The Stacks project

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.

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 $A$.

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. $\square$

reference

Lemma 37.18.2. 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:

  1. $S'$ is the spectrum of a valuation ring with generic point $t'$ and closed point $s'$,

  2. $g(t') = t$ and $g(s') = s$,

  3. 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 $g'(y') = y$,

  4. 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 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

  1. 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

  2. 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 37.18.5. Here are some cases where the material above, especially Lemma 37.18.4, 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

  1. $X$ is integral with generic point $\xi $,

  2. the transcendence degree of $\kappa (\xi )$ over $\kappa (f(\xi ))$ is $d$,

  3. $f$ is locally of finite type, and

  4. 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:

  1. $S$ is irreducible with generic point $\eta $,

  2. $X_\eta $ is dense in $X$,

  3. every irreducible component of $X_\eta $ has dimension $d$,

  4. $f$ is locally of finite type, and

  5. 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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