The Stacks project

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.

Lemma 31.18.1. Let $f : X \to S$ be a morphism of schemes. Let $D \subset X$ be a closed subscheme. Assume

  1. $D$ is an effective Cartier divisor, and

  2. $D \to S$ is a flat morphism.

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

\[ B = k[t, x, y, x^{-1}y, x^{-2}y, \ldots ]/(ty, tx^{-1}y, tx^{-2}y, \ldots ) \]

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 $\mathop{\mathrm{Spec}}(A)$.

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 $s \in S$.

  1. 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 $Z_ s \cap U = D_ s$.

  2. 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 $Z \cap U = D$.

  3. 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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 056P. Beware of the difference between the letter 'O' and the digit '0'.