
## 15.21 Descent of flatness along integral maps

First a few simple lemmas.

Lemma 15.21.1. Let $R$ be a ring. Let $P(T)$ be a monic polynomial with coefficients in $R$. Let $\alpha \in R$ be such that $P(\alpha ) = 0$. Then $P(T) = (T - \alpha )Q(T)$ for some monic polynomial $Q(T) \in R[T]$.

Proof. By induction on the degree of $P$. If $\deg (P) = 1$, then $P(T) = T - \alpha$ and the result is true. If $\deg (P) > 1$, then we can write $P(T) = (T - \alpha )Q(T) + r$ for some polynomial $Q \in R[T]$ of degree $< \deg (P)$ and some $r \in R$ by long division. By assumption $0 = P(\alpha ) = (\alpha - \alpha )Q(\alpha ) + r = r$ and we conclude that $r = 0$ as desired. $\square$

Lemma 15.21.2. Let $R$ be a ring. Let $P(T)$ be a monic polynomial with coefficients in $R$. There exists a finite free ring map $R \to R'$ such that $P(T) = (T - \alpha )Q(T)$ for some $\alpha \in R'$ and some monic polynomial $Q(T) \in R'[T]$.

Proof. Write $P(T) = T^ d + a_1T^{d - 1} + \ldots + a_0$. Set $R' = R[x]/(x^ d + a_1x^{d - 1} + \ldots + a_0)$. Set $\alpha$ equal to the congruence class of $x$. Then it is clear that $P(\alpha ) = 0$. Thus we win by Lemma 15.21.1. $\square$

Lemma 15.21.3. Let $R \to S$ be a finite ring map. There exists a finite free ring extension $R \subset R'$ such that $S \otimes _ R R'$ is a quotient of a ring of the form

$R'[T_1, \ldots , T_ n]/(P_1(T_1), \ldots , P_ n(T_ n))$

with $P_ i(T) = \prod _{j = 1, \ldots , d_ i} (T - \alpha _{ij})$ for some $\alpha _{ij} \in R'$.

Proof. Let $x_1, \ldots , x_ n \in S$ be generators of $S$ over $R$. For each $i$ we can choose a monic polynomial $P_ i(T) \in R[T]$ such that $P(x_ i) = 0$ in $S$, see Algebra, Lemma 10.35.3. Say $\deg (P_ i) = d_ i$. By Lemma 15.21.2 (applied $\sum d_ i$ times) there exists a finite free ring extension $R \subset R'$ such that each $P_ i$ splits completely:

$P_ i(T) = \prod \nolimits _{j = 1, \ldots , d_ i} (T - \alpha _{ij})$

for certain $\alpha _{ik} \in R'$. Let $R'[T_1, \ldots , T_ n] \to S \otimes _ R R'$ be the $R'$-algebra map which maps $T_ i$ to $x_ i \otimes 1$. As this maps $P_ i(T_ i)$ to zero, this induces the desired surjection. $\square$

Lemma 15.21.4. Let $R$ be a ring. Let $S = R[T_1, \ldots , T_ n]/J$. Assume $J$ contains elements of the form $P_ i(T_ i)$ with $P_ i(T) = \prod _{j = 1, \ldots , d_ i} (T - \alpha _{ij})$ for some $\alpha _{ij} \in R$. For $\underline{k} = (k_1, \ldots , k_ n)$ with $1 \leq k_ i \leq d_ i$ consider the ring map

$\Phi _{\underline{k}} : R[T_1, \ldots , T_ n] \to R, \quad T_ i \longmapsto \alpha _{ik_ i}$

Set $J_{\underline{k}} = \Phi _{\underline{k}}(J)$. Then the image of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is equal to $V(\bigcap J_{\underline{k}})$.

Proof. This lemma proves itself. Hint: $V(\bigcap J_{\underline{k}}) = \bigcup V(J_{\underline{k}})$. $\square$

The following result is due to Ferrand, see [Ferrand].

