Lemma 37.18.1. Let A be a valuation ring. Let A \to B is a local homomorphism of local rings which is essentially of finite type. Let u : N \to M be a map of finite B-modules. Assume M is flat over A and \overline{u} : N/\mathfrak m_ A N \to M/\mathfrak m_ A M is injective. Then u is injective and M/u(N) is flat over A.
37.18 Flat modules and relative assassins
In this section we will prove that the support of a flat module is (in some sense) equidimensional over the base in geometric situations. For the Noetherian case we refer the reader to [IV Proposition 12.1.1.5, EGA]. First, we prove two helper lemmas.
Proof. We will deduce this lemma from Algebra, Lemma 10.128.4 (please note that we exchanged the roles of M and N). To do the reduction we will use More on Algebra, Lemma 15.25.7 to reduce to the finitely presented case.
By assumption we can write B as a quotient of the localization of a polynomial algebra P = A[x_1, \ldots , x_ n] at a prime ideal \mathfrak q. Then we can think of u : N \to M as a map of finite P_\mathfrak q-modules. Hence we may and do assume that B is essentially of finite presentation over A.
Next, the B-module N is finite but perhaps not of finite presentation. Write N = \mathop{\mathrm{colim}}\nolimits N_\lambda as a filtered colimit of finitely presented B-modules with surjective transition maps. For example choose a presentation 0 \to K \to B^{\oplus r} \to N \to 0, write K as the union of its finite submodules K_\lambda , and set N_\lambda = \mathop{\mathrm{Coker}}(K_\lambda \to B^{\oplus r}). The module N/\mathfrak m_ A N is of finite presentation over the Noetherian ring B/\mathfrak m_ A B. Hence for \lambda large enough we have N_\lambda /\mathfrak m_ A N_\lambda = N/\mathfrak m_ A N. Now, if we can show the lemma for the composition u_\lambda : N_\lambda \to M, then we conclude that N_\lambda = N and the result holds for u. Hence we may and do assume N is of finite presentation over B.
By More on Algebra, Lemma 15.25.7 the module M is of finite presentation over B. Thus all the assumptions of Algebra, Lemma 10.128.4 hold and we conclude. \square
Lemma 37.18.2.reference Let f : X \to S be a morphism of schemes. Let y \in X be a point with image t \in S. Denote Y \subset X the closure of \{ y\} viewed as an integral closed subscheme of X. Let s \in S and let x \in Y_ s be a generic point of an irreducible component of Y_ s. There exists a cartesian diagram
\xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ S' \ar[r]^ g & S }
with the following properties:
S' is the spectrum of a valuation ring with generic point t' and closed point s',
g(t') = t and g(s') = s,
there exists a point y' \in X'_{t'} which is a generic point of an irreducible component of (S' \times _ S Y)_{t'} = Y_ t \times _ t t' and satisfies g'(y') = y,
denoting Y' \subset X' the closure of \{ y'\} viewed as an integral closed subscheme of X' there exists a point x' \in Y'_{s'} which is a generic point of an irreducible component of Y'_{s'} with g'(x') = x.
Proof. We choose a valuation ring R, we set S' = \mathop{\mathrm{Spec}}(R) with generic point t' and closed point s', and we choose a morphism h : S' \to X with h(t') = y and h(s') = x. See Schemes, Lemma 26.20.4. Set g = f \circ h so that g(t') = t and g(s') = s. Consider the base change
\xymatrix{ X' \ar[r]_{g'} \ar[d] & X \ar[d] \\ S' \ar@/^1em/[u]^\sigma \ar[r]^-g & S }
We obtain a section \sigma of the base change such that h = g' \circ \sigma .
