(This question was originally posted on Math.SE here with no response.)
Let $Q$ be the subvariety of $\mathbb{P}^3$ given by $\{x_0x_1=x_2x_3\}$. Given a prime divisor $Y$ of $\mathbb{P}^3$, Hartshorne (II, Example 6.6.2) defines a divisor $Y\cdot Q$ on $Q$ as follows:
[O]n each standard open set $U_i$ of $\mathbb{P}^3$, $Y$ is defined by a single function $f$; we can take the value of this function (restricted to $Q$) for each valuation of a prime divisor of $Q$ to define the divisor $Y\cdot Q$.
He then extends this linearly to each divisor $D$ of $\mathbb{P}^3$ (ignoring the $Q$-components) .
My Question:
Hartshorne then says that this "clearly" takes principal divisors to principal divisors. However, I'm unable to see why this doesn't require a proof. Maybe the proof I came up with (see below) is making things to difficult?
My Proof:
(I write $\operatorname{div}(f)$ to mean the divisor of $f$. For any divisor $D$ and any prime divisor $Z$, I denote by $D(Z)$ the coefficient of $Z$ in $D$.)
Let $F\in K^\times$, where $K$ is the function field of $\mathbb{P}^3$. Assume $v_Q(F)=0$, so that $F$ restricts to a well-defined non-0 element $f$ of $K(Q)$.
Claim: For any prime divisor $Z$ of $Q$,
$$ \operatorname{div}(f)(Z) = (\operatorname{div}(F)\cdot Q)(Z).$$
$\because$) Have $Z\cap Q_i\neq \emptyset$ for some $i$, where $Q_i=Q\cap U_i$. Write $A=\mathcal{O}(U_i)$. Without loss of generality, assume $F\in A$. Let $I$ be the ideal of $Q_i$ in $A$, and let $\mathfrak{p}$ be the ideal of $Z$ in $A/I$. Let $Y_1,\ldots,Y_r$ be the (distinct) prime divisors of $\mathbb{P}^3$ containing $Z$ such that $n_j=v_{Y_j}(F)\neq 0$. Note that since $Z\cap Q_i\neq \emptyset$, each $Y_j$ intersects $U_i$; write $Y_j\cap U_i=\{G_j=0\}$, where $G_j\in A$ is a nonconstant irreducible. Set $g_j=G_j|_{Q_j}=G_j+I$. Then
$$ (\operatorname{div}(F)\cdot Q)(Z) = \sum_{j=1}^r n_j v_Z(g_j) = v_Z(g),$$
where $g=g_1^{n_1}\cdots g_r^{n_r}$. On the other hand,
$$ \operatorname{div}(f)(Z) = v_Z(f).$$
Thus, we just need to show that $v_Z(f/g)=0$, i.e. that $f/g\in \mathfrak{p}$.
Consider the rational function $G=G_1^{n_1}\cdots G_r^{n_r}$. By definition of the $G_j$'s and $n_j$'s,
$$ F= HG $$
for some $H\in A$. Since $F\notin I$ and $I$ is prime, $H\notin I$; so, $h=H|_{Q_i}$ is a well-defined non-0 regular function and is equal to $f/g$. If $H\in A^\times$, we're done. Otherwise, if $h\in \mathfrak{p}$, then $H\in P$, and so $P$ contains a minimal prime $P'$ of $(H)$. By Krull's PID Theorem, $P'$ must have height 1. But then $Y=V(P')$ is a prime divisor of $\mathbb{P}^3$ containing $Z$ with $v_Y(F)\neq 0$. Since $Y\neq Y_j$ for all $j$ by construction, we have a contradiction. Thus, $f/g=h\notin \mathfrak{p}$.
Best Answer
By linearity it's enough to check the statement for rational functions of the form $g/x_i^n$, where $\deg g = n$, $g\neq x_i$, $V(g)\neq Q$.
Again, by linearity $(g/x_i^n)\cdot Q = (V(g)-nH_i)\cdot Q = V(g)\cdot Q - n (H_i\cdot Q)$. On the other hand, the right-hand side is exactly how you compute the divisor of $(g/x_i^n)$ as a rational function on $Q$.