I'm reading Fulton's algebraic curves book.

I'm trying to understand this solution which I found online of the question 4.17 on page 97.

What I didn't understand is why $V(J_z)$ are exactly those points where $z$ is not defined (see the last line the solution).

Maybe I'm missing something, but I couldn't understand why this is true.


  • $\begingroup$ Please don't delete and repost questions - to get attention, you can make a small edit to bump back to the front page (like I've just done), or you can offer a bounty. $\endgroup$ May 23 '15 at 16:49
  • $\begingroup$ @ZevChonoles I didn't know it's forbidden. It won't happen again. Thanks $\endgroup$
    – user42912
    May 23 '15 at 16:55

Write $z = a/b$, with $a, b \in \Gamma_{h}(V)$ forms of the same degree $d$. Note that $\overline{b}z = \overline{a} \in \Gamma_{h}(V)$, and therefore $b \in J_{z}$.

Let $P \in V(J_{z})$. Then $F(P) = 0$, for every polynomial $F \in k[X_{1}, \ldots, X_{n+1}]$ such that $\overline{F}z \in \Gamma_{h}(V)$. Since $b \in J_{z}$, we have $b(P) = 0$, and thus $z$ is not defined at $P$. This shows that $V(J_{z})$ is contained in the pole set of $z$.

Conversely, let $P$ be a pole of $z$. Then $a(P) \neq 0$ and $b(P) = 0$. Let $F \in J_{z}$. Then, $\overline{F}z = \overline{G}$, for some $\overline{G} \in \Gamma_{h}(V) = k[X_{1}, \ldots, X_{n+1}]/I(V)$. That is, $Fz - G \in I(V)$. Since $I(V)$ is an ideal, $b(Fz - G) = Fa - bG \in I(V)$. Evaluating at $P$, we obtain $0 = F(P)a(P) - b(P)G(P) = F(P)a(P)$, because $b(P) = 0$. Since $a(P) \neq 0$, we conclude that $F(P) = 0$. This shows that the pole set of $z$ is contained in $V(J_{z})$.

Therefore, the pole set of $z$ is exactly $V(J_{z})$; in particular, it is an algebraic subset of $V$, as the solution posted shows.

  • $\begingroup$ In the first line why are you writing $\overline bz=\overline a$, instead of $bz=a$? Thank you for your answer! $\endgroup$
    – user42912
    May 23 '15 at 22:45
  • $\begingroup$ Because $\Gamma_{h}(V)$ is the quotient of $k[X_{1}, \ldots, X_{n+1}]$ by the ideal $I(V)$, so its elements are classes, denoted by the overline. $\endgroup$
    – rla
    May 23 '15 at 22:51
  • $\begingroup$ Yes, but $a$ and $b$ are already in $\Gamma_h(V)$, no? $\endgroup$
    – user42912
    May 23 '15 at 22:52
  • $\begingroup$ Yes! And that is not a problem. $\endgroup$
    – rla
    May 23 '15 at 23:00
  • 1
    $\begingroup$ It is Ok! But for me, it is essentially what I wanted to say. $\endgroup$
    – rla
    May 23 '15 at 23:38

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