Lemma 15.21.5. Let $R \to S$ be a finite injective homomorphism of Noetherian rings. Let $M$ be an $R$-module. If $M \otimes _ R S$ is a flat $S$-module, then $M$ is a flat $R$-module.

Proof. Let $M$ be an $R$-module such that $M \otimes _ R S$ is flat over $S$. By Algebra, Lemma 10.38.8 in order to prove that $M$ is flat we may replace $R$ by any faithfully flat ring extension. By Lemma 15.21.3 we can find a finite locally free ring extension $R \subset R'$ such that $S' = S \otimes _ R R' = R'[T_1, \ldots , T_ n]/J$ for some ideal $J \subset R'[T_1, \ldots , T_ n]$ which contains the elements of the form $P_ i(T_ i)$ with $P_ i(T) = \prod _{j = 1, \ldots , d_ i} (T - \alpha _{ij})$ for some $\alpha _{ij} \in R'$. Note that $R'$ is Noetherian and that $R' \subset S'$ is a finite extension of rings. Hence we may replace $R$ by $R'$ and assume that $S$ has a presentation as in Lemma 15.21.4. Note that $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is surjective, see Algebra, Lemma 10.35.17. Thus, using Lemma 15.21.4 we conclude that $I = \bigcap J_{\underline{k}}$ is an ideal such that $V(I) = \mathop{\mathrm{Spec}}(R)$. This means that $I \subset \sqrt{(0)}$, and since $R$ is Noetherian that $I$ is nilpotent. The maps $\Phi _{\underline{k}}$ induce commutative diagrams

$\xymatrix{ S \ar[rr] & & R/J_{\underline{k}} \\ & R \ar[lu] \ar[ru] }$

from which we conclude that $M/J_{\underline{k}}M$ is flat over $R/J_{\underline{k}}$. By Lemma 15.16.1 we see that $M/IM$ is flat over $R/I$. Finally, applying Algebra, Lemma 10.100.5 we conclude that $M$ is flat over $R$. $\square$

Lemma 15.21.6. Let $R \to S$ be an injective integral ring map. Let $M$ be a finitely presented module over $R[x_1, \ldots , x_ n]$. If $M \otimes _ R S$ is flat over $S$, then $M$ is flat over $R$.

Proof. Choose a presentation

$R[x_1, \ldots , x_ n]^{\oplus t} \to R[x_1, \ldots , x_ n]^{\oplus r} \to M \to 0.$

Let's say that the first map is given by the $r \times t$-matrix $T = (f_{ij})$ with $f_{ij} \in R[x_1, \ldots , x_ n]$. Write $f_{ij} = \sum f_{ij, I} x^ I$ with $f_{ij, I} \in R$ (multi-index notation). Consider diagrams

$\xymatrix{ R \ar[r] & S \\ R_\lambda \ar[u] \ar[r] & S_\lambda \ar[u] }$

where $R_\lambda$ is a finitely generated $\mathbf{Z}$-subalgebra of $R$ containing all $f_{ij, I}$ and $S_\lambda$ is a finite $R_\lambda$-subalgebra of $S$. Let $M_\lambda$ be the finite $R_\lambda [x_1, \ldots , x_ n]$-module defined by a presentation as above, using the same matrix $T$ but now viewed as a matrix over $R_\lambda [x_1, \ldots , x_ n]$. Note that $S$ is the directed colimit of the $S_\lambda$ (details omitted). By Algebra, Lemma 10.162.1 we see that for some $\lambda$ the module $M_\lambda \otimes _{R_\lambda } S_\lambda$ is flat over $S_\lambda$. By Lemma 15.21.5 we conclude that $M_\lambda$ is flat over $R_\lambda$. Since $M = M_\lambda \otimes _{R_\lambda } R$ we win by Algebra, Lemma 10.38.7. $\square$

Comment #3555 by Laurent Moret-Bailly on

Shouldn't the title be "Descent of flatness..."?

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