Of course \sigma factors through the base change S' \times _ S Y of Y as h factors through Y. Let y' \in X'_{t'} \subset X' be the generic point of an irreducible component of the fibre
(S' \times _ S Y)_{t'} = Y_ t \times _ t t'
containing the point \sigma (t'), i.e., such that y' \leadsto \sigma (t'). Since g'(y') \in Y_ t and g(y') \leadsto g(\sigma (t')) = y we find that g'(y') = y because y is the generic point of the fibre Y_ t. Denote Y' \subset X' the closure of \{ y'\} in X' viewed as an integral closed subscheme. Then \sigma factors through Y' as \sigma (t') \in Y'. Choose a generic point x' \in Y'_{s'} of an irreducible component of Y'_{s'} which contains \sigma (s'), i.e., we get x' \leadsto \sigma (s') and hence g'(x') \leadsto g'(\sigma (s')) = x. Again as x is a generic point of an irreducible component of Y_ s by assumption and as g'(Y') \subset Y we conclude that g'(x') = x. \square
Lemma 37.18.3. Let f : X \to S be a morphism of schemes which is locally of finite type. Let \mathcal{F} be a quasi-coherent finite type \mathcal{O}_ X-module. Let y \in \text{Ass}_{X/S}(\mathcal{F}) with image t \in S. Denote Y \subset X the closure of \{ y\} in X viewed as an integral closed subscheme. Let s \in S and let x \in Y_ s be a generic point of an irreducible component of Y_ s. If \mathcal{F} is flat over S at x, then x \in \text{Ass}_{X/S}(\mathcal{F}) and \dim _ x(Y_ s) = \dim (Y_ t).
Proof. Choose a diagram as in Lemma 37.18.2. Set \mathcal{F}' = (g')^*\mathcal{F}. Divisors, Lemma 31.7.3 implies that y' \in \text{Ass}_{X'/S'}(\mathcal{F}'). By our choice of y' we also see that \dim (Y'_{t'}) = \dim (Y_ t), see for example Algebra, Lemma 10.116.7. By Algebra, Lemma 10.125.9 we see that Y'_{s'} is equidimensional of dimension equal to \dim (Y_ t). Since \mathcal{F} is flat at x over S we see that \mathcal{F}' is flat at x' over S', see Morphisms, Lemma 29.25.7.
Suppose that we can show x' \in \text{Ass}_{X'/S}(\mathcal{F}'). Then Divisors, Lemma 31.7.3 implies that x \in \text{Ass}_{X/S}(\mathcal{F}) and that the irreducible component C' of Y'_{s'} containing x' is an irreducible component of C \times _ s s' where C \subset Y_ s is the irreducible component containing x. Whence \dim (C) = \dim (C') = \dim (Y_ t) (see above) and the proof is complete. This reduces us to the case discussed in the next paragraph.
Assume S = \mathop{\mathrm{Spec}}(A) where A is a valuation ring and t and s are the generic and closed points of S. We will assume x \not\in \text{Ass}_{X/S}(\mathcal{F}) in order to get a contradiction. In other words, we assume x \not\in \text{Ass}_{X_ s}(\mathcal{F}_ s) where \mathcal{F}_ s is the pullback of \mathcal{F} to X_ s. Consider the ring map
A \longrightarrow \mathcal{O}_{X, x} = B
and the module N = \mathcal{F}_ x over B = \mathcal{O}_{X, x}. Then B/\mathfrak m_ A B = \mathcal{O}_{X_ s, x} and N/\mathfrak m_ A N is the stalk of \mathcal{F}_ s at the point x. Denote \mathfrak q \subset B the prime ideal corresponding to the point y, see Schemes, Lemma 26.13.2. Since x is a generic point of Y_ s we see that the radical of \mathfrak q + \mathfrak m_ A B is \mathfrak m_ B. Then \text{Ass}_{B/\mathfrak m_ A B}(N/\mathfrak m_ A N) is a finite set of prime ideals (Algebra, Lemma 10.63.5) which doesn't contain the maximal ideal of B/\mathfrak m_ A B since x \not\in \text{Ass}_{X/S}(\mathcal{F}). Thus the image of of \mathfrak q in B/\mathfrak m_ A B is not contained in any of those prime ideals. Hence by prime avoidance (Algebra, Lemma 10.15.2) we can find an element g \in \mathfrak q whose image in B/\mathfrak m_ A B is a nonzerodivisor on N/\mathfrak m_ A N (this uses the description of zerodivisors in Algebra, Lemma 10.63.9). Since N = \mathcal{F}_ x is A-flat by Lemma 37.18.1 we see that
