The Étale Fundamental Group — Faithfully flat morphisms

A Faithfully flat morphisms

In this appendix, we document some important facts about flat and faithfully flat morphisms.


Let f:ABf: A \to B be a ring homomorphism. We say ff is flat if the functor

AB:A-ModB-Mod -\otimes _ A B: A\text{-Mod} \to B\text{-Mod}

is exact.


A morphism p:YXp: Y \to X is flat if for all yYy \in Y and x=f(y)x = f(y), the map OX,xOY,y\mathcal{O}_{X, x} \to \mathcal{O}_{Y, y} is flat. This in particular implies the pullback functor

p:QCoh(Y)QCoh(X) p^*: \operatorname{QCoh}(Y) \to \operatorname{QCoh}(X)

is exact.

If XX is quasi-compact and quasi-separated, then pp^* being exact implies pp being flat.


Compositions and pullbacks of flat maps are flat.

We only have to show that the pullback of flat maps is flat. Since flatness is local, it suffices to show that if f:ABf: A \to B is a flat map of rings and g:ACg:A \to C is any map of rings, then CBACC \to B \otimes _ A C is flat. But if MM_\bullet is a exact sequence of chain complexes, then BACCM=BAM, B \otimes _ A C \otimes _ C M_\bullet = B \otimes _ A M_\bullet , where we think of MM_\bullet as an AA-module via gg. So this is exact.


Let p:YXp: Y \to X be flat. Then the following are equivalent:

  1. pp^* is faithful, i.e. if h:FFh: \mathcal{F} \to \mathcal{F}' is a morphism of quasi-coherent sheaves over YY, and ph=0p^*h = 0, then h=0h = 0.

  2. If pF=0p^* \mathcal{F} = 0, then F=0\mathcal{F} = 0.

  3. pp^* reflects exactness, i.e. if a sequence F\mathcal{F}_\bullet is such that pFp^* \mathcal{F}_\bullet is exact, then so is F\mathcal{F}_\bullet .

  4. pp is surjective.

When these hold, we say pp is faithfully flat.

  • (1) \Rightarrow (2): Take h=id:FFh = \mathrm{id}: \mathcal{F} \to \mathcal{F}.

  • (2) \Rightarrow (3): Apply (2) to the homology groups of F\mathcal{F}_\bullet .

  • (3) \Rightarrow (1): h=0h = 0 iff FhF1F\mathcal{F} \overset {h}{\to } \mathcal{F}' \overset {1}{\to } \mathcal{F}' is exact.

  • (4) \Rightarrow (2): Take F̸=0\mathcal{F} \not= 0. We may assume F\mathcal{F} is in fact coherent, for F\mathcal{F} contains a coherent subsheaf G\mathcal{G} and pp^* preserves subsheaves by flatness. So if pG̸=0p^* \mathcal{G} \not= 0, then pF̸=0p^* \mathcal{F} \not= 0.

    Pick xXx \in X such that Fx̸=0\mathcal{F}_ x \not= 0. By surjectivity, there is a field kk and a map x~:SpeckX\tilde{x}: \operatorname{Spec}k \to X that sends the unique point to xx and has a lift to YY (e.g. by first picking a map to YY that hits a preimage of xx). This means the pullback Y×SpeckXY \times _{\operatorname{Spec}k} X is non-empty. Moreover, x~F̸=0\tilde{x}^* \mathcal{F} \not= 0 by Nakayama, and is free since Speck\operatorname{Spec}k is a field. So the pullback of F\mathcal{F} to Y×SpeckXY \times _{\operatorname{Spec}k} X is non-zero. Hence pF̸=0p^* \mathcal{F} \not= 0.

  • (2) \Rightarrow (4): Let pX\mathfrak {p} \in X, and SpecAX\operatorname{Spec}A \subseteq X an affine open containing p\mathfrak {p}. Set FSpecA=Ap/pAp~\mathcal{F}|_{\operatorname{Spec}A} = \widetilde{A_\mathfrak {p}/\mathfrak {p} A_\mathfrak {p}} and extend by zero. Then pF̸=0p^* \mathcal{F} \not= 0 implies there is some affine open SpecBY\operatorname{Spec}B \subseteq Y such that BAAppAp=Bp/pBp̸=0B \otimes _ A \frac{A_\mathfrak {p}}{\mathfrak {p}A_\mathfrak {p}} =B_\mathfrak {p}/\mathfrak {p}B_\mathfrak {p} \not= 0. Then a prime of Bp/pBpB_\mathfrak {p}/\mathfrak {p}B_\mathfrak {p} is a prime of BB that gets mapped to p\mathfrak {p} under pp.

Using (4), it is clear that


Compositions and pullbacks of faithfully flat maps are faithfully flat.

We say a property P of morphisms satisfies fpqc descent if whenever we have a pullback diagram

      Y \times_X X' \ar[d, "p'"] \ar[r, "f'"] & Y \ar[d, "p"]\\
      X' \ar[r, "f"] & X

with ff faithfully flat, then pp has property P iff pp' does.


Flat morphisms satisfy fpqc descent.

Suppose pp' is flat. If we have a sequence F\mathcal{F}_\bullet of quasi-coherent sheaves on XX, then pfFp'^* f^* \mathcal{F}_\bullet is exact since ff and pp' are flat, and since pf=fpp'^* f^* = f'^* p^*, we know pFp^* \mathcal{F}_\bullet is exact by faithfulness.


Finite morphisms satisfy fpqc descent.

Proof sketch

The pullback of a finite morphism is clearly finite. For the other direction, We will prove the affine version. The gluing step part (which is the hard step) is annoying and will be omitted.

Suppose that f:RSf: R \to S is faithfully flat and MM is an RR-module. We want to show that MRSM \otimes _ R S being finitely-generated implies MM is finitely generated.

Suppose y1,ymy_1, \ldots y_ m generate MRSM \otimes _ R S, and yj=xi,jfi,jy_ j = \sum x_{i, j} \otimes f_{i, j}. Then the xi,jx_{i, j} generate MM, since they generate MRSM \otimes _ R S as an SS-module and RSR \to S is faithfully flat, hence reflects surjectivity.