Lemma 31.18.1. Let $f : X \to S$ be a morphism of schemes. Let $D \subset X$ be a closed subscheme. Assume
$D$ is an effective Cartier divisor, and
$D \to S$ is a flat morphism.
31.18 Relative effective Cartier divisors
The following lemma shows that an effective Cartier divisor which is flat over the base is really a “family of effective Cartier divisors” over the base. For example the restriction to any fibre is an effective Cartier divisor.
Then for every morphism of schemes $g : S' \to S$ the pullback $(g')^{-1}D$ is an effective Cartier divisor on $X' = S' \times _ S X$ where $g' : X' \to X$ is the projection.
Proof. Using Lemma 31.13.2 we translate this as follows into algebra. Let $A \to B$ be a ring map and $h \in B$. Assume $h$ is a nonzerodivisor and that $B/hB$ is flat over $A$. Then
\[ 0 \to B \xrightarrow {h} B \to B/hB \to 0 \]
is a short exact sequence of $A$-modules with $B/hB$ flat over $A$. By Algebra, Lemma 10.39.12 this sequence remains exact on tensoring over $A$ with any module, in particular with any $A$-algebra $A'$. $\square$
This lemma is the motivation for the following definition.
Definition 31.18.2. Let $f : X \to S$ be a morphism of schemes. A relative effective Cartier divisor on $X/S$ is an effective Cartier divisor $D \subset X$ such that $D \to S$ is a flat morphism of schemes.
We warn the reader that this may be nonstandard notation. In particular, in [IV, Section 21.15, EGA] the notion of a relative divisor is discussed only when $X \to S$ is flat and locally of finite presentation. Our definition is a bit more general. However, it turns out that if $x \in D$ then $X \to S$ is flat at $x$ in many cases (but not always).
Lemma 31.18.3. Let $f : X \to S$ be a morphism of schemes. If $D_1, D_2 \subset X$ are relative effective Cartier divisor on $X/S$ then so is $D_1 + D_2$ (Definition 31.13.6).
Proof. This translates into the following algebra fact: Let $A \to B$ be a ring map and $h_1, h_2 \in B$. Assume the $h_ i$ are nonzerodivisors and that $B/h_ iB$ is flat over $A$. Then $h_1h_2$ is a nonzerodivisor and $B/h_1h_2B$ is flat over $A$. The reason is that we have a short exact sequence
\[ 0 \to B/h_1B \to B/h_1h_2B \to B/h_2B \to 0 \]
where the first arrow is given by multiplication by $h_2$. Since the outer two are flat modules over $A$, so is the middle one, see Algebra, Lemma 10.39.13. $\square$
Lemma 31.18.4. Let $f : X \to S$ be a morphism of schemes. If $D_1, D_2 \subset X$ are relative effective Cartier divisor on $X/S$ and $D_1 \subset D_2$ as closed subschemes, then the effective Cartier divisor $D$ such that $D_2 = D_1 + D$ (Lemma 31.13.8) is a relative effective Cartier divisor on $X/S$.
Proof. This translates into the following algebra fact: Let $A \to B$ be a ring map and $h_1, h_2 \in B$. Assume the $h_ i$ are nonzerodivisors, that $B/h_ iB$ is flat over $A$, and that $(h_2) \subset (h_1)$. Then we can write $h_2 = h h_1$ where $h \in B$ is a nonzerodivisor. We get a short exact sequence
\[ 0 \to B/hB \to B/h_2B \to B/h_1B \to 0 \]
where the first arrow is given by multiplication by $h_1$. Since the right two are flat modules over $A$, so is the middle one, see Algebra, Lemma 10.39.13. $\square$
Lemma 31.18.5. Let $f : X \to S$ be a morphism of schemes. Let $D \subset X$ be a relative effective Cartier divisor on $X/S$. If $x \in D$ and $\mathcal{O}_{X, x}$ is Noetherian, then $f$ is flat at $x$.
