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Much before Bezout's theorem the following exercise is given:

If $Y$ is a curve of degree $d$ in ${\mathbb{P}}^2$ and if $L$ is a line in ${\mathbb{P}}^2$, $L \neq Y$, show that $(L \cdot Y) = d$. The definition of $(L \cdot Y)$ is given by $(L \cdot Y) = \sum (L \cdot Y)_P$ taken over all points $P \in L \cap Y$, where $(L \cdot Y)_P$ is defined using a suitable affine cover of ${\mathbb{P}}^2$.

Of course the definition $(L \cdot Y)_P$ in an affine space is simply $\ell_{\mathcal{O}_P}({\mathcal{O}_P}/{\langle f, g \rangle})$, where $L = Z(f), Y = Z(g)$. And even it is easy to verify $\ell_{\mathcal{O}_P}({\mathcal{O}_P}/{\langle f, g \rangle}) = {\mathrm{dim}}_k~{\mathcal{O}_P}/{\langle f, g \rangle}$.

The solution can be found here : Problem I.5.4(c) in Hartshorne and to the sources quoted as well.

My question: is the term $(L \cdot Y)_P$ depends on the choice of the affine cover we take?

My perception is "yes" as the equation need not be symmetric towards the dehomogenization taken for the corresponding cover. However, I can't see get a precise counterexample.

Here is another reason why I am asking this: the answer to the exercise begin with the assumption that $(L \cdot Y)_P$ is invariant under global linear change and hence we assume $L = Z(y)$. However, does dehomogenization respect this global linear change?

Siddhartha
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    No, the term $(L,Y)P$ is independent of the affine open set containing $P$. In fact it depends only on $\mathcal{O}{\mathbb{P}^2,P}$. – Mohan Jun 22 '20 at 18:39
  • I am not worried about the ring. But rather the dehomogenized elements which differs for a different cover. – Siddhartha Jun 26 '20 at 07:58

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No, the quantity $(L\cdot Y)_P$ does not depend on the open neighborhood of $P$ one uses to calculate this. Recall Hartshorne's definition of $\mathcal{O}_{P,X}$: this is the ring of equivalence classes $(U,f)$ where $U\subset X$ is an open neighborhood of $P$ and $f$ is a regular function defined on $U$ where we say that $(U,f)\sim (V,g)$ if $f-g=0$ on $U\cap V$. It is trivial to see that if we instead calculate this from some open $X'\subset X$ which contains $P$, we still get the same ring: just pick representatives of each equivalence class which are entirely contained inside $X'$.

It is also not hard to see that an automorphism $\varphi:X\to X$ induces an isomorphism of $\mathcal{O}_{P,X}$ with $\mathcal{O}_{\varphi(P),X}$: this is Hartshorne exercise I.3.3(b), for instance. So since a global linear change of coordinates is an automorphism, we get that the quantity described is invariant under such actions (because it's an isomorphism invariant). In short: everything is alright and all the statements you're worried about are okay.

KReiser
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  • Just a follow up question: here $(L \cdot Y)P$ is the length of the ${\mathcal{O}}{P,U_i}$-module ${\mathcal{O}}{P,U_i}/{\langle F,G \rangle}$, where $F,G$ are the dehomogenization of $f,g$. The ring ${\mathcal{O}}{P,U_i}$ does not change for a smaller neighbourhood, surely. But the elements $F,G$ are different if we take a different $U_i$. – Siddhartha Jun 26 '20 at 07:56
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    The worst thing that could possibly happen in a situation like that is that the different dehomogenizations would be associates of each other, thus generating the same ideal. There is nothing going wrong here - intersection multiplicities are difficult to define correctly and thus there has been much blood and ink spilled about getting it right. As a subject, we're pretty confident that things are fine here. – KReiser Jun 26 '20 at 08:19
  • So if $f, g$ are the old homogeneous polynomials (of degrees $d, e$) and $f_i, g_i$ are its dehomogenizations in $U_i$, then for the ring ${\mathcal{O}}_{P, U_i \cap U_j}$ we have $f_i$ and $f_j$ are associates because $u := {x^d_i}/{x^d_j}$ is a unit and $u f_j = f_i$. Finally this answers my question. – Siddhartha Jun 26 '20 at 10:05