I'm very interested in this interesting approach to the problem whether $\Bbb Q[x,y]/(x^2+y^2-1)$ is PID or not.
$(1-x,y)$ is not principal in $\Bbb Q[x,y]/(x^2＋y^2-1)$
In the last paragraph the aswerer uses a geometrical approach to this problem.
I want to ask about the answer here:
When $Z＝0$, why can we say that the intersection multiplicity is exactly one?
A slight correction to your question first: The claim is that the intersection multiplicity is $1$ at the point $(1:0:1) \in \mathbb{P}^2_{\mathbb{C}}$ (which has $Z \neq 0$, not $Z = 0$).
I'll use Hartshorne's definition of intersection multiplicity (see section 1.7 of Hartshorne's Algebraic Geometry) and I'll use without mentioning below the standard properties of localizations (that they "commute" with "modding out by ideals" and with each other, etc.). I'll also be abusing the notation slightly, e.g. I'll say stuff like "$Z$" is invertible in a ring when in fact it's the image of $Z$ under the natural underlying map from the polynomial ring $\mathbb{C}[X,Y,Z]$ that is the subject of the discussion.
The two varieties in question are cut out by $X^2 + Y^2 - Z^2$ and $F = Z^d f(X/Z, Y/Z)$ where $d$ is the (total) degree of $f$, in $\mathbb{P}^2_{\mathbb{C}}$. The point $(1:0:1)$ corresponds to the ideal $(X-Z, Y) \subset \mathbb{C}[X,Y,Z]$, so we need to check that the intersection multiplicity, which is the length of the algebra $B = \frac{A}{(X^2 + Y^2 - Z^2, F)A}$ over $A = \mathbb{C}[X,Y,Z]_{(X-Z,Y)}$, the localization of the polynomial ring $\mathbb{C}[X,Y,Z]$ at the prime ideal $(X-Z, Y)$, is $1$. That is to say, $B$ is a field (a priori we know that $B$ is Artinian, so $l_A(B) = l_B(B)$ is finite).
Since $Z \notin (X- Z,Y)\mathbb{C}[X,Y,Z]$, $Z$ is invertible in $A$, so $A \cong \mathbb{C}[X,Y,Z,1/Z]_{(X-Z,Y)} \cong \mathbb{C}[X/Z,Y/Z, Z, 1/Z]_{(X/Z - 1, Y/Z)} = \mathbb{C}[x,y,Z,1/Z]_{(x-1,y)}$, where $x = X/Z$ and $y = Y/Z$. (Actually we can replace the isomorphism symbols here with equalities if we understand the ring $A$ to be naturally embedded in the function field $\mathbb{C}(X,Y,Z)$)
So then, $(X^2 + Y^2 - Z^2, F)A = (x^2 + y^2 - 1, Z^d f(X/Z, Y/Z))A = (x^2 + y^2 - 1, f(x,y))A$,
and so $B \cong \frac{\mathbb{C}[x,y,Z,1/Z]_{(x-1,y)}}{(x^2 + y^2 - 1, f(x,y))} \cong \frac{\mathbb{C}[x,y,Z,1/Z]_{(x-1,y)}}{(x-1,y)}$ (this is where we use $(x^2 + y^2 - 1, f(x,y)) = (x-1,y)$ in $\mathbb{C}[x,y]$). Noting that $x,y,Z$ are algebraically independent over $\mathbb{C}$ (the first two are inhomogeneous coordinates $X/Z, Y/Z$ and the latter $Z$ is a homogeneous coordinate), $\mathfrak{p} = (x-1,y)$ remains a prime ideal in $C = \mathbb{C}[x,y,Z,1/Z]$, so $B \cong C_{\mathfrak{p}} / \mathfrak{p}C_{\mathfrak{p}}$ is in fact a field, so the intersection multiplicity is $1$.