I am confused by a proof from Fulton's Algebraic Curves.
Still confused after staring it for 2 hours (3 after writing this) =(
Can someone help me out on some ideas that confuses me? This is the proof in text:

(1) The pole set of a rational function is an algebraic subset of $V$.
(2) $\Gamma (V)=\cap_{P\in V}\mathcal{O}_P(V)$
Proof. Suppose $V\subset A^n$. For $G\in k[X_1,\dots ,X_n]$, denote the residue of $G$ in $\Gamma (V)$ by $\bar G$.
Let $f\in k(V)$. Let $J_f = \lbrace G \in k[X_1,\dots ,X_n]\;|\; \bar Gf \in \Gamma (V)\rbrace$. Note that $J_f$ is an ideal in $k[X_1,\dots ,X_n]$ containing $I(V)$, and the points of $V(J_f)$ are exactly those points where $f$ is not defined. This proves (1). If $f\in \cap_{P\in V}\mathcal{O}_P(V), V(J_f)=\emptyset$, so $1\in J_f$ (Nullstellensatz!), i.e., $1\cdot f= f\in \Gamma (V)$, which proves (2).

Firstly, since $\Gamma(V)$ is not a rational function, I assume the goal of $J_f$ is to "remove" the denominator of $f$.
If that is the case, suppose $f=a/b$, where $a,b\in \Gamma(V)$.
Then $J_f=\lbrace b\cdot g\;|\;g\in k[X_1,\dots,X_n]\rbrace$. Is this right? (For simplicity I assume $a,b$ has no common factor)
If $b\cdot g\in I(V)$, then $\bar {(bg)} = 0$ and $\bar {(bg)}f=0\in\Gamma(V)$. I suppose this is why $I(V)\subset J_f$.

I have no idea why $V(J_f)$ describes the points where $f$ is not defined.
Clearly if $f$ is not defined at some $P$, $b(P)=0$ and hence $(bg)(P)=0\implies P\in V(J_f)$.

But on the other hand, let $P\in V(J_f)$ and $h=b\cdot g\in J_f$.
Then $h(P)=0$ and $h(P) = (bg)(P)\implies b(P)g(P)=0$
So the proof implies that $b(P)=0$, which means $g(P)\neq 0$
I am guessing that if $g(P)=0$ then $\bar {(bg)}f\not\in\Gamma(V)$, but I do not know why.

Finally, I am also unable to see why this process proves (1).

P.S. I am also not quite sure how the part on "residue of $G$ in $\Gamma(V)$ by $\bar G$" comes into play.
Thank you for your time. =)

  • 1
    $\begingroup$ Dear Yong, you may not say that you remove common factors from $a,b$ because $\Gamma (V)$ needn't be a UFD. $\endgroup$ Oct 27, 2012 at 9:35
  • $\begingroup$ Ahh that's right! Noted. $\endgroup$ Oct 27, 2012 at 10:07

1 Answer 1


The goal is to find points where $f$ is not defined. Let $Z$ be the set of such points. Let $P \in V(J_f)$. Let $\bar G$ be a denominator of $f$, where $G \in k[X_1,\dots,X_n]$. Since $\bar G f \in \Gamma(V)$, $G \in J_f$. Hence $G(P) = \bar G(P) = 0$. Hence $P \in Z$. Hence $V(J_f) \subset Z$.

Conversely suppose $P \in Z$. Let $G \in J_f$. Then $\bar G$ is a denominator of $f$. Hence $G(P) = \bar G(P) = 0$. Hence $P \in V(J_f)$. Hence $Z \subset V(J_f)$. Therefore $V(J_f) = Z$.

  • $\begingroup$ That's correct:+1. Could you maybe compute, as an illustration, $J(\frac{1}{1-x^2})$ where $V$ is the circle $V(X^2+Y^2-1)\subset \mathbb A^2_k$ ? $\endgroup$ Oct 27, 2012 at 9:29
  • $\begingroup$ Thanks for the answer! This also helped me realize that my confusion stems for the following: $G(P)=0\Longleftrightarrow \bar G(P)=0$. More precisely, does the equality $G\equiv\bar G$ hold on points $P\not\in V$? $\endgroup$ Oct 27, 2012 at 10:06
  • $\begingroup$ Dear Yong, for a point $P\notin V$, the notation $\bar G(P)$ doesn't even make sense! $\endgroup$ Oct 27, 2012 at 10:12
  • $\begingroup$ I see! I get it now. Thanks~ =D $\endgroup$ Oct 27, 2012 at 10:21

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