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I'm trying to understand Shafarevich's definition of intersection numbers:

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By definition, "$D_1,...,D_n$ general position at $x$" means that $\bigcap_{i=1}^n\text{Supp}(D_i)$ has finitely many points, say $P_1,...,P_k$ (one of them is $x$, say $P_k=x$).

I assume the neighbourhood $U$ is can be taken as the complement of $\{P_1,...,P_{k-1}\}$ so, by construction, $Z(f_1,...,f_n)=\{P\}$.

I don't understand how we can apply Nullstellensatz, since we are dealing with the local ring $\mathcal{O}_x$, and not the usual polynomial ring. I would get it if $f_1,...,f_n$ were polynomials, but that's not necessarily the case, so how do we justify that?

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Firstly, any open set is the union of open affine sets.

[Without loss of generality, assume $U$ is open in the affine variety ${\rm Spec \ }A$. Then it is possible to express $U$ as a union of fundamental open sets of the form $D(f) := {\rm Spec \ }A \setminus V(f)$. (We say that the fundamental open sets form a basis for the Zariski topology on ${\rm Spec \ }A$.) The standard way to prove this is to note that, since $U$ is open, it must be of the form ${\rm Spec \ } A \setminus V(I)$ for some ideal $I$. So for any point $x \in U$, there exists some $f \in I$ such that $f$ does not vanish at $x$, i.e. $x \in D(f) \subset U$. Finally, note that $D(f)$ is actually affine: it is isomorphic to ${\rm Spec\ } A_f$.]

Anyway, the point is that by making the open neighbourhood of $x$ small enough, we can assume that $f_1, \dots, f_n$ are polynomials in some polynomial ring $B := k[x_1, \dots, x_m ] / J$.

Then all that remains is to observe that $x_1, \dots, x_m$ generate the maximal ideal $\mathfrak m_x$, so each $x_i$ is in $\sqrt{(f_1, \dots, f_m)}$, i.e. $x_i^{k_i} \in (f_1, \dots, f_n)$ for a suitably high power $k_i$, by the Nullstellensatz. Then take $k$ to be the maximum of these $k_i$'s.

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