ⊗ ec)
= db ⊗ ecn
= (d ⊗ e) · (b ⊗ cn)
= (d ⊗ e) · q_{1}(n ⊗ b ⊗ c).

Let’s repeat this computation (being slightly masochistic) for the other map. First, the identification along p_{2} gives

n ⊗ (b ⊗ c) = n ⊗ (b ⊗ 1)(1 ⊗ c) = bn ⊗ (1 ⊗ c).

Next, the B ⊗_{A} B structure is given by

(d ⊗ e) · (n ⊗ (b ⊗ c)) = n ⊗ (db ⊗ ec) = dbn ⊗ (1 ⊗ ec).

Again, the definition is clear:

q_{2}(n ⊗ b ⊗ c) = bn ⊗ c.

Now the check that this map preserves the B ⊗_{A} B structures reduces to

q_{2}((d ⊗ e) · (n ⊗ b ⊗ c)) = q_{(}n ⊗ db ⊗ ec)
= dbn ⊗ ec
= (d ⊗ e) · (bn ⊗ c)
= (d ⊗ e) · q_{2}(n ⊗ b ⊗ c).

Remark 3. We then asserted that similar things happen if you use the various maps p_{i} : B ⊗_{A} B → B ⊗_{A} B ⊗_{A} B that keep the elements in order, but insert 1 in the i^{th} slot. (We did not prove this. Masochism is one thing; stupidity is something else entirely.) We finally agreed that I was allowed to put B ⊗_{A} B ⊗_{A} B module structures on a triple tensor product of B modules N ⊗_{A} M ⊗_{A} P by letting each copy of B act on the corresponding factor in the tensor product, and we then used these structures to complete the discussion given in Remark 15 of the second report. The main point that needed to be added was the proof of the commutativity of the following diagram:

N ⊗_{A} B \overrightarrow{⟶}_{β⊗1}^{α⊗1} B ⊗_{A} N ⊗_{A} B
↓^{φ}  ↓^{φ_{1}}
B ⊗_{A} N \overrightarrow{⟶}_{p_{2}⊗1}^{p_{1}⊗1} B ⊗_{A} B ⊗_{A} N

where α(n) = 1 ⊗ n, β(n) = φ(n ⊗ 1), p_{1}(b) = 1 ⊗ b, and p_{2}(b) = b ⊗ 1. The underlying assumptions are (i) that φ is an isomorphism of B ⊗_{A} B-modules, and (ii) that φ induces maps φ_{i} (by putting the identity map on B into the i^{th} slot) which satisfy φ_{2} = φ_{1}φ_{3}.

Commutativity using the top arrows is given by the computation

φ_{1}((α ⊗ 1)(n ⊗ b)) = φ_{1}(1 ⊗ n ⊗ b) = 1 ⊗ φ(n ⊗ b) = (p_{1} ⊗ 1)(φ(n ⊗ b)).

Commutativity using the bottom arrows is more involved.

φ_{1}((β ⊗ 1)(n ⊗ b)) = φ_{1}(φ(n ⊗ 1) ⊗ b)

= φ_{1}(φ_{3}(n ⊗ 1 ⊗ b)) = φ_{2}(n ⊗ 1 ⊗ b) = (p_{2} ⊗ 1)(φ(n ⊗ b)).

Remark 4. We finally began what I had originally intended to be the main subject for the day: what does it mean for a map to be unramified? The simplest case of an unramified map is a separable field extension. Recall that a finite extension of fields is called separable if it is generated by separable elements, and an element is separable if its minimal polynomial has distinct roots. In other words, if f(x) is an irreducible polynomial over a field F, then the extension F [x]/(f(x)) is separable if and only if gcd(f(x), f′(x)) = 1. (This is an easy computation.)

Definition 5. Let F be a field. An F -algebra A is called separable if Ā = A ⊗ F̄ has trivial Jacobson radical.

Proposition 6. Let A be a finite F -algebra. Then the following are equivalent

(i) A is a separable F -algebra. (ii) Ā is isomorphic to a finite product of copies of F̄. (iii) A is isomorphic to a finite product of finite separable field extensions of F. (iv) The discriminant of any F -basis of A is nonzero.

