Divisor of two relatively prime
0
Let $X=mathbb{P}^1$, let $f,gin k[t]$ relatively prime, where $t=X_1/X_2$, two coordinates. Then what can we say $operatorname{div}(f/g)$? I can just tell that $operatorname{div}(f/g)=operatorname{div}(f)-operatorname{div}(g)$. Is there any further result more than that ?
algebraic-geometry algebraic-curves
edited 2 days ago
Bernard
118k639112
asked 2 days ago
Peter Liu
315114
add a comment |
0
Let $X=mathbb{P}^1$, let $f,gin k[t]$ relatively prime, where $t=X_1/X_2$, two coordinates. Then what can we say $operatorname{div}(f/g)$? I can just tell that $operatorname{div}(f/g)=operatorname{div}(f)-operatorname{div}(g)$. Is there any further result more than that ?
algebraic-geometry algebraic-curves
edited 2 days ago
Bernard
118k639112
asked 2 days ago
Peter Liu
315114
1
What's motivating this question? I'm curious as to why you expect to be able to say more than this? And by the way, the formula ${rm div}(f/g) = {rm div}(f) - {rm div}(g)$ is true even if $f$ and $g$ are not relatively prime: for example, if $f = (t-a)^2(t - b)$ and $g = (t-a)(t - c)$, then $(f) = 2a + b$, $(g) = a + c$ and $(f/g) = a + b - c$.
– Kenny Wong
2 days ago
@KennyWong This is an exercise in a textbook (Fultons's Algebraic curve). They simply ask to calculate $div(f/g)$ if they are relatively prime. I am not quite sure what is the motivation.
– Peter Liu
2 days ago
So yeah, I don't think there's any more to say here...
– Kenny Wong
2 days ago
You can express it in terms of the zeroes of $f$ and $g$. You should really include the whole problem: the point of this exercise is showing that a principal divisor (on $mathbb{P}^1$, at least) has degree zero.
– André 3000
yesterday
add a comment |
0
0
0
Let $X=mathbb{P}^1$, let $f,gin k[t]$ relatively prime, where $t=X_1/X_2$, two coordinates. Then what can we say $operatorname{div}(f/g)$? I can just tell that $operatorname{div}(f/g)=operatorname{div}(f)-operatorname{div}(g)$. Is there any further result more than that ?
algebraic-geometry algebraic-curves
edited 2 days ago
Bernard
118k639112
asked 2 days ago
Peter Liu
315114
Let $X=mathbb{P}^1$, let $f,gin k[t]$ relatively prime, where $t=X_1/X_2$, two coordinates. Then what can we say $operatorname{div}(f/g)$? I can just tell that $operatorname{div}(f/g)=operatorname{div}(f)-operatorname{div}(g)$. Is there any further result more than that ?
algebraic-geometry algebraic-curves
algebraic-geometry algebraic-curves
edited 2 days ago
Bernard
118k639112
asked 2 days ago
Peter Liu
315114
edited 2 days ago
Bernard
118k639112
asked 2 days ago
Peter Liu
315114
edited 2 days ago
Bernard
118k639112
edited 2 days ago
Bernard
118k639112
edited 2 days ago
Bernard
118k639112
118k639112
asked 2 days ago
Peter Liu
315114
asked 2 days ago
Peter Liu
315114
asked 2 days ago
Peter Liu
315114
315114
1
What's motivating this question? I'm curious as to why you expect to be able to say more than this? And by the way, the formula ${rm div}(f/g) = {rm div}(f) - {rm div}(g)$ is true even if $f$ and $g$ are not relatively prime: for example, if $f = (t-a)^2(t - b)$ and $g = (t-a)(t - c)$, then $(f) = 2a + b$, $(g) = a + c$ and $(f/g) = a + b - c$.
– Kenny Wong
2 days ago
@KennyWong This is an exercise in a textbook (Fultons's Algebraic curve). They simply ask to calculate $div(f/g)$ if they are relatively prime. I am not quite sure what is the motivation.
– Peter Liu
2 days ago
So yeah, I don't think there's any more to say here...
