Factorization of ideals in a coordinate ring (Dedekind domain)

Question

Consider $f \in \mathbb{C}[X,Y]$ an irreducible curve non singular. Let $A = \mathbb{C}[X,Y] / (f)$ be the coordinate ring of $f$ and choose a curve $g \in \mathbb{C}[X,Y]$ with no component in common with $f$. Let $(x_i,y_i)$ be a common zero of $f$ and $g$ with multiplicity $r_i$, where $1 \leq i \leq n = | Z(f,g)|$. Then, it's true that we can factorize the ideal $(\overline{g})$ of $A$ as $\prod_{i \leq n} (\overline{X}-x_i,\overline{Y} - y_i)^{r_i}$?

Answer 1

This is correct.

Denote the smooth plane curve considered by $C\equiv f(x,y)=0.$ Then ${\mathfrak m}_{C,p}=\frac{{\mathfrak m}_p}{(f)}$ is the maximal ideal of $C$ at $p$, where ${\mathfrak m}_p=(X-x_0,Y-y_0).$

Consider now the ideal $(g)\subset k[x,y]$. The radical of $(f,g)$ is the intersection of maximals of points in the intersection, ${\mathfrak m}_{p_i}$. Localising yields $\left(\frac{(f,g)}{(f)}\right)_{{\mathfrak m}_p}={\mathfrak m}_{C,p}^{r_p}=\left(\frac{{\mathfrak m}_p^r+(f)}{(f)}\right)_{{\mathfrak m}_p}.$

Thus, in the case where $C$ is smooth one has, after applying the Chinese remainder theorem:

$$(f,g)=(f)+\prod_{f(p)=g(p)=0} {\mathfrak m}_p^{r_p},$$ where $r_p$ denotes the order of $(g(x,y)) (mod\ f)$ at $p$. The equality holds by localising at each maximal of $k[x,y],$ see Atiyah-Macdonald.

One needs to take care not to forget $(f)$ when writing the above (it's important!).

Locally at a point $p$, one may obtain a local coordinate (one of $x$ or $y$ will work in the case of a smooth curve $C$), so it suffices to write, in case $x$ is a local coordinate of $f$ at $p$: ${\mathfrak m}_{C,p}=(x)$ (`implicit function theorem'), though this will depend on the point.