The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

41.18 Principal divisors and pushforward

The first lemma implies that the pushforward of a principal divisor along a generically finite morphism is a principal divisor.

Lemma 41.18.1. Let $(S, \delta )$ be as in Situation 41.7.1. Let $X$, $Y$ be locally of finite type over $S$. Assume $X$, $Y$ are integral and $n = \dim _\delta (X) = \dim _\delta (Y)$. Let $p : X \to Y$ be a dominant proper morphism. Let $f \in R(X)^*$. Set

\[ g = \text{Nm}_{R(X)/R(Y)}(f). \]

Then we have $p_*\text{div}(f) = \text{div}(g)$.

Proof. Let $Z \subset Y$ be an integral closed subscheme of $\delta $-dimension $n - 1$. We want to show that the coefficient of $[Z]$ in $p_*\text{div}(f)$ and $\text{div}(g)$ are equal. We may apply Lemma 41.16.2 to the morphism $p : X \to Y$ and the generic point $\xi \in Z$. Hence we may replace $Y$ by an affine open neighbourhood of $\xi $ and assume that $p : X \to Y$ is finite. Write $Y = \mathop{\mathrm{Spec}}(R)$ and $X = \mathop{\mathrm{Spec}}(A)$ with $p$ induced by a finite homomorphism $R \to A$ of Noetherian domains which induces an finite field extension $L/K$ of fraction fields. Now we have $f \in L$, $g = \text{Nm}(f) \in K$, and a prime $\mathfrak p \subset R$ with $\dim (R_{\mathfrak p}) = 1$. The coefficient of $[Z]$ in $\text{div}_ Y(g)$ is $\text{ord}_{R_\mathfrak p}(g)$. The coefficient of $[Z]$ in $p_*\text{div}_ X(f)$ is

\[ \sum \nolimits _{\mathfrak q\text{ lying over }\mathfrak p} [\kappa (\mathfrak q) : \kappa (\mathfrak p)] \text{ord}_{A_{\mathfrak q}}(f) \]

The desired equality therefore follows from Algebra, Lemma 10.120.8. $\square$

An important role in the discussion of principal divisors is played by the “universal” principal divisor $[0] - [\infty ]$ on $\mathbf{P}^1_ S$. To make this more precise, let us denote

\[ D_0, D_\infty \subset \mathbf{P}^1_ S = \underline{\text{Proj}}_ S(\mathcal{O}_ S[T_0, T_1]) \]

the closed subscheme cut out by the section $T_1$, resp. $T_0$ of $\mathcal{O}(1)$. These are effective Cartier divisors, see Divisors, Definition 30.13.1 and Lemma 30.14.10. The following lemma says that loosely speaking we have “$\text{div}(T_1/T_0) = [D_0] - [D_1]$” and that this is the universal principal divisor.

Lemma 41.18.2. Let $(S, \delta )$ be as in Situation 41.7.1. Let $X$ be locally of finite type over $S$. Assume $X$ is integral and $n = \dim _\delta (X)$. Let $f \in R(X)^*$. Let $U \subset X$ be a nonempty open such that $f$ corresponds to a section $f \in \Gamma (U, \mathcal{O}_ X^*)$. Let $Y \subset X \times _ S \mathbf{P}^1_ S$ be the closure of the graph of $f : U \to \mathbf{P}^1_ S$. Then

  1. the projection morphism $p : Y \to X$ is proper,

  2. $p|_{p^{-1}(U)} : p^{-1}(U) \to U$ is an isomorphism,

  3. the pullbacks $Y_0 = q^{-1}D_0$ and $Y_\infty = q^{-1}D_\infty $ via the morphism $q : Y \to \mathbf{P}^1_ S$ are defined (Divisors, Definition 30.13.12),

  4. we have

    \[ \text{div}_ Y(f) = [Y_0]_{n - 1} - [Y_\infty ]_{n - 1} \]
  5. we have

    \[ \text{div}_ X(f) = p_*\text{div}_ Y(f) \]
  6. if we view $Y_0$ and $Y_\infty $ as closed subschemes of $X$ via the morphism $p$ then we have

    \[ \text{div}_ X(f) = [Y_0]_{n - 1} - [Y_\infty ]_{n - 1} \]

Proof. Since $X$ is integral, we see that $U$ is integral. Hence $Y$ is integral, and $(1, f)(U) \subset Y$ is an open dense subscheme. Also, note that the closed subscheme $Y \subset X \times _ S \mathbf{P}^1_ S$ does not depend on the choice of the open $U$, since after all it is the closure of the one point set $\{ \eta '\} = \{ (1, f)(\eta )\} $ where $\eta \in X$ is the generic point. Having said this let us prove the assertions of the lemma.

For (1) note that $p$ is the composition of the closed immersion $Y \to X \times _ S \mathbf{P}^1_ S = \mathbf{P}^1_ X$ with the proper morphism $\mathbf{P}^1_ X \to X$. As a composition of proper morphisms is proper (Morphisms, Lemma 28.39.4) we conclude.

It is clear that $Y \cap U \times _ S \mathbf{P}^1_ S = (1, f)(U)$. Thus (2) follows. It also follows that $\dim _\delta (Y) = n$.

