Divisor of two relatively prime












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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







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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
















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







share|cite|improve this question




















  • 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














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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Bernard

118k639112




118k639112










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














  • 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










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