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

enter image description here

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?


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.