Proof. Set $A = \mathcal{O}_{S, f(x)}$ and $B = \mathcal{O}_{X, x}$. Let $h \in B$ be an element which generates the ideal of $D$. Then $h$ is a nonzerodivisor in $B$ such that $B/hB$ is a flat local $A$-algebra. Let $I \subset A$ be a finitely generated ideal. Consider the commutative diagram
\[ \xymatrix{ 0 \ar[r] & B \ar[r]_ h & B \ar[r] & B/hB \ar[r] & 0 \\ 0 \ar[r] & B \otimes _ A I \ar[r]^ h \ar[u] & B \otimes _ A I \ar[r] \ar[u] & B/hB \otimes _ A I \ar[r] \ar[u] & 0 } \]
The lower sequence is short exact as $B/hB$ is flat over $A$, see Algebra, Lemma 10.39.12. The right vertical arrow is injective as $B/hB$ is flat over $A$, see Algebra, Lemma 10.39.5. Hence multiplication by $h$ is surjective on the kernel $K$ of the middle vertical arrow. By Nakayama's lemma, see Algebra, Lemma 10.20.1 we conclude that $K= 0$. Hence $B$ is flat over $A$, see Algebra, Lemma 10.39.5. $\square$
The following lemma relies on the algebraic version of openness of the flat locus. The scheme theoretic version can be found in More on Morphisms, Section 37.15.
Lemma 31.18.6. Let $f : X \to S$ be a morphism of schemes. Let $D \subset X$ be a relative effective Cartier divisor. If $f$ is locally of finite presentation, then there exists an open subscheme $U \subset X$ such that $D \subset U$ and such that $f|_ U : U \to S$ is flat.
Proof. Pick $x \in D$. It suffices to find an open neighbourhood $U \subset X$ of $x$ such that $f|_ U$ is flat. Hence the lemma reduces to the case that $X = \mathop{\mathrm{Spec}}(B)$ and $S = \mathop{\mathrm{Spec}}(A)$ are affine and that $D$ is given by a nonzerodivisor $h \in B$. By assumption $B$ is a finitely presented $A$-algebra and $B/hB$ is a flat $A$-algebra. We are going to use absolute Noetherian approximation.
Write $B = A[x_1, \ldots , x_ n]/(g_1, \ldots , g_ m)$. Assume $h$ is the image of $h' \in A[x_1, \ldots , x_ n]$. Choose a finite type $\mathbf{Z}$-subalgebra $A_0 \subset A$ such that all the coefficients of the polynomials $h', g_1, \ldots , g_ m$ are in $A_0$. Then we can set $B_0 = A_0[x_1, \ldots , x_ n]/(g_1, \ldots , g_ m)$ and $h_0$ the image of $h'$ in $B_0$. Then $B = B_0 \otimes _{A_0} A$ and $B/hB = B_0/h_0B_0 \otimes _{A_0} A$. By Algebra, Lemma 10.168.1 we may, after enlarging $A_0$, assume that $B_0/h_0B_0$ is flat over $A_0$. Let $K_0 = \mathop{\mathrm{Ker}}(h_0 : B_0 \to B_0)$. As $B_0$ is of finite type over $\mathbf{Z}$ we see that $K_0$ is a finitely generated ideal. Let $A_1 \subset A$ be a finite type $\mathbf{Z}$-subalgebra containing $A_0$ and denote $B_1$, $h_1$, $K_1$ the corresponding objects over $A_1$. By More on Algebra, Lemma 15.31.3 the map $K_0 \otimes _{A_0} A_1 \to K_1$ is surjective. On the other hand, the kernel of $h : B \to B$ is zero by assumption. Hence every element of $K_0$ maps to zero in $K_1$ for sufficiently large subrings $A_1 \subset A$. Since $K_0$ is finitely generated, we conclude that $K_1 = 0$ for a suitable choice of $A_1$.
