Faithful Flatness

Definition 1.3. Let Ag>s be the full subcategory of Ag,d consisting of those objects (5,X, A) of A9id such that there exists a faithfully flat T — S such that Ker(A)r zz K( )t-... 

We put Z = G/S principal homogeneous space under S(a)z over Z. Thus we can use EGA IV 2.7.1 to see that the morphism G Z is of finite presentation. This allows us to use SGA Expose V Proposition 9.1 to conclude that Z is of finite type over Spec(Z). Finally, we leave it to the reader to show that Z is separated. 

The ring extension Ap — R is faithfully flat (since R is the completion of the strict henselization of, 4p) and we know by Proposition 1.7.2 and the result of Crick mentioned above that the map... 

Proof. If there exists a faithfully flat morphism of schemes T S such that Mt satisfies the conclusions of Proposition 7.6 then this is true for Ms also. Thus we may assume S is the spectrum of a local ring and the next lemma finishes the proof. ... 

If the residue field of R is not perfect then there exists a faithfully flat homomorphism of... 

The A-modu e M is said to be faithfully flat if for every sequence of A-modules N N — N" the sequence... 

X) if B in an A-algebra and if there exists a B-module fi which is faithfully flat, then the morphism Spec(B) — Spec(A) is surjective. 

The last morphism is an isomorphism, and Tor,(6,It) (0) because 6 is A-flat this follows immediately from the flatness of A — B and of B — 6, and from the faithful flatness of A — A. Hence k - (0). Applying Nakayama s lemma as before we deduce that I = (0), equivalently that... 

Theorem. Let A- Bbe faithfully flat. Then the image ofMinM B consists of those elements having the same image under the two maps M B- M B B sending m b to m b 1 and m l b respectively. 

Suppose for illustration that A and B are rings of functions on closed sets in k", with k = k. The maximal ideals P in A then correspond to points x in the set. If PB 4 B, some maximal ideal of B contains P, and the corresponding point maps to x. Thus when A - B is flat, the extra condition involved in faithful flatness is precisely surjectivity on the closed sets. Condition (2) is the generalization of that to arbitrary rings. 

Theorem. Let k be a field, A B finitely generated k-algebras with A an integral domain. Then there are nonzero elements a in A and b in B such that that the map of localizations Aa -> Bb is faithfully flat. 

Proof. We proceed by successive localizations, eliminating at each step a proper closed set on which something goes wrong eventually we reach an extension with structure known so explicitly that faithful flatness will be obvious. 

The first and last stages here are free module extensions, while the middle one is a localization thus Aa - Bb is flat. Let P now be a maximal ideal of A . Then PAjxlt..., xr] does not contain g, since one coefficient is invertible in A . Hence PAa[xu xr], is a proper ideal, and there is a maximal ideal of Ajxu xr]g lying over P. Thus Aa-+ 4a[xt,..., xr]g is faithfully flat. And the last stage of the extension is faithfully flat because it is free as a module. ... 

---