Ideals with Finite Zeros

Fulton Algebraic Curves

We cover the following theorem in Fulton:

Theorem) Let $I$ be an ideal in $k[X_1,X_2,\dots ,X_n]$ ($k$ algebraically closed), and suppose $V(I)=\{P_1,\dots ,P_N\}$ is finite. Let ${\mathcal{O}}_i={\mathcal{O}}_{P_i}({\mathbb{A}}^n)$. Then there is a natural isomorphism of $k[X_1,\dots ,X_n]/I$ with ${\prod }_{i=1}^N{\mathcal{O}}_i/I{\mathcal{O}}_i$.

Some technical observations to start off: (1) Let $I_i=I(\{P_i\})$. Then, $\mathrm{Rad}(I)={\bigcap }_{i=1}^N{I}_i$, and there exists some $d>0$ such that $({\bigcap }_{i=1}^N{I}_i{)}^d\subseteq I$ as $k[X_1,\dots ,X_n]$ is Noetherian. Furthermore, as $I_i$ is comaximal with ${\bigcap }_{j\ne i}{I}_j$ (as $V(I_i)\cap V({\bigcap }_{j\ne i}{I}_j)=\varnothing$) for all $i$, we have that $({\bigcap }_{i=1}^N{I}_i{)}^d=I_1^d\cap \dots \cap I_N^d=(I_1{I}_2\dots I_N{)}^d$. (2) If $V$ is a variety and $I$ is an ideal in $\Gamma (V)$, for any $P\in V$, $I{\mathcal{O}}_P(V)$ precisely consists of elements of the form $[f/g]$ ($f,g\in \Gamma (V)$) where $g(P)\ne 0$ and $f\in I$.

As $k[X_1,\dots ,X_n]$ is a subring of ${\mathcal{O}}_i$ for all $i$, there is a natural homomorphism $k[X_1,\dots ,X_n]/I\to {\mathcal{O}}_i/I{\mathcal{O}}_i$. This induces a natural homomorphism $\phi :k[X_1,\dots ,X_n]/I\to {\prod }_{i=1}^N{\mathcal{O}}_i/I{\mathcal{O}}_i$. We claim $\phi$ is an isomorphism.

We begin by finding elements $E_i\in k[X_1,\dots ,X_n]/I$ such that $\phi (E_i)$ is $1$ in the $i$th ring and zero elsewhere. First, we can find polynomials $F_i\in k[X_1,\dots ,X_n]$ such that $F_i(P_j)={\delta }_i^j$. Then, $F_i\equiv 1(\mathrm{mod}{I}_i)$ and $F_i\equiv 0(\mathrm{mod}{I}_j)$ for $j\ne i$. Note this is insufficient to guarantee $\phi ([F_i])$ is of the desired form as $\mathcal{O}/I{\mathcal{O}}_i$ might retain information about higher-order zeros. However, as $I_i^d{\mathcal{O}}_i\subseteq I{\mathcal{O}}_i$ for all $i$, a sufficient condition is for $E_i\equiv {\delta }_i^j(\mathrm{mod}{I}_j^d)$.

How do we achieve this? A natural first try is taking $F_i^d$, which achieves $F_i^d\in I_j^d$ for all $j\ne i$ but gives no further information for $I_i^d$ other than $F_i^d\equiv 1(\mathrm{mod}{I}_i)$. For this, we have to take $(1-F_i^d)\in I_i$ and $1-(1-F_i^d{)}^d$, which satisfies $1-(1-F_i^d{)}^d\equiv 1(\mathrm{mod}{I}_i^d)$ and $1-(1-F_i^d{)}^d\equiv 0(\mathrm{mod}{I}_j^d)$ for $j\ne i$ as it is a product of $F_i^d$. Let $E_i=1-(1-F_i^d{)}^d$ and $e_i=[E_i]\in k[X_1,\dots ,X_n]$. $e_i$ satisfies $\phi (e_i)=(0,\dots ,0,1,0,\dots ,0)$ with $1$ in the $i$th position. We also have that ${\sum }_{i=1}^N{e}_i=1$, as $1-\sum (1-(1-F_i^d{)}^d)$ is in $I_i^d$ for all $i$, and thus is in $I$.

To prove surjectivity, we must show that for any $f,g\in k[X_1,\dots ,X_n]$ such that $g(P_i)\ne 0$, there is some $w\in k[X_1,\dots ,X_n]/I$ such that $\phi (w)=(0,\dots ,f/g,\dots ,0)$ with $f/g$ in the $i$th position. Although we can make $f$ in the $i$th position, we need further results to make $1/g$ and thus $f/g$. A reasonable claim to make is that, for any $g\in k[X_1,\dots ,X_n]$ such that $g(P_i)\ne 0$, there is some $h\in k[X_1,\dots ,X_n]$ such that $e_i=[gh]$.

We prove this claim is true. WLOG, assume $g(P_i)=1$; then, $g^{\prime }:=1-g\in I_i$. We have that $g{e}_i(1+g^{\prime }+g^{\prime 2}+\dots +g^{\prime d-1})=e_i(1-g^{\prime d})=e_i-e_i{g}^{\prime d}$. As $e_i\in I_j^d$ for all $j\ne i$ and $g^{\prime d}\in I_i^d$, $e_i{g}^{\prime d}\in I$ and $e_i-e_i{g}^{\prime d}\equiv e_i(\mathrm{mod}I)$, as desired. This establishes surjectivity.

It remains to show $\phi$ is injective. Say that $F\in k[X_1,\dots ,X_n]$ satisfies $\phi (F)=0$. For each $1\le i\le N$, $F$ must be in $I{\mathcal{O}}_i$, so $F$ has a representation $F=f_i/g_i$ where $g_i(P_i)\ne 0$ and $f_i\in I$. Thus, $F{g}_i=f_i\in I$, and as $g_i(P_i)\ne 0$, there is some $g_i^{\prime }$ such that $[g_i{g}_i^{\prime }]=e_i$. We have that

$$[F]=[F\sum _{i=1}^N{e}_i]=[\sum _{i=1}^N{g}_i{g}_i^{\prime }F]=[\sum _{i=1}^N{g}_i^{\prime }{f}_i]=0$$