# Help me with this solution of the exercise 4.17 from Fulton's Algebraic Curves

I'm studying Fulton's algebraic curves book and in order to prove the well-definiteness of the divisor $div(z)$ on page 97 I'm trying to understand this solution which I found online. I didn't understand these points:

1. Why $J_z$ is an ideal containing $I(V)$? if we take $F\in I(V)$, $\overline F$ is identity in $\Gamma_h(V)=k[X_1,\ldots,X_{n+1}]/I(V)$, right? Then $\overline Fz$ is not necessarily in $\Gamma_h(V)$, because $z$ is in the fraction field $k(V)$.

2. Why the fact that $J_z$ is an homogeneous ideal containing $I(V)$ implies $J_z$ is algebraic subset of $V$?

I'm sorry for these basic questions, I really have problems with this subject. If anyone could help me with one of these questions or help me with a direct proof of the well-definiteness of $div(z)$, I would be grateful.

Thanks

• Where did you find this on line? I'm working through Fulton as well and have not been able to find answers or hints for those exercises that I can't solve myself. Apr 2 '18 at 20:21
• @rogerl I've found online, I don't remember where. Apr 8 '18 at 22:58

1. If $F\in I(V)$, then $\overline{F} = 0$ in $\Gamma_h(V)$, so $\overline{F}z = 0$ too.
2. Because $J_z\supseteq I(V)$, we have $V(J_z)\subseteq V\bigl(I(V)\bigr)$. By the Nullstellensatz, $V\bigl(I(V)\bigr) = V$.
• Thank you for your answer. I didn't understand why $V(I(V))=V$ using the Hilbertstellensatz. What I know is $I(V(I))=\sqrt I$ Thank you very much again! May 20 '15 at 23:12
• @user42912 Because $V$ is a variety, $V = V(J)$ for some ideal $J$. Thus, $V(I(V)) = V(I(V(J))) = V(\sqrt{J}) = V(J) = V$.
• I have another question, why $V(J_z)$ are exactly those points where $z$ is not defined? Thank you! May 22 '15 at 5:20