Proof: (i ⇒ ii) Look at the map Ā → ∏ Ā/M over maximal ideals of Ā. This map has no kernel (since the Jacobson radical vanishes) and is onto by the Chinese Remainder Theorem.

(ii ⇒ iii) Let J be the Jacobson radical of A. By the Chinese Remainder Theorem, A/J is isomorphic to a product of finite field extensions K_{i}. Let [K_{i} : F]_{s} denote the separable degree of the extension. Then

\begin{aligned}[Ā : F̄] & = |HomF̄(Ā, F̄)| = |Hom_{F} (A, F̄)| \\ & = ∑[K_{i} : F]_{s} ≤ ∑[K_{i} : F] ≤ [A : F] \end{aligned}

Here the first equality uses (ii). Since the extreme terms are equal, all the intermediate inequalites are actually equalities. In particular, each of the fields in the product is separable, and J must be trivial.

(iii ⇒ iv) The discriminant of A is the product of the discriminants of the finite separable fields extensions that are factors in the product.

(iv ⇒ i) A and Ā have the same discriminant. If x is in the Jacobson radical of Ā, then all xa are nilpotent, so T r(xa) = 0 and thus x = 0.

Definition 7. A local homomorphism A → B of local rings is called unramified if the maximal ideals satisfy M_{A}B = M_{B} and the residue field extension A/M_{A} → B/M_{B} is a finite separable field extension.

Definition 8. A morphism of schemes f : X → Y, which is locally of finite type, is called unramified if each of the local homomorphisms O_{Y,y} → O_{X,f(y)} is unramified.

Proposition 9. Let f : X → Y be a morphsim of schems that is locally of finite type. Then the following are equivalent

(i) f is unramified (ii) All the fibres of f are unramified. (iii) All the geometric fibres of f are unramified.

(iv) For all x ∈ X, the fibre Y_{x} is covered by spectra of finite separable k(x)-algebras.

(v) For all x ∈ X, the fibre Y_{x} is a disjoint union of spectra of finite separable field extensions of k(x).

Proof: These are local properties. Reduce to the affine case, and mimic the proofs in the last proposition.

Remark 10. It is a fundamental fact of life that the correct way to think about unramified extensions is in terms of differentials. So, we need some more definitions.

Definition 11. Let A → B be a ring homomorphism, and let M be a B-module. An A-derivation of B with values in M is a map D : B → M satisfying

(i) D(a) = 0 for all a ∈ A (ii) D(b + c) = D(b) + D(c) for all b, c ∈ B. (iii) D(bc) = bD(c) + cD(b) for all b, c ∈ B.

Remark 12. (Trite) These relations are clearly chosen so they look like the elementary properties of the usual derivative.

Definition 13. Let A → B be a ring homomorphism. Define the module of differential forms of B relative to A, usually written Ω_{B/A}, to be the universal object for A-derivations to B.

Remark 14. Obviously, we should add the phrase “if it exists” to the end of the previous definition. (But we won’t.) The easy way to see that it exists is to start with the free B-module generated by symbols db for each b ∈ B, and kill the submodule generated by relations (i), (ii), and (iii). A more interesting way to see the existence of the module of differentials is as follows.

Proposition 15. Let B be an A-algebra, and define I to be the kernel of the map B ⊗_{A} B → B by multiplication. Then I/I^{2} ≈ Ω_{B/A}.

Proof: Define a map δ : B → I/I^{2} by δ(b) = b ⊗ 1 − 1 ⊗ b. It is straightforward to check that this is a well-defined A-derivation of B. So, by the universal property, there is a natural map Ω_{B/A} → I/I^{2}.

