The field $k$ is algebraically closed throughout.
First, a definition coming from exercise 1.5.3. Let $Y \subset \mathbf{A}^2$ be a curve defined by $f(x, y) = 0$, where $f$ is an irreducible polynomial. Let $P = (a, b) \in \mathbf{A}^2$. Apply a translation $T$ on $\mathbf{A}^2$ to send $P$ to $(0, 0)$. Now define the multiplicity of $P$ on $Y$, $\mu_p(Y)$, to be the lowest degree of a monomial in the polynomial $f \circ T$.
Here is the problem statement. Let $Y, Z \subset \mathbf{A}^2$ be two distinct curves defined by the equations $f = 0$, $g = 0$, where $f$ and $g$ are polynomials (and I believe both are implicitly presumed to be irreducible). If $P \in Y \cap Z$, the intersection multiplicity $(Y \cdot Z)_P$ is defined as the length of the $\mathcal{O}_P$-module $\mathcal{O}_P/(f, g)$. The exercise asks to show that (i) $(Y \cdot Z)_p$ is finite, and (ii) that $(Y \cdot Z)_P \ge \mu_p(Y) * \mu_p(Z)$.
I found a proposed solution to 1.5.4a(i) here, which I will call solution 1. To quote:
$f$ and $g$ both vanish at only a finite number of points, so we can find a polynomial $h(y)$ which vanishes whenever $f$ and $g$ both vanish, so $h^n ∈ (f, g)$ for some n, so we can assume $n = 1$. The submodules of $\mathcal{O}_P /(f, g)$ correspond to ideals of $\mathcal{O}_P$ containing $f$ and $g$, so it is sufficient to show that $k[x, y]/(f, g)$ is finite dimensional (as its dimension is at least the length of $\mathcal{O}_P /(f, g)$). But if we have polynomials $h_1(x)$ and $h_2(y)$ of degrees $m$ and $n$ in $(f, g)$ then $k[x, y]/(f, g)$ has dimension at most that of $k[x, y]/(h_1, h_2)$ which is $mn$ which is finite.
The problem I see with this is saying that $\dim k[x, y] / (f, g) \ge \textrm{length } \mathcal{O}_P (f, g)$. The quantity on the left counts only the maximal length of a chain of prime ideals in $k[x, y] / (f, g)$, whereas the quantity on the right counts the maximal length of a chain of (potentially non-prime) ideals of $\mathcal{O}_P$ containing $(f, g)$. Therefore it is not immediately obvious why this inequality should hold. Is there some unstated lemma which justifies it?
I found another solution to (i) here. Call it solution 2. To quote:
The ring $\mathcal O_P$ is a two-dimensional, catenarian, local, integral, noetherian ring. Therefore $f$ is a non-zero-divisor (and not a unit). Therefore every minimal prime $\mathfrak q$ over $(f)$ in $\mathcal O_P$ is of height $1$. (Krull's Theorem). Because $\mathcal O_p$ is catenarian, it follows that $\dim \mathcal O_P/(f) = 1$.
As $V(f)$ is a variety the ring $A=\mathcal O_P/(f)$ is integral. Therefore $g$ is also a non-zero-divisor in $A$ (and not a unit). Therefore every minimal prime $\mathfrak q'$ in $A$ over $g$ is of height $1$. As $\mathcal O_P$ is catenarian of dimension two it follows, that $\dim A/(g) = \dim \mathcal O_P/(f,g) = 0$. So $\mathcal O_P/(f,g)$ is a zero-dimensional noetherian ring, therefore artinian and of finite length.
To me it is much clearer that solution 2 is valid than solution 1. My questions are: can solution 1 be made to work? And can anyone provide a solution to part (ii) of the question? Thanks.
