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].
15.21 Descent of flatness along integral maps
First a few simple lemmas.
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_ i(x_ i) = 0 in S, see Algebra, Lemma 10.36.3. 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 10.39.8 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 10.36.17. 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 10.101.5 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 10.168.1 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 10.39.7. \square
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