g : N \longrightarrow N
is injective. In particular, if K = \text{Frac}(A) is the fraction field of A, then we see that
g : N \otimes _ A K \longrightarrow N \otimes _ A K
is injective. Observe that \mathfrak q corresponds to a prime ideal of B \otimes _ A K. Denote \mathcal{F}_ t the restriction of \mathcal{F} to the generic fibre X_ t. We have (B \otimes _ A K)_{\mathfrak q} = \mathcal{O}_{X_ t, y} and (N \otimes _ A K)_\mathfrak q is the stalk at y of \mathcal{F}_ t. Hence we find that g \in \mathfrak m_ y \subset \mathcal{O}_{X_ t, y} is a nonzerodivisor on the stalk (\mathcal{F}_ t)_ y which contradicts our assumption that y \in \text{Ass}_{X/S}(\mathcal{F}). \square
Lemma 37.18.4. Let f : X \to S be a morphism of schemes which is locally of finite type. Let \mathcal{F} be a finite type, quasi-coherent \mathcal{O}_ X-module flat over S. Assume S is irreducible with generic point \eta . If \dim (\text{Supp}(\mathcal{F}_\eta )) \leq r then for all s \in S we have \dim (\text{Supp}(\mathcal{F}_ s)) \leq r.
Proof. Let x \in \text{Supp}(\mathcal{F}_ s) be a generic point of an irreducible component of \text{Supp}(\mathcal{F}_ s). By Algebra, Lemma 10.41.12 we can find a specialization y \leadsto x in \text{Supp}(\mathcal{F}) with f(y) = \eta . Of course we may assume y is a generic point of an irreducible component of \text{Supp}(\mathcal{F}_\eta ). We conclude from Lemma 37.18.3 that the dimension of \overline{\{ x\} } is at most r. \square
Lemma 37.18.5. Let f : X \to S be a morphism of schemes which is locally of finite type. Let \mathcal{F} be a quasi-coherent \mathcal{O}_ X-module of finite type. Let y \in \text{Ass}_{X/S}(\mathcal{F}). Denote Y \subset X the closure of \{ y\} in X viewed as an integral closed subscheme. Denote T \subset S the closure of \{ f(y)\} viewed as an integral closed subscheme. We obtain a commutative diagram
\xymatrix{ Y \ar[r] \ar[d] & X \ar[d] \\ T \ar[r] & S }
where Y \to T is dominant. Assume \mathcal{F} is flat over S at all generic points of irreducible components of fibres of Y \to T (for example if \mathcal{F} is flat over S). Then
if s \in S and x \in Y_ s is the generic point of an irreducible component of Y_ s, then x \in \text{Ass}_{X/S}(\mathcal{F}), and
there is an integer d \geq 0 such that Y \to T is of relative dimension d, see Morphisms, Definition 29.29.1.
Proof. This follows immediately from the pointwise version Lemma 37.18.3. Note that to compute the dimension of the locally algebraic schemes Y_ s it suffices to look near the generic points, see Varieties, Section 33.20. \square
Remark 37.18.6. Here are some cases where the material above, especially Lemma 37.18.5, allows one to conclude that a morphism f : X \to S of schemes has relative dimension d as defined in Morphisms, Definition 29.29.1. For example, this is true if
X is integral with generic point \xi ,
the transcendence degree of \kappa (\xi ) over \kappa (f(\xi )) is d,
f is locally of finite type, and
there exists a quasi-coherent \mathcal{O}_ X-module \mathcal{F} of finite type which is flat over S with \text{Supp}(\mathcal{F}) = X.
Another set of hypotheses that work are the following:
S is irreducible with generic point \eta ,
X_\eta is dense in X,
every irreducible component of X_\eta has dimension d,
f is locally of finite type, and
there exists a quasi-coherent \mathcal{O}_ X-module \mathcal{F} of finite type which is flat over S with \text{Supp}(\mathcal{F}) = X.
Of course, we can relax the flatness condition on \mathcal{F} and require only that \mathcal{F} is flat over S in codimension 0, i.e., that \mathcal{F} is flat over S at every generic point of every fibre. If we ever need these results, we will carefully state and prove them here.
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