Set $f_1 : X_1 \to S_1$ equal to $\mathop{\mathrm{Spec}}$ of the ring map $A_1 \to B_1$. Set $D_1 = \mathop{\mathrm{Spec}}(B_1/h_1B_1)$. Since $B = B_1 \otimes _{A_1} A$, i.e., $X = X_1 \times _{S_1} S$, it now suffices to prove the lemma for $X_1 \to S_1$ and the relative effective Cartier divisor $D_1$, see Morphisms, Lemma 29.25.7. Hence we have reduced to the case where $A$ is a Noetherian ring. In this case we know that the ring map $A \to B$ is flat at every prime $\mathfrak q$ of $V(h)$ by Lemma 31.18.5. Combined with the fact that the flat locus is open in this case, see Algebra, Theorem 10.129.4 we win. $\square$
There is also the following lemma (whose idea is apparently due to Michael Artin, see [Nobile]) which needs no finiteness assumptions at all.
Lemma 31.18.7. Let $f : X \to S$ be a morphism of schemes. Let $D \subset X$ be a relative effective Cartier divisor on $X/S$. If $f$ is flat at all points of $X \setminus D$, then $f$ is flat.
Proof. This translates into the following algebra fact: Let $A \to B$ be a ring map and $h \in B$. Assume $h$ is a nonzerodivisor, that $B/hB$ is flat over $A$, and that the localization $B_ h$ is flat over $A$. Then $B$ is flat over $A$. The reason is that we have a short exact sequence
\[ 0 \to B \to B_ h \to \mathop{\mathrm{colim}}\nolimits _ n (1/h^ n)B/B \to 0 \]
and that the second and third terms are flat over $A$, which implies that $B$ is flat over $A$ (see Algebra, Lemma 10.39.13). Note that a filtered colimit of flat modules is flat (see Algebra, Lemma 10.39.3) and that by induction on $n$ each $(1/h^ n)B/B \cong B/h^ nB$ is flat over $A$ since it fits into the short exact sequence
\[ 0 \to B/h^{n - 1}B \xrightarrow {h} B/h^ nB \to B/hB \to 0 \]
Some details omitted. $\square$
Example 31.18.8. Here is an example of a relative effective Cartier divisor $D$ where the ambient scheme is not flat in a neighbourhood of $D$. Namely, let $A = k[t]$ and Then $B$ is not flat over $A$ but $B/xB \cong A$ is flat over $A$. Moreover $x$ is a nonzerodivisor and hence defines a relative effective Cartier divisor in $\mathop{\mathrm{Spec}}(B)$ over $\mathop{\mathrm{Spec}}(A)$.
\[ B = k[t, x, y, x^{-1}y, x^{-2}y, \ldots ]/(ty, tx^{-1}y, tx^{-2}y, \ldots ) \]
If the ambient scheme is flat and locally of finite presentation over the base, then we can characterize a relative effective Cartier divisor in terms of its fibres. See also More on Morphisms, Lemma 37.23.1 for a slightly different take on this lemma.
Lemma 31.18.9. Let $\varphi : X \to S$ be a flat morphism which is locally of finite presentation. Let $Z \subset X$ be a closed subscheme. Let $x \in Z$ with image $s \in S$.
If $Z_ s \subset X_ s$ is a Cartier divisor in a neighbourhood of $x$, then there exists an open $U \subset X$ and a relative effective Cartier divisor $D \subset U$ such that $Z \cap U \subset D$ and $Z_ s \cap U = D_ s$.
If $Z_ s \subset X_ s$ is a Cartier divisor in a neighbourhood of $x$, the morphism $Z \to X$ is of finite presentation, and $Z \to S$ is flat at $x$, then we can choose $U$ and $D$ such that $Z \cap U = D$.
If $Z_ s \subset X_ s$ is a Cartier divisor in a neighbourhood of $x$ and $Z$ is a locally principal closed subscheme of $X$ in a neighbourhood of $x$, then we can choose $U$ and $D$ such that $Z \cap U = D$.
In particular, if $Z \to S$ is locally of finite presentation and flat and all fibres $Z_ s \subset X_ s$ are effective Cartier divisors, then $Z$ is a relative effective Cartier divisor. Similarly, if $Z$ is a locally principal closed subscheme of $X$ such that all fibres $Z_ s \subset X_ s$ are effective Cartier divisors, then $Z$ is a relative effective Cartier divisor.