– Kenny Wong
2 days ago
You can express it in terms of the zeroes of $f$ and $g$. You should really include the whole problem: the point of this exercise is showing that a principal divisor (on $mathbb{P}^1$, at least) has degree zero.
– André 3000
yesterday
add a comment |
1
What's motivating this question? I'm curious as to why you expect to be able to say more than this? And by the way, the formula ${rm div}(f/g) = {rm div}(f) - {rm div}(g)$ is true even if $f$ and $g$ are not relatively prime: for example, if $f = (t-a)^2(t - b)$ and $g = (t-a)(t - c)$, then $(f) = 2a + b$, $(g) = a + c$ and $(f/g) = a + b - c$.
– Kenny Wong
2 days ago
@KennyWong This is an exercise in a textbook (Fultons's Algebraic curve). They simply ask to calculate $div(f/g)$ if they are relatively prime. I am not quite sure what is the motivation.
– Peter Liu
2 days ago
So yeah, I don't think there's any more to say here...
– Kenny Wong
2 days ago
You can express it in terms of the zeroes of $f$ and $g$. You should really include the whole problem: the point of this exercise is showing that a principal divisor (on $mathbb{P}^1$, at least) has degree zero.
– André 3000
yesterday
1
1
What's motivating this question? I'm curious as to why you expect to be able to say more than this? And by the way, the formula ${rm div}(f/g) = {rm div}(f) - {rm div}(g)$ is true even if $f$ and $g$ are not relatively prime: for example, if $f = (t-a)^2(t - b)$ and $g = (t-a)(t - c)$, then $(f) = 2a + b$, $(g) = a + c$ and $(f/g) = a + b - c$.
– Kenny Wong
2 days ago
What's motivating this question? I'm curious as to why you expect to be able to say more than this? And by the way, the formula ${rm div}(f/g) = {rm div}(f) - {rm div}(g)$ is true even if $f$ and $g$ are not relatively prime: for example, if $f = (t-a)^2(t - b)$ and $g = (t-a)(t - c)$, then $(f) = 2a + b$, $(g) = a + c$ and $(f/g) = a + b - c$.
– Kenny Wong
2 days ago
@KennyWong This is an exercise in a textbook (Fultons's Algebraic curve). They simply ask to calculate $div(f/g)$ if they are relatively prime. I am not quite sure what is the motivation.
– Peter Liu
2 days ago
@KennyWong This is an exercise in a textbook (Fultons's Algebraic curve). They simply ask to calculate $div(f/g)$ if they are relatively prime. I am not quite sure what is the motivation.
– Peter Liu
2 days ago
So yeah, I don't think there's any more to say here...
– Kenny Wong
2 days ago
So yeah, I don't think there's any more to say here...
– Kenny Wong
2 days ago
You can express it in terms of the zeroes of $f$ and $g$. You should really include the whole problem: the point of this exercise is showing that a principal divisor (on $mathbb{P}^1$, at least) has degree zero.
– André 3000
yesterday
You can express it in terms of the zeroes of $f$ and $g$. You should really include the whole problem: the point of this exercise is showing that a principal divisor (on $mathbb{P}^1$, at least) has degree zero.
– André 3000
yesterday
add a comment |
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1
What's motivating this question? I'm curious as to why you expect to be able to say more than this? And by the way, the formula ${rm div}(f/g) = {rm div}(f) - {rm div}(g)$ is true even if $f$ and $g$ are not relatively prime: for example, if $f = (t-a)^2(t - b)$ and $g = (t-a)(t - c)$, then $(f) = 2a + b$, $(g) = a + c$ and $(f/g) = a + b - c$.
– Kenny Wong
2 days ago
@KennyWong This is an exercise in a textbook (Fultons's Algebraic curve). They simply ask to calculate $div(f/g)$ if they are relatively prime. I am not quite sure what is the motivation.
– Peter Liu
2 days ago
So yeah, I don't think there's any more to say here...
– Kenny Wong
2 days ago
You can express it in terms of the zeroes of $f$ and $g$. You should really include the whole problem: the point of this exercise is showing that a principal divisor (on $mathbb{P}^1$, at least) has degree zero.
– André 3000
yesterday