Question about Correspondences from Mumford’s Complex Projective Varieties

Question

I study David Mumford's Algebraic Geometry I - Complex Projective Varieties and have some problems to understand a step in the proof of Lemma 6.7 (b). Firstly, the general setting & preparations around the problem I intend to present below (pages 97/98):

Some remarks on used notations: For a rational map $Z: X \dashrightarrow \mathbb{P}^m$ Mumford uses the notation "$Z$" ambiguously. On the one hand for rational map itself, on the other hand he associates to the rational map a closed subvariety $Z \subset X \times \mathbb{P}^n$ which he calls "correspondence", noted also by $Z$: (See also Def 2.15 page 29)

I think that if $Z: X \to \mathbb{P}^m= \operatorname{Proj} \mathbb{C} [Y_0,...,Y_m]$ is moreover a regular map defined by $x \to [f_0(x):...f_m(x)]$ then $Z \subset X \times \mathbb{P}^m$ coinsides with vanishing locus $V(..., f_i Y_j - f_j Y_i,... ) \subset X \times \mathbb{P}^m$. First question is does this closed subvariety $V$ coinside exactly with the closure of graph $\Gamma_Z$ of $Z$ of we assume that $Z$ is a regular map?

Another probably not well known notation Mumford uses is $Z[S]$ for closed $S \subset X$. Here the definition (see Cor. 2.26 page 35):

Having the background now we now come to my actual problem:

PROBLEM: I not understand following argument from proof on 6.7 (b) (p 98):

For $H:=V(l)= V(\sum \alpha_i Y_i)$ and $H_i := V(Y_i)$, we observe that because of

$$((Y_i/l) \circ Z) = Z^*(H_i) -Z^*(H)$$

all $f_i=(Y_i/l) \circ Z$ have over $X \backslash \operatorname{Supp} Z^*(H)$ no poles. (because over this subset the divisor $(f_i)$ is positive)

But why does this observation imply that over $X \backslash \operatorname{Supp} Z^*(H)$ the correspondence $Z$ (considered again as subset of $X \times \mathbb{P}^n$) is contained in $V(..., f_i Y_j - f_j Y_i,...) \subset X \times \mathbb{P}^m$? That is in the locus of zeros of $f_i Y_j - f_j Y_i$ for $0 \le i,j \le m$?

In other words why the condition that $f_i$ have no poles over $X \backslash \operatorname{Supp} Z^*(H)$ suffice, to conclude that $Z \subset V(..., f_i Y_j - f_j Y_i,...)$? Why is this assumption on the $f_i$ it neccessary?

Answer 1

Let $U$ be the open subset of $X$ where $Z$ defines a regular map. Let's consider the map: $U \cap (X \ \backslash \ \mathrm{Supp} \ Z^*H) \longrightarrow \mathbb{P}^n$ defined by: $$ x \longrightarrow \left[ \dfrac{Y_0 \circ Z}{l \circ Z}(x), \ldots, \dfrac{Y_m \circ Z}{l \circ Z}(x) \right] $$

Let us denote by $\tilde{Z} \subset (U \cap (X \ \backslash \ \mathrm{Supp} \ Z^*H)) \times \mathbb{P}^m$ the correspondance defining this map. Using the remark you made about the equtions for the graph of a regular map $U \cap (X \ \backslash \ \mathrm{Supp} \ Z^*H) \longrightarrow \mathbb{P}^m$, we find that over $(U \cap X \ \backslash \ \mathrm{Supp} \ Z^*H)$, $\tilde{Z}$ is included in the zero locus of $(f_iY_j-f_jY_i)_{0 \leq i,j \leq m}$ where $f_i = \dfrac{Y_i \circ Z}{l \circ Z}$.

Since $Y_i$ is the projection on the $i$-th coordinate and the denominator $l \circ Z$ does not vanish on $X \ \backslash \ \mathrm{Supp} \ Z^*H$ (by definition of $Z^*H$), the regular map $U \cap (X \ \backslash \ \mathrm{Supp} \ Z^*H) \longrightarrow \mathbb{P}^n$ defined by: $$ x \longrightarrow \left[ \dfrac{Y_0 \circ Z}{l \circ Z}(x), \ldots, \dfrac{Y_m \circ Z}{l \circ Z}(x) \right] $$ is equal to the regular map $Z : U \cap (X \ \backslash \ \mathrm{Supp} \ Z^*H) \longrightarrow \mathbb{P}^m$. As a consequence, we have :

$$ Z \cap \bigg(U \cap (X \ \backslash \ \mathrm{Supp} \ Z^*H) \times \mathbb{P}^n \bigg) = \tilde{Z}.$$

As $Z$ is assumed to be irreducible, $Z \cap \bigg(X \ \backslash \ \mathrm{Supp} \ Z^*H \times \mathbb{P}^n \bigg)$ is also irreducible and we deduce from the above equality that $Z$ is also included in the vanishing locus of $(f_iY_j-f_jY_i)_{0 \leq i,j \leq m}$ over $X \ \backslash \ \mathrm{Supp} \ Z^*H$.