To define the inverse map, we need more convenient generators for I. The slickest way to get them is to note that there is an exact sequence

⋯ B^{⊗3} → B^{⊗2} → B → 0

where the map B^{⊗i+2} → B^{⊗i+1} is given by the formula

δ(b_{0} ⊗ ⋯ ⊗ b_{i+1}) = ∑_{k=0}^{i}(−1)^{k}b_{0} ⊗ ⋯ ⊗ b_{k}b_{k+1} ⊗ ⋯ ⊗ b_{i+1}

Exactness follows using the contracting homotopy s(b_{0} ⊗ ⋯ ⊗ b_{i}) = b_{0} ⊗ ⋯ ⊗ b_{i} ⊗ 1. (This exact sequence is called the acyclic Hochschild complex.)

From this exact sequence, we see that I is generated by elements of the form xy ⊗ z − x ⊗ yz. Using the B-module structure on I/I^{2}, we can assumer x = z = 1 and send the resulting generator y ⊗ 1 − 1 ⊗ y to dy in the module of differentials. It is straightforward to check that the two maps so defined are inverses.

Proposition 16. Let B and C be A-algebras, and let D = B ⊗_{A} C. Then there is an isomorphism Ω_{E/C} ≈ Ω_{B/A} ⊗_{B} E.

Proof: The natural map goes from Ω_{B/A} ⊗_{B} E to Ω_{E/C}, and takes db ⊗ e to ed(b ⊗ 1). This is surjective since d(b ⊗ c) = (b ⊗ 1)d(1 ⊗ c) + (1 ⊗ c)d(b ⊗ 1) and d(1 ⊗ c) is zero in the module of differentials relative to C. Injectivity follows from the description of the generators of I in the previous proof. (Well, to be more precise, you check that the acyclic Hochschild sequence for E as a C-algebra is obtained by tensoring the sequence for B as an A-algebra with C, and then use the right exactness of tensor products.)

Remark 17. Let A be a ring, and take B = A[x_{1}, …, x_{n}] be a polynomial ring. It is fairly easy to show that

Ω_{B/A} = Bdx_{1} ⊕ Bdx_{2} ⊕ ⋯ ⊕ Bdx_{n},

so the module of differentials is free. The universal derivation B → Ω_{B/A} is given by partial derivatives. The fundamental importance of this computation can be seen from the following exact sequence.

Proposition 18. Let B be an A-algebra, and let C = B/J be a quotient ring of B. Then there is an exact sequence

J/J^{2} → Ω_{B/A} ⊗ C → Ω_{C/A} → 0

Proof: The first map sends j ↦ dj ⊗ 1. Exactness is a straightforward computation that you should do for yourself.

Remark 19. For historical (and logical) reasons, this result is always called the second exact sequence. I put it first out of perversity. Well, actually, I did this because it’s the one that gets used in real computations. We can now compute the module of differentials on any finitely generated A-algebra C; just write it as a quotient of a polynomial algebra B. Then the middle term of the second exact sequence is a free C module generated by the dx_{i}. To get the moduile we want, we have to kill the differentials coming from the polynomial equations that define C. In particular, we get the following result.

Corollary 20. Let f : Y → X be a morphism of schemes that is locally of finite type. Then the following are equivalent:

(i) f is unramified (ii) Ω_{Y/X} = 0

Proof: Well, I’ve been lackadaisical; I haven’t told you what differentials are for schemes. So, you take the diagonal morphism Δ : Y → Y ×_{X} Y. This is locally a closed immersion. So, take an open subscheme U ⊂ Y ×_{X} Y in which Y sits as a closed subscheme defined by some ideal sheaf I, and take Ω = Δ^{∗}(I/I^{2}). This definition is compatible with the definition used in the affine case (since the diagonal morphism is the geometric version of the multiplication map). Now differentials localize properly, so the result eventually reduces to the computation of the module of differentials in a finite separable field extension. Using the Principal Element Theorem, we reduce to the case of F [x]/(f(x)) as an extension of F, where the module of differentials vanishes if and only if gcd(f(x), f′(x)) = 1.

For completeness, we should also write down the first exact sequence.

Proposition 21. Let A → B → C be any composable pair of ring homomorphisms. Then there is an exact sequence

Ω_{B/A} ⊗_{B} C → Ω_{C/A} → Ω_{C/B} → 0.

Proof: The maps are obvious; the computation is easy; but it’s late and I’m tired. Do it yourself.