Hartshorne Exercise II.6.2.
6.2. Varieties in Projective Space. Let $k$ be an algebraically closed field, and let $X$ be a closed subvariety of $\PP{k}{n}$ which is nonsingular in codimension one (hence satisfies $(*)$). For any divisor $D = \sum_i n_i Y_i$ on $X$, we define the degree of $D$ to be $\sum_i n_i \deg Y_i$, where $\deg Y_i$ is the degree of $Y_i$, considered as a projective variety itself (I, §7).
(a) Let $V$ be an irreducible hypersurface in $\PP{k}{n}$ which does not contain $X$, and let $Y_i$ be the irreducible components of $V\cap X$. They all have codimension $1$ by (I, Ex. 1.8). For each $i$, let $f_i$ be a local equation for $v$ on some open set $U_i$ of $\PP{k}{n}$ for which $Y_i\cap U_i\ne 0$, and let $n_i=v_{Y_i}(\bar f_i)$ where $\bar f_i$ is the restriction of $f_i$ to $U_i\cap X$. Then we define the $V.X$ to be $\sum n_iY_i$. Extend by linearity, and show that this gives a well-defined homomorphism from the subgroup of $\op{Div} \PP{k}{n}$ consisting of divisors, none of whose components contain $X$, to $\op{Div} X$.
(b) If $D$ is a principal divisor on $\PP{k}{n}$ for which $D.X$ is defined as in (a),show that $D.X$ is principal on $X$. Thus we get a homomorphism $\op{Cl}\PP{k}{n}\ra\op{Cl}X$.
(c) Show that the integer $n_i$ defined in (a) is the same as the intersection multiplicity $i(X,V; Y)$ defined in (I, §7). Then use the generalized Bezout theorem (I, 7.7) to show that for any divisor $D$ on $\PP{k}{n}$, none of whose components contain $X$,
$$\deg (D.X) = (\deg D)\cdot (\deg X).$$
(d) If $D$ is a principal divisor on $X$, show that there is a rational function $f$ on $\PP{k}{n}$ such that $D=\op{div}f.X$. Conclude that $\deg D = 0$. Thus the degree function defines a homomorphism $\deg: \op{Cl}X\ra\Z$. (This gives another proof of (6.10), since any complete nonsingular curve is projective.) Finally, there is a commutative diagram
$$\op{Cl}\PP{k}{n}\quad\longrightarrow\quad\op{Cl}X$$
$$\:\:\cong \downarrow \deg\quad\quad\:\:\downarrow \deg$$
$$\Z\xrightarrow{\cdot (\deg X)}\Z$$
and in particular, we see that the map $\op{Cl}\PP{k}{n}\ra\op{Cl}X$ is injective.
(a) What even is there to check?
(b) Let $D=\op{div}f$ for $f\in k(X_0,…,X_n)$ such that $X$ is not contained in $\op{supp}D$. Write $\op{div}f=\sum_ja_jH_j$ where the $H_j=\{f_j=0\}$ are prime and principle divisors that do note contain $X$, i.e., irreducible hypersurfaces on $\PP{k}{n}$ whose ideal is given by an irreducible homogeneous polynomial $f_j\in k[X_0,…,X_n]$. Consider the irreducible components $Y_i$ of $H\cap X$, all of codimension 1 on $X$ (indeed they all have dimension at least $\dim X-1$ and so it must be exactly that since they are proper closed subschemes of $X$), where we have dropped the subscript on $H$ for notational simplicity. Let $U_i\cong\A{k}{n}$ be an open subset of $\PP{k}{n}$ that meets $Y_i$. Then the equation $f_i$ for $H$ on $U_i$ is just the “de-homogenization” of $f$ and the images $\bar f_i\in \OO_{X}(U_i\cap X)$ of the $f_i$ are identical on overlaps $U_i\cap U_j\cap X$ (regardless of choice of $U_i$ since $H$ is fixed), and thus glue to give a global section $g\in \Bbb K(X)$ (actually $g\in\OO_X(1)$ is simply the pullback of $f\in\OO_{\PP{k}{n}}$). The divisor of $g$ is exactly $\sum_iv_{Y_i}(\bar f_i) Y_i$ because $\{\bar f_i=0\}=U_i\cap X\cap H=Y_i\cap U_i$, so $g$ vanishes only on the prime divisors $Y_i\subset X$. That is to say,
$$\op{div}(f_j).X=\op{div}g_j.$$
By linearity of this “intersection form” and the taking of divisors, we conclude that $D.X$ is principle on $X$.
