Characterizing fppf-algebras

abstract-algebraalgebraic-geometrycommutative-algebrahomological-algebra

Background:

All rings are commutative with $1$. Let $R$ be a ring. We say that an $R$-algebra $\varphi : R \to S$ is an fppf-$R$-algebra (see e.g. Jantzen "Representations of Algebraic Groups") if it is faithfully flat (as an $R$-module) and finitely presented (as an $R$-algebra). The latter condition means that (as an $R$-algebra) $S$ is isomorphic to $R[x_1, \ldots, x_n] / (f_1 , \ldots, f_m)$ for some $n$ and $m$, and some polynomials $f_i$.

Question:

Is there a nice characterization of when a finitely presented $R$-algebra is faithfully flat in terms of the polynomials $f_i$ (or the ideal they generate)? Or if that's too much to ask, what can be said about these polynomials?

Here are some easy observations along these lines:

Observations:

  1. Since faithfully flat ring homomorphisms are injective, we must have $R \cap (f_1 , \ldots, f_m) = 0$ inside $R[x_1, \ldots, x_n]$.
  2. (A re-phrasing of the previous observation.) Since a quotient $R \to R / I$ is faithfully flat if and only if $I = 0$ ($I \otimes_R R / I = 0$ hence by faithful flatness $I = 0$), we cannot have $(x_1, \ldots, x_n) \subsetneq (f_1, \ldots, f_m)$.
  3. If $f$ is monic, $R[x] / (f)$ is an fppf-$R$-algebra.
  4. $R[x_1, \ldots, x_n]$ is an fppf-$R$-algebra.

Best Answer

Answer: You may have seen the "flatness criteria" in Milnes "Etale cohomology" (2.7d): Let $f:A\rightarrow B$ be be a flat map of rings with $A \neq 0$. It follows $f$ is faithfully flat iff for any maximal ideal $\mathfrak{m}\subseteq A$ it follows $f(\mathfrak{m})B \subsetneq B$ is a strict ideal. In your case let

$$0 \rightarrow I:=(f_1,..,f_m) \rightarrow R[x_1,..,x_n] \rightarrow B:=R[x_i]/I \rightarrow 0$$

and assume $f:R \rightarrow B$ is flat. Use the functor $-\otimes_R R/\mathfrak{m}$ with $\mathfrak{m} \subseteq R$ a maximal ideal. It follows you get the sequence

$$I\otimes_R R/\mathfrak{m}:=(\overline{f_1},..,\overline{f_m}) \rightarrow (R/\mathfrak{m})[x_1,..,x_n] \rightarrow R/\mathfrak{m}\otimes_R B \rightarrow 0$$

hence there is an isomorphism

$$ R/\mathfrak{m}\otimes_R B \cong (R/\mathfrak{m})[x_1,..,x_n]/(\overline{f_1},..,\overline{f_m}).$$

Here the notation $(\overline{f_1},..,\overline{f_m})$ means you reduce the coefficients of the polynomials $f_i$ modulo the maximal ideal.

Question: "Is there a nice characterization of when a finitely presented R-algebra is faithfully flat in terms of the polynomials fi (or the ideal they generate)?"

The map $f$ is faithfully flat iff for any maximal ideal $\mathfrak{m} \subseteq R$ it follows the ideal

$$(\overline{f_1},..,\overline{f_m})\subseteq (R/\mathfrak{m})[x_1,..,x_n] $$

is not the unit ideal, which is a criterion on the ideal $I$.

Example: Let $A:=\mathbb{Z}[x,y]$ and let $f(x,y)\in A$ be any "non constant polynomial and let $F(x,y):=pf(x,y)-1$ where $p\in \mathbb{Z}$ is any prime number. It follows the map $f:\mathbb{Z} \rightarrow B:=A/(F)$ has the propety that

$$\mathbb{Z}/p\mathbb{Z} \otimes_{\mathbb{Z}} B \cong (0)$$

is the zero ring. If $\mathfrak{m}:=(p) \subseteq \mathbb{Z}$ it follows the ideal $f(\mathfrak{m})B:=(p)B \subseteq B$ contains the element

$$pf(x,y) \cong -1$$

which is a unit. Hence $f(\mathfrak{m})B =B$. Hence the corresponding map

$$\pi: Spec(B) \rightarrow Spec(\mathbb{Z})$$

is not surjective: $(p) \notin Im(\pi)$.

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