Let $K/F$ be a function field of genus $ggeq 2$, and $deg(P)=1$. If $0leq kleq 2g - 2$ show there are $g$...
$begingroup$
Let $K/F$ be a function field. First some notation:
$D_K$ is the group of divisors of $K$ (i.e. the free abelian group generated by the primes of $K$).
For $x in K^{times}$, $(x) = sum_P ord_P(x)P $
$L(D) = {x in K^{times} ,:, (x) + D geq 0} cup {0}$ for $D in D_K$
$l(D) = dim_F(L(D))$
Here is the problem:
Suppose $K/F$ is of genus $ggeq 2$, and $P$ a prime of degree $1$. For all integers $k$ we have $l(kP) leq l((k+1)P)$. If we restrict $k$ to the range $0leq kleq 2g - 2$ show there are exactly $g$ values of $k$ where $l(kP) = l((k+1)P)$.
I am able to show that $l((k+1)P) - l(kP) leq 1$ for $k geq 0$ (this seems like it should help, but maybe not). I have also observed that by Riemann-Roch, $l(kP) = l((k+1)P)$ (with $0leq k leq 2g-2$) if and only if $l(C - kP) = l(C - (k+1)P) + 1$ where $C$ is a divisor in the canonical class $mathcal{C}$.
Edit:
I'm not sure where to go from here. I tried to gain some intuition by looking at the case when $k = 0$. Of course $l(0P) = l(0) = 1$. But $l(P)$ can be either $1$ or $2$, and I don't see why $l(P)$ has to be $1$.
number-theory function-fields
asked Dec 2 '18 at 21:45
matt stokesmatt stokes
582310
$endgroup$
add a comment |
$begingroup$
Let $K/F$ be a function field. First some notation:
$D_K$ is the group of divisors of $K$ (i.e. the free abelian group generated by the primes of $K$).
For $x in K^{times}$, $(x) = sum_P ord_P(x)P $
$L(D) = {x in K^{times} ,:, (x) + D geq 0} cup {0}$ for $D in D_K$
$l(D) = dim_F(L(D))$
Here is the problem:
Suppose $K/F$ is of genus $ggeq 2$, and $P$ a prime of degree $1$. For all integers $k$ we have $l(kP) leq l((k+1)P)$. If we restrict $k$ to the range $0leq kleq 2g - 2$ show there are exactly $g$ values of $k$ where $l(kP) = l((k+1)P)$.
I am able to show that $l((k+1)P) - l(kP) leq 1$ for $k geq 0$ (this seems like it should help, but maybe not). I have also observed that by Riemann-Roch, $l(kP) = l((k+1)P)$ (with $0leq k leq 2g-2$) if and only if $l(C - kP) = l(C - (k+1)P) + 1$ where $C$ is a divisor in the canonical class $mathcal{C}$.
Edit:
I'm not sure where to go from here. I tried to gain some intuition by looking at the case when $k = 0$. Of course $l(0P) = l(0) = 1$. But $l(P)$ can be either $1$ or $2$, and I don't see why $l(P)$ has to be $1$.
number-theory function-fields
asked Dec 2 '18 at 21:45
matt stokesmatt stokes
582310
$endgroup$
add a comment |
$begingroup$
Let $K/F$ be a function field. First some notation:
$D_K$ is the group of divisors of $K$ (i.e. the free abelian group generated by the primes of $K$).
For $x in K^{times}$, $(x) = sum_P ord_P(x)P $
$L(D) = {x in K^{times} ,:, (x) + D geq 0} cup {0}$ for $D in D_K$
$l(D) = dim_F(L(D))$
Here is the problem:
Suppose $K/F$ is of genus $ggeq 2$, and $P$ a prime of degree $1$. For all integers $k$ we have $l(kP) leq l((k+1)P)$. If we restrict $k$ to the range $0leq kleq 2g - 2$ show there are exactly $g$ values of $k$ where $l(kP) = l((k+1)P)$.
I am able to show that $l((k+1)P) - l(kP) leq 1$ for $k geq 0$ (this seems like it should help, but maybe not). I have also observed that by Riemann-Roch, $l(kP) = l((k+1)P)$ (with $0leq k leq 2g-2$) if and only if $l(C - kP) = l(C - (k+1)P) + 1$ where $C$ is a divisor in the canonical class $mathcal{C}$.
Edit:
I'm not sure where to go from here. I tried to gain some intuition by looking at the case when $k = 0$. Of course $l(0P) = l(0) = 1$. But $l(P)$ can be either $1$ or $2$, and I don't see why $l(P)$ has to be $1$.
number-theory function-fields
asked Dec 2 '18 at 21:45
matt stokesmatt stokes
582310
$endgroup$
Let $K/F$ be a function field. First some notation:
$D_K$ is the group of divisors of $K$ (i.e. the free abelian group generated by the primes of $K$).
For $x in K^{times}$, $(x) = sum_P ord_P(x)P $
$L(D) = {x in K^{times} ,:, (x) + D geq 0} cup {0}$ for $D in D_K$
$l(D) = dim_F(L(D))$
Here is the problem:
Suppose $K/F$ is of genus $ggeq 2$, and $P$ a prime of degree $1$. For all integers $k$ we have $l(kP) leq l((k+1)P)$. If we restrict $k$ to the range $0leq kleq 2g - 2$ show there are exactly $g$ values of $k$ where $l(kP) = l((k+1)P)$.
I am able to show that $l((k+1)P) - l(kP) leq 1$ for $k geq 0$ (this seems like it should help, but maybe not). I have also observed that by Riemann-Roch, $l(kP) = l((k+1)P)$ (with $0leq k leq 2g-2$) if and only if $l(C - kP) = l(C - (k+1)P) + 1$ where $C$ is a divisor in the canonical class $mathcal{C}$.