Note that $q(\eta ') = f(\eta )$ is not contained in $D_0$ or $D_\infty $ since $f \in R(X)^*$. Hence (3) by Divisors, Lemma 30.13.13. We obtain $\dim _\delta (Y_0) = n - 1$ and $\dim _\delta (Y_\infty ) = n - 1$ from Lemma 41.16.1.

Consider the effective Cartier divisor $Y_0$. At every point $\xi \in Y_0$ we have $f \in \mathcal{O}_{Y, \xi }$ and the local equation for $Y_0$ is given by $f$. In particular, if $\delta (\xi ) = n - 1$ so $\xi $ is the generic point of a integral closed subscheme $Z$ of $\delta $-dimension $n - 1$, then we see that the coefficient of $[Z]$ in $\text{div}_ Y(f)$ is

\[ \text{ord}_ Z(f) = \text{length}_{\mathcal{O}_{Y, \xi }} (\mathcal{O}_{Y, \xi }/f\mathcal{O}_{Y, \xi }) = \text{length}_{\mathcal{O}_{Y, \xi }} (\mathcal{O}_{Y_0, \xi }) \]

which is the coefficient of $[Z]$ in $[Y_0]_{n - 1}$. A similar argument using the rational function $1/f$ shows that $-[Y_\infty ]$ agrees with the terms with negative coefficients in the expression for $\text{div}_ Y(f)$. Hence (4) follows.

Note that $D_0 \to S$ is an isomorphism. Hence we see that $X \times _ S D_0 \to X$ is an isomorphism as well. Clearly we have $Y_0 = Y \cap X \times _ S D_0$ (scheme theoretic intersection) inside $X \times _ S \mathbf{P}^1_ S$. Hence it is really the case that $Y_0 \to X$ is a closed immersion. It follows that

\[ p_*\mathcal{O}_{Y_0} = \mathcal{O}_{Y'_0} \]

where $Y'_0 \subset X$ is the image of $Y_0 \to X$. By Lemma 41.12.3 we have $p_*[Y_0]_{n - 1} = [Y'_0]_{n - 1}$. The same is true for $D_\infty $ and $Y_\infty $. Hence (6) is a consequence of (5). Finally, (5) follows immediately from Lemma 41.18.1. $\square$

The following lemma says that the degree of a principal divisor on a proper curve is zero.

Lemma 41.18.3. Let $K$ be any field. Let $X$ be a $1$-dimensional integral scheme endowed with a proper morphism $c : X \to \mathop{\mathrm{Spec}}(K)$. Let $f \in K(X)^*$ be an invertible rational function. Then

\[ \sum \nolimits _{x \in X \text{ closed}} [\kappa (x) : K] \text{ord}_{\mathcal{O}_{X, x}}(f) = 0 \]

where $\text{ord}$ is as in Algebra, Definition 10.120.2. In other words, $c_*\text{div}(f) = 0$.

Proof. Consider the diagram

\[ \xymatrix{ Y \ar[r]_ p \ar[d]_ q & X \ar[d]^ c \\ \mathbf{P}^1_ K \ar[r]^-{c'} & \mathop{\mathrm{Spec}}(K) } \]

that we constructed in Lemma 41.18.2 starting with $X$ and the rational function $f$ over $S = \mathop{\mathrm{Spec}}(K)$. We will use all the results of this lemma without further mention. We have to show that $c_*\text{div}_ X(f) = c_*p_*\text{div}_ Y(f) = 0$. This is the same as proving that $c'_*q_*\text{div}_ Y(f) = 0$. If $q(Y)$ is a closed point of $\mathbf{P}^1_ K$ then we see that $\text{div}_ X(f) = 0$ and the lemma holds. Thus we may assume that $q$ is dominant. Suppose we can show that $q : Y \to \mathbf{P}^1_ K$ is finite locally free of degree $d$ (see Morphisms, Definition 28.46.1). Since $\text{div}_ Y(f) = [q^{-1}D_0]_0 - [q^{-1}D_\infty ]_0$ we see (by definition of flat pullback) that $\text{div}_ Y(f) = q^*([D_0]_0 - [D_\infty ]_0)$. Then by Lemma 41.15.2 we get $q_*\text{div}_ Y(f) = d([D_0]_0 - [D_\infty ]_0)$. Since clearly $c'_*[D_0]_0 = c'_*[D_\infty ]_0$ we win.

It remains to show that $q$ is finite locally free. (It will automatically have some given degree as $\mathbf{P}^1_ K$ is connected.) Since $\dim (\mathbf{P}^1_ K) = 1$ we see that $q$ is finite for example by Lemma 41.16.2. All local rings of $\mathbf{P}^1_ K$ at closed points are regular local rings of dimension $1$ (in other words discrete valuation rings), since they are localizations of $K[T]$ (see Algebra, Lemma 10.113.1). Hence for $y\in Y$ closed the local ring $\mathcal{O}_{Y, y}$ will be flat over $\mathcal{O}_{\mathbf{P}^1_ K, q(y)}$ as soon as it is torsion free (More on Algebra, Lemma 15.22.11). This is obviously the case as $\mathcal{O}_{Y, y}$ is a domain and $q$ is dominant. Thus $q$ is flat. Hence $q$ is finite locally free by Morphisms, Lemma 28.46.2. $\square$

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 02RS. Beware of the difference between the letter 'O' and the digit '0'.