(c) In the situation of (a), we must show that for any $i$, $v_{Y_i}(\bar f_i)=\mu_{\mathfrak{p}_i}(S/(I_X+I_V))$ where $S=k[X_0,…,X_n]$, $\mathfrak p_i=\Gamma(\mathcal{I}_{Y_i})=\bigoplus_{n\in\Z}\Gamma(X,\mathcal{I}_{Y_i}(n))\subset S/I_X$ is the homogeneous prime ideal of $Y_i$, $I_X=\Gamma_*(\mathcal{I}_X)$ is the homogeneous ideal of $X$, and $I_V=\Gamma_*(\mathcal{I}_V)$ is the homogeneous ideal of $V$. Note that $\mathcal{I}_{\mathfrak p_i}=\tilde{\mathfrak p}_i$ on $X$, and likewise for $X$ and $V$ on $\PP{k}{n}$. Recall that if $\mathfrak p$ is a minimal prime of a graded $S$-algebra $M$, then $\mu_\mathfrak p(M)=\op{length}_{S_\mathfrak p}M_\mathfrak p$. We know that $v_{Y_i}(\bar f_i)$ is just $\op{length}_{\OO_{X\cap U_i,\eta_i}}\big(\OO_{X\cap U_i,\eta_i}/(\bar f_i)\big)=\op{length}_{\OO_{X,\mathfrak p_i}}\big(\OO_{X,\mathfrak p_i}/(\bar f_i)\big)$ where $\eta_i=\mathfrak p_i\in X$ is the generic point of $Y_i$. Observe that $\mathfrak p_i\subset S/(I_X+I_V)$ is a minimal prime and that $\big( S/(I_X+I_V)\big )_{\mathfrak p_i}=\OO_{X\cap V,\mathfrak p_i}=\OO_{X,\mathfrak p_i}/(\bar f_i)$. It is easy that $(S/I_X)_{\mathfrak p_i}=\OO_{X,\mathfrak p_i}$, and since, for any $R$-module $M$ that is also an $R/I$-module we have $\op{length}_RM=\op{length}_{R/I}M$, we get that
$$\op{length}_{\OO_{X,\mathfrak p_i}}\big(\OO_{X,\mathfrak p_i}/(\bar f_i)\big)=\op{length}_{\OO_{\PP{k}{n},\mathfrak p_i}}\big(\OO_{X,\mathfrak p_i}/(\bar f_i)\big)=n_i=i(X,V; Y_i).$$
Now let $D$ be a divisor on $\PP{k}{n}$, none of whose components contain $X$. To show that $\deg (D.X)=(\deg D)(\deg X)$ it suffices to show this when $D$ is a prime divisor, by linearity of this intersection form and by linearity of $\deg$. Recall that the degree of a projective scheme is $r!$ times the leading coefficient of the Hilbert polynomial $P$ of the coherent sheaf $\iota_*\OO_X\cong\OO_{\PP{k}{n}}/\mathcal{I}_X$ (see Problem 7), where $\deg P=r=\dim \op{Supp}(\iota\OO_X)$ and $P(n)=\chi(\iota_*\OO_X(n))=\chi(\OO_X(n))$. But then $D$ is just an irreducible hypersurface that does not contain $X$, and so if $Y_1,…,Y_s$ are the irreducible components of $D\cap X$, all of codimension $1$ on $X$, then by Bezout (Hartshorne I.7.7) we have,
$$(\deg D)(\deg X)=\sum_{i=1}^si(X,D;Y_i)\cdot \deg Y_i=\sum_{i=1}^sn_i\cdot \deg Y_i=\deg (D.X).$$