Edit:
I'm not sure where to go from here. I tried to gain some intuition by looking at the case when $k = 0$. Of course $l(0P) = l(0) = 1$. But $l(P)$ can be either $1$ or $2$, and I don't see why $l(P)$ has to be $1$.
number-theory function-fields
number-theory function-fields
asked Dec 2 '18 at 21:45
matt stokesmatt stokes
582310
asked Dec 2 '18 at 21:45
matt stokesmatt stokes
582310
edited Dec 2 '18 at 22:17
matt stokes
asked Dec 2 '18 at 21:45
matt stokesmatt stokes
582310
asked Dec 2 '18 at 21:45
matt stokesmatt stokes
582310
asked Dec 2 '18 at 21:45
matt stokesmatt stokes
582310
582310
add a comment |
add a comment |
1 Answer
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$begingroup$
Notice that $l(0P) = 1$ and $l((2g-1)P) = deg((2g - 1)P) - g + 1 = g$.
Denote $n_0 = |K_0|$, where $K_0 = {k ,:,0leq k leq 2g-2 ; l(kP) = l((k+1)P) }$ and let $K_1 = ([0,2g-2]cap mathbb{Z})setminus K_0$. Then since for each $0leq kleq 2g-2$ we have $l((k+1)P)- l(kP) leq 1$, for each $k in K_1$, we must add $1$ to $l(0) = 1$ in order to reach $l((2g-1)P) = g$ (having a real hard time expressing this idea). Since $|K_1| = 2g-1 - n_0$, we get that $g = 2g-1 - n_0 + 1$ and rearranging terms we arrive at $n_0 = g$.
answered Dec 3 '18 at 2:36
matt stokesmatt stokes
582310
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Notice that $l(0P) = 1$ and $l((2g-1)P) = deg((2g - 1)P) - g + 1 = g$.
Denote $n_0 = |K_0|$, where $K_0 = {k ,:,0leq k leq 2g-2 ; l(kP) = l((k+1)P) }$ and let $K_1 = ([0,2g-2]cap mathbb{Z})setminus K_0$. Then since for each $0leq kleq 2g-2$ we have $l((k+1)P)- l(kP) leq 1$, for each $k in K_1$, we must add $1$ to $l(0) = 1$ in order to reach $l((2g-1)P) = g$ (having a real hard time expressing this idea). Since $|K_1| = 2g-1 - n_0$, we get that $g = 2g-1 - n_0 + 1$ and rearranging terms we arrive at $n_0 = g$.
answered Dec 3 '18 at 2:36
matt stokesmatt stokes
582310
$endgroup$
add a comment |
$begingroup$
Notice that $l(0P) = 1$ and $l((2g-1)P) = deg((2g - 1)P) - g + 1 = g$.
Denote $n_0 = |K_0|$, where $K_0 = {k ,:,0leq k leq 2g-2 ; l(kP) = l((k+1)P) }$ and let $K_1 = ([0,2g-2]cap mathbb{Z})setminus K_0$. Then since for each $0leq kleq 2g-2$ we have $l((k+1)P)- l(kP) leq 1$, for each $k in K_1$, we must add $1$ to $l(0) = 1$ in order to reach $l((2g-1)P) = g$ (having a real hard time expressing this idea). Since $|K_1| = 2g-1 - n_0$, we get that $g = 2g-1 - n_0 + 1$ and rearranging terms we arrive at $n_0 = g$.
answered Dec 3 '18 at 2:36
matt stokesmatt stokes
582310
$endgroup$
add a comment |
$begingroup$
Notice that $l(0P) = 1$ and $l((2g-1)P) = deg((2g - 1)P) - g + 1 = g$.
Denote $n_0 = |K_0|$, where $K_0 = {k ,:,0leq k leq 2g-2 ; l(kP) = l((k+1)P) }$ and let $K_1 = ([0,2g-2]cap mathbb{Z})setminus K_0$. Then since for each $0leq kleq 2g-2$ we have $l((k+1)P)- l(kP) leq 1$, for each $k in K_1$, we must add $1$ to $l(0) = 1$ in order to reach $l((2g-1)P) = g$ (having a real hard time expressing this idea). Since $|K_1| = 2g-1 - n_0$, we get that $g = 2g-1 - n_0 + 1$ and rearranging terms we arrive at $n_0 = g$.
answered Dec 3 '18 at 2:36
matt stokesmatt stokes
582310
$endgroup$
Notice that $l(0P) = 1$ and $l((2g-1)P) = deg((2g - 1)P) - g + 1 = g$.
Denote $n_0 = |K_0|$, where $K_0 = {k ,:,0leq k leq 2g-2 ; l(kP) = l((k+1)P) }$ and let $K_1 = ([0,2g-2]cap mathbb{Z})setminus K_0$. Then since for each $0leq kleq 2g-2$ we have $l((k+1)P)- l(kP) leq 1$, for each $k in K_1$, we must add $1$ to $l(0) = 1$ in order to reach $l((2g-1)P) = g$ (having a real hard time expressing this idea). Since $|K_1| = 2g-1 - n_0$, we get that $g = 2g-1 - n_0 + 1$ and rearranging terms we arrive at $n_0 = g$.
answered Dec 3 '18 at 2:36
matt stokesmatt stokes
582310
edited Dec 3 '18 at 3:15
answered Dec 3 '18 at 2:36
matt stokesmatt stokes
582310
answered Dec 3 '18 at 2:36
matt stokesmatt stokes
582310
answered Dec 3 '18 at 2:36
matt stokesmatt stokes
582310
582310
add a comment |
add a comment |
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