Reputation
Next privilege 10,000 Rep.
Access moderator tools
Badges
3 10 42
Impact
~148k people reached

18m
asked Why this set is the pole set of $z$?
16h
comment Proof in Fulton's *Algebraic Curves*
yes, it's true!!! Thanks
20h
comment I need help in this proof of this exercise from Fulton's book
Using global representations and using your solution I think I can show that $\overline F(P)=\overline 0$, i.e., $F(P)\in I(V)$. I don't know how to get $F(P)=0$. Would you have any idea how to proceed?
22h
comment I need help in this proof of this exercise from Fulton's book
Thank you again! you helped me a lot!
22h
comment I need help in this proof of this exercise from Fulton's book
I'm thinking about an alternative what do you think? "$z=\overline a/\overline b$, where $\overline a$ and $\overline b$ are the images of the forms of the same degree $a$ and $b$ respectively. 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 then $b(P)\in I(V)$. Therefore $b(P)=\overline 0$ and $z$ isn't defined at $P$. "
22h
comment I need help in this proof of this exercise from Fulton's book
Your proof is really great. It's the only part I couldn't understand. See $a=a'+I(V)$, where $a'$ is a polynomial. You can't say $\overline a$, because $a$ is not a polynomial. Thank you again.
22h
comment I need help in this proof of this exercise from Fulton's book
Yes, but $a$ and $b$ are already in $\Gamma_h(V)$, no?
23h
comment I need help in this proof of this exercise from Fulton's book
In the first line why are you writing $\overline bz=\overline a$, instead of $bz=a$? Thank you for your answer!
1d
comment I need help in this proof of this exercise from Fulton's book
@ZevChonoles I didn't know it's forbidden. It won't happen again. Thanks
1d
asked I need help in this proof of this exercise from Fulton's book
1d
comment Why is $div(z)$ well-defined?
@JohnBrevik No problems. Thank you very mucho for helping me!
1d
comment Proof in Fulton's *Algebraic Curves*
I think you made a mistake: you've wrote: $$F=\sum a_{ij}X^{pi}Y^{pj} = \left(\sum \sqrt[p]{a_{ij}} X^iY^j\right)^p.$$ instead of $$F=\sum a_{ij}X^{pi}Y^{pj} = \sum\left(\sqrt[p]{a_{ij}} X^iY^j\right)^p.$$
2d
comment Proof in Fulton's *Algebraic Curves*
Yes, of course! Thank you again!
2d
comment Proof in Fulton's *Algebraic Curves*
Why $F(x,y)=0$ in the second part?
2d
comment Why is $div(z)$ well-defined?
@JohnBrevik I think I got it! Let's call $P$ the pole set of $z$. In the exercise (second box) we have $z\in k(V)\implies P\subset V$. In our case, we have $V=C$, where $C$ is a curve. Since $z\in k(C)$, then by the exercise we have $P\subset C$, Am I right?
2d
comment Why is $div(z)$ well-defined?
@JohnBrevik I'm sorry, but I'm still having problems to understand your reasoning. In our case $V$ is any variety (not necessarily an algebraic set of a curve). The fact of the pole set of $z$ is an algebraic subset of $V$ doesn't say anything to me.
2d
accepted Proof in Fulton's *Algebraic Curves*
2d
accepted Example of a divisor of a function
2d
comment Proof in Fulton's *Algebraic Curves*
Thank you very much!
2d
comment Why is $div(z)$ well-defined?
@JohnBrevik I think I didn't understand what you meant by "The algebraic sets of a curve are the finite sets". Let's pick for example this curve $C:Y=X$. Obviously, $V(C)$ is not finite.