The Stacks project

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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. $\square$


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