(d) Now let $D=\op{div}g$ be a divisor on $X$ for some rational function $g=\frac{g_1}{g_2}\in \Bbb K(X)$. WLOG $X$ is contained in no hyperplane. For each $1\le i\le n$ We have $g|_{U_i}:=g_i\in \Bbb K(X\cap U_0)\cong \op{Frac}(k[T_1,…,T_n]/\mathcal {I}_X(U_i))$ where $T_j=\frac{X_j}{X_i}$. Hence, writing $g_i=\frac{g_{i,1}}{g_{i,2}}$ where $g_{i,1},g_{i,2}\in k[T_1,…,T_n]/\mathcal {I}_X(U_0)=\OO_X(U_0\cap X)$, $\op{div}(g_i)=\op{div}g_{i,1}-\op{div}g_{i,2}$ is a difference of effective divisors. Each $g_{i,l}$ is the image of some $f_{i,l}\in k[T_1,…,T_n]$ so write $f_i=\frac{f_{i,1}}{f_{i,2}}$. On any overlap $U_i\cap U_j$ we see that $f_i$ and $f_j$ may differ by some $\alpha\in\mathcal{I}_X(U_i\cap U_j)$, but we can always make it so that $\alpha=0$ since this choice only affects each $f_i$ by something only in $\mathcal{I}_X(U_i)$. From here we obtain a global rational function, $f=\frac{f_1}{f_2}\in\Bbb K(X)$. To see that $\op{div}f.D=\op{div}g$, I claim that $\op{div}f_1.X=\op{div}g_1$, and likewise for $g_2$. To see this, for each irreducible hypersurface $H_s=\{p_s=0\}$ appearing in $\op{supp}\op{div}f_1$, where $p_s\in k[X_0,…,X_n]$ is a prime homogeneous factor of $f_1$ appearing with multiplicity $a_s$, let $Y_{1,s},…,Y_{r,s}$ be all the irreducible components of $H_s\cap X$. Once and for all associate to each open set $U_i$ a collection $\mathcal{A}_i$ of the $Y_{j,s}$ such that $Y_{j,s}\in\mathcal{A}_i\implies U_i\cap Y_{j,s}=\varnothing$ and $\coprod_i\mathcal{A}_i=\{Y_{j,s}\}_{j,s}$. Then
$$H_s.X=\sum_{i=0}^n\sum_{Y_{j,s}\in\mathcal{A}_i}v_{Y_{j,s}}(\bar p_{i,s})$$
Where $\bar p_{i,s}$ is the image of $p_1$ in $\OO_X(U_i\cap X)$. By linearity of this intersection form, we see that,
$$\op{div}f_1.X=\sum_sa_s\sum_{i=0}^n\sum_{Y_{j,s}\in\mathcal{A}_i}v_{Y_{j,s}}(\bar p_{i,s}).$$
Note that the image of $f_1$ in the DVR $\OO_{X,\eta_{Y_{j,s}}}$ after restricting to $\OO_X(X\cap U_i)$ is the unique factorization $\prod_s\bar p_{i,s}^{a_s}$. But the image of $f_1$ in $\OO_X(X\cap U_i)$ is exactly $g_{i,1}$ by construction, and moreover all the irreducible components $Y_{j,s}$ are just the irreducible components in the support of $\op{div}g_1$. Hence using the linearity of the valuation on any $O_{X,\eta_{Y_{j,s}}}$ we see that
$$\op{div}f_1.X=\sum_{i=0}^n\sum_{Y_{j,s}\in\mathcal{A}_i}v_{Y_{j,s}}(g_{i,1}).$$
The sum on the right is exactly the divisor $\op{div}g_1$. The same logic applies to $f_2$ and $g_2$, so the desired result that $\op{div}f.X=D$ is proved. It is immediate from part (c) that $\deg D=0$. The only thing that remains to check is that the composition $\op{Cl}\PP{k}{n}\xrightarrow{\deg}\Z\xrightarrow{\cdot (\deg X)}\Z$ is injective. But this is immediate because if $\deg X= 0$ then $\dim \op{Supp}(\iota_*\OO_X)=0$ so $X= \op{Supp}(\iota_*\OO_X)$ is a point and this problem makes no sense to discuss (we should assume $\dim X>0$.
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