Proof. Choose affine open neighbourhoods $\mathop{\mathrm{Spec}}(A)$ of $s$ and $\mathop{\mathrm{Spec}}(B)$ of $x$ such that $\varphi (\mathop{\mathrm{Spec}}(B)) \subset \mathop{\mathrm{Spec}}(A)$. Let $\mathfrak p \subset A$ be the prime ideal corresponding to $s$. Let $\mathfrak q \subset B$ be the prime ideal corresponding to $x$. Let $I \subset B$ be the ideal corresponding to $Z$. By the initial assumption of the lemma we know that $A \to B$ is flat and of finite presentation. The assumption in (1) means that, after shrinking $\mathop{\mathrm{Spec}}(B)$, we may assume $I(B \otimes _ A \kappa (\mathfrak p))$ is generated by a single element which is a nonzerodivisor in $B \otimes _ A \kappa (\mathfrak p)$. Say $f \in I$ maps to this generator. We claim that after inverting an element $g \in B$, $g \not\in \mathfrak q$ the closed subscheme $D = V(f) \subset \mathop{\mathrm{Spec}}(B_ g)$ is a relative effective Cartier divisor.
By Algebra, Lemma 10.168.1 we can find a flat finite type ring map $A_0 \to B_0$ of Noetherian rings, an element $f_0 \in B_0$, a ring map $A_0 \to A$ and an isomorphism $A \otimes _{A_0} B_0 \cong B$. If $\mathfrak p_0 = A_0 \cap \mathfrak p$ then we see that
\[ B \otimes _ A \kappa (\mathfrak p) = \left(B_0 \otimes _{A_0} \kappa (\mathfrak p_0)\right) \otimes _{\kappa (\mathfrak p_0))} \kappa (\mathfrak p) \]
hence $f_0$ is a nonzerodivisor in $B_0 \otimes _{A_0} \kappa (\mathfrak p_0)$. By Algebra, Lemma 10.99.2 we see that $f_0$ is a nonzerodivisor in $(B_0)_{\mathfrak q_0}$ where $\mathfrak q_0 = B_0 \cap \mathfrak q$ and that $(B_0/f_0B_0)_{\mathfrak q_0}$ is flat over $A_0$. Hence by Algebra, Lemma 10.68.6 and Algebra, Theorem 10.129.4 there exists a $g_0 \in B_0$, $g_0 \not\in \mathfrak q_0$ such that $f_0$ is a nonzerodivisor in $(B_0)_{g_0}$ and such that $(B_0/f_0B_0)_{g_0}$ is flat over $A_0$. Hence we see that $D_0 = V(f_0) \subset \mathop{\mathrm{Spec}}((B_0)_{g_0})$ is a relative effective Cartier divisor. Since we know that this property is preserved under base change, see Lemma 31.18.1, we obtain the claim mentioned above with $g$ equal to the image of $g_0$ in $B$.
At this point we have proved (1). To see (2) consider the closed immersion $Z \to D$. The surjective ring map $u : \mathcal{O}_{D, x} \to \mathcal{O}_{Z, x}$ is a map of flat local $\mathcal{O}_{S, s}$-algebras which are essentially of finite presentation, and which becomes an isomorphisms after dividing by $\mathfrak m_ s$. Hence it is an isomorphism, see Algebra, Lemma 10.128.4. It follows that $Z \to D$ is an isomorphism in a neighbourhood of $x$, see Algebra, Lemma 10.126.6. To see (3), after possibly shrinking $U$ we may assume that the ideal of $D$ is generated by a single nonzerodivisor $f$ and the ideal of $Z$ is generated by an element $g$. Then $f = gh$. But $g|_{U_ s}$ and $f|_{U_ s}$ cut out the same effective Cartier divisor in a neighbourhood of $x$. Hence $h|_{X_ s}$ is a unit in $\mathcal{O}_{X_ s, x}$, hence $h$ is a unit in $\mathcal{O}_{X, x}$ hence $h$ is a unit in an open neighbourhood of $x$. I.e., $Z \cap U = D$ after shrinking $U$.
The final statements of the lemma follow immediately from parts (2) and (3), combined with the fact that $Z \to S$ is locally of finite presentation if and only if $Z \to X$ is of finite presentation, see Morphisms, Lemmas 29.21.3 and 29.21.11. $\square$
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