In this problem we consider a closed irreducible subset of $X\subset \mathbb{P}^n_k$ which is regular in codimension 1, where $k$ is an algebraically closed field.
I'm stuck on parts b and c and much would appreciate any long or short hint or full solution!
In part a) we define a group homomorphism out of the subgroup of $\operatorname{div}\mathbb{P}^n_k$ generated by the prime divisors $Z\subset \mathbb{P}^n_k$ that don't contain $X$ into $\operatorname{div}X$. This map is defined by
$$Z\mapsto Z.X$$
where $Z.X:=\sum n_iY_i$, $Y_i$ are the irreducible components of $X\cap Z$ (which are automatically of codim 1 in $X$). The numbers $n_i$ are defined as follows: let $r=r(i)$ be such that $D_+(x_{r})\cap Y_i\neq \emptyset$. Then $Y_i\cap D_+(X_{r})=Y_i\cap \operatorname{Spec}(k[x_0/x_{r},…,x_n/x_r])$ is cut out by some $f_i\in \operatorname{Spec}(k[x_0/x_{r},…,x_n/x_r])$. If $\eta_i$ is the generic point of $Y_i$ we know $\mathcal{O}_{X,\eta_i}$ is a DVR with valuation say $v_i$ and we define $n_i:=v_i(\bar{f})$.
In part b) we have to show that if D is principal in $\operatorname{div}\mathbb{P}^n_k$ and if $D.X$ is defined then D.X is principal in $\operatorname{div}X$. Any hints on how to do this?
In part c) we have to show that the integers $n_i$ equal the intersection multiplicities $i(X,Z;Y_i)$. Here too I would appreciate any help!
After some hints I've made a little bit of progress on part b:
Since $D$ is principal there exists
$$f=g/h\in K(\mathbb{P}^n)=k[x_0,…,x_n]_{((0))}$$
such that $D=(f)=\sum_{Z_i\subset \mathbb{P}^n}v_{Z_i}(f)\cdot Z_i$. Here, the sum is over all irreducible subsets of codim 1 for which $v_Z(f)\neq 0$ and $v_Z$ denotes the valuation defining the local ring associated to the generic point of $Z$.
Since $X$ is a closed irreducible subscheme of $\mathbb{P}^n$ we have $X=\text{Proj}(k[x_0,…,x_n]/I_X)$ for some prime homogeneous ideal $I_X$. I claim that $g,h\notin I_X$. If $h\in I_X$, then let us consider any point $p\in X$. We have $h\in I_X\subset p$. Let $q\subset k[x_0,…,x_n]$ be a minimal prime such that $h\in q\subset p$. By Krull's Hauptidealsatz $q$ has height one and there for $\overline{\{q\}}\subset \mathbb{P}^n$ has codimension 1. Therefore $\mathcal{O}_{\mathbb{P}^n,q}$ is a DVR and $h\in q$ means $g/h\notin \mathcal{O}_{\mathbb{P}^n,q} $ so $v_q(f)<0$. This means $ \overline{\{q\}}=Z$ for one of the $Z$ occurring in $D$. Since $p\in \overline{\{q\}}$ and $p$ was an arbitrary point of $X$ we have $X\subset D$ which is a contradiction. A similar argument shows that $g\notin I_X$. Therefore $\bar{g}/\bar{h}$ is an element (non-zero) of the field
$$(k[x_0,…,x_n]/I_X)_{((0))}=K(X).$$
We may note also that $\bar{g}/\bar{h}$ is the image of $f$ under the composite
$$\mathcal{O}_{\mathbb{P}^n}(D_+(h))\to\mathcal{O}_X(D_+(h)\cap X)\to K(X).$$
Next I want to show that $D.X=(\bar{f})$ where $\bar{f}:=\bar{g}/\bar{h}$ and this is where I am currently stuck. What I have tried so far is to write $D.X$ and $(\bar{f})$ as:
$$D.X=\Big(\sum_{Z_i\subset \mathbb{P}^n}v_{Z_i}(f)\cdot Z_i\Big).X=\sum_{ij}v_{Z_i}v_{ij}(f_j)Y_{ij}$$
(where the $Y_{ij}'$s are the components of $X\cap Z_i$) and
$$(\bar{f})=\sum_{Y\subset X}v_Y(\bar{f})\cdot Y$$
and show that the same prime divisors appear in these two expressions but I'm not really getting anywhere.
Best Answer
Claim: Let $X\subset \Bbb P^n$ be a closed subvariety, regular in codimension one. If $D=(f)$ is a principal divisor on $\Bbb P^n$, none of who's components contain $X$, then $D.X=(f|_X)$.
Proof: Write $D=\sum v_{Y_i}(f)Y_i$, and for each $Y_i$ let $f_i$ be a local equation for $Y_i$ on some neighborhood which intersects $X$. Then by definition of $D.X$, we have that $$D.X=\sum_{Y_i} v_{Y_i}(f)\left(\sum_{Y_{ij}\subset Y_i\cap X} v_{Y_{ij}}(f_i|_X)Y_{ij}\right).$$ On the other hand, for each $Y_i$ we can find an open set of $\Bbb P^n$ which intersects $Y_i$ and $X$ on which $f=u\cdot f_i^{v_{Y_i}(f)}$ for $f_i$ as before and $u$ a unit: this is true in the local ring $\mathcal{O}_{\Bbb P^1,Y_i}$, so it is true in an open neighborhood. Then restricting to $X$, we get that $f|_X$ has valuation $v_{Y_i}(f)\cdot v_{Y_{ij}}(f_i|_X)$ along $Y_{ij}$. Therefore $(f|_X)=\sum_{Y_{ij}\subset X} v_{Y_i}(f)\cdot v_{Y_{ij}}(f_i|_X) Y_{ij}$ and we're done.
To attack part (c), the key idea is to sheafify the filtration coming from the definition of intersection multiplicity in chapter I. I'll write a full proof, but I'll leave most of it under spoilers so you can access it at your own pace.
The first step is to prove that the functor sending a graded module $M$ over a graded ring $S$ to the sheaf $\widetilde{M}$ on $\operatorname{Proj} S$ is exact:
With that proven, we can actually follow our main idea.