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Norm map on function fields of curves

exercise 2.10 in silverman AECRamification of a prime in a Dedekind ring and curvesShowing that the map on $mboxDiv^0(E)$ induced by an isogeny takes principal divisors to principal divisors.Map of smooth curves and its separability degreeWhy should the fibers of a surjective morphism of curves be finite?The degree of a principal divisorPushforward on principal divisorsProof of Chapter 2 Proposition 2.6a in Silverman Arithmetic of Elliptic CurvesWhy is the degree of a rational map of projective curves equal to the degree of the homogeneous polynomials?Why does a mapping of curves have degree equal to size of kernel?composition of rational maps between two curves in Silverman

2

2

$begingroup$

For a curve $C$, $fin barK(C)^*$ and a divisor $D = sum n_P(P) in textDiv(C)$ such that $D$ and $textdiv(f)$ have disjoint supports (support of a divisor is the set of points with non-zero coefficients in its sum), we define :
$$f(D) = prod_Pin Cf(P)^n_P$$

Let $phi:C_1rightarrow C_2$ be a non-constant map of smooth curves. Then show that $f(phi^*D) = (phi_*f)(D)$ for all $fin barK(C_1)^*$ and all $Din textDiv(C_2)$ whenever both sides are defined. Here, $phi^*:barK(C_2)rightarrow barK(C_1)$ is the induced map on function fields and $phi_*:barK(C_1)rightarrowbarK(C_2)$ is the map defined as $$phi_* = (phi^*)^-1circ N_barK(C_1)/phi^*barK(C_2)$$
where $N$ denotes the usual norm of field extensions. This is question 2.10(a) from Silverman's Arithmetic of Elliptic Curves. So far, I have managed to reduce it to the case where $phi$ is separable, but I don't understand how to proceed further, since I don't really have a good understanding of what the norm map does in this context. I would prefer if the solution uses only the algebraic geometry discussed so far in Silverman.

This question has been asked before at exercise 2.10 in silverman AEC, but it received no responses.

algebraic-geometry algebraic-curves

share|cite|improve this question

asked Mar 22 at 10:15

Devang AgarwalDevang Agarwal

182

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $P in C_1,Q=phi(P)in C_2, f(X) in k(C_1), g(X) = N_k(C_1)/phi^* k(C_2)(f(X))$ you want to show $g(P) = prod_Rin phi^-1(Q) f(R)^e(R)$ where $e(R)$ is the ramification index defined by $e(R)=ord_R h(phi(X))$ for $h(Y) in k(C_2)$ with a simple zero at $Q$. If $k(C_1)/phi^* k(C_2)$ is a normal extension this is easy because $N_k(C_1)/phi^* k(C_2)(f(X)) = (prod_j=1^n f(psi_j(X)))^d$ where $f(X) mapsto f(psi_j(X))$ are the automorphisms $in Aut(k(C_1)/phi^* k(C_2))$ and $n = [k(C_1):phi^* k(C_2)]_s$ and $d =[k(C_1):phi^*k(C_2)]_i=frac[k(C_1):phi^* k(C_2)]n$.
  $endgroup$
  – reuns
  Mar 23 at 11:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The $psi_j$ are some birational maps $C_1 to C_1$ such that $phi( psi_j(X))=phi(X)$, equality of rational maps so there are a few problematic points at the poles of the $psi_j$. Now if $k(C_1)/phi^* k(C_2)$ is not a normal extension (for example in the elliptic curve/isogeny case) then the $psi_j$ are replaced by the embeddings from $k(C_1)$ to $L$ the normal closure of $k(C_1)/phi^* k(C_2)$ and I'm not sure how to deal with it.
  $endgroup$
  – reuns
  Mar 23 at 11:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Even in the case when $k(C_1)/phi^*k(C_2)$ is a normal extension, I don't see why the ramification indices should appear in the product expression for the norm.
  $endgroup$
  – Devang Agarwal
  Mar 23 at 21:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Let $u(X)$ with a unique zero of order $l$ at $P$ then $N_k(C_1)/phi^* k(C_2)(u(X)) = (prod_j u(psi_j(X))^d=v(X)=w(phi(X))$ where the zeros of $v(X)$ are at the $R in phi^-1(phi(P)$ so $w(Y)$ has a unique zero of order $m$ at $phi(P)$. If necessary replace $u(X)$ by $1/(1/u(X)+c)$ so there is no cancellation in the zeros/poles of the $u(psi_j(X))$ so that $l = m$ obtaining $e(R) /d$ is the number of $psi_j$ sending $P$ to $R$
  $endgroup$
  – reuns
  Mar 24 at 13:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! I don't really see what to do in case of a non-normal extension either. Assuming a result from Hartshorne, that given any finitely generated field of trancendence degree one over $k$ is a function field of some curve, using that on the normal closure of the extension $k(C_1)/phi^*k(C_2)$ and that $N_K/M = N_L/Mcirc N_K/L$, I found that $f(phi^*D)^d = ((phi_*f)(D))^d$, where $d = [K:k(C_1)]_s$ and $K$ is the normal closure of $k(C_1)/phi^*k(C_2)$, but I am not sure how to get rid of the exponent.
  $endgroup$
  – Devang Agarwal
  Mar 24 at 19:29
  
  

  
  
  
  

add a comment | 

2

2

$begingroup$

For a curve $C$, $fin barK(C)^*$ and a divisor $D = sum n_P(P) in textDiv(C)$ such that $D$ and $textdiv(f)$ have disjoint supports (support of a divisor is the set of points with non-zero coefficients in its sum), we define :
$$f(D) = prod_Pin Cf(P)^n_P$$

Let $phi:C_1rightarrow C_2$ be a non-constant map of smooth curves. Then show that $f(phi^*D) = (phi_*f)(D)$ for all $fin barK(C_1)^*$ and all $Din textDiv(C_2)$ whenever both sides are defined. Here, $phi^*:barK(C_2)rightarrow barK(C_1)$ is the induced map on function fields and $phi_*:barK(C_1)rightarrowbarK(C_2)$ is the map defined as $$phi_* = (phi^*)^-1circ N_barK(C_1)/phi^*barK(C_2)$$
where $N$ denotes the usual norm of field extensions. This is question 2.10(a) from Silverman's Arithmetic of Elliptic Curves. So far, I have managed to reduce it to the case where $phi$ is separable, but I don't understand how to proceed further, since I don't really have a good understanding of what the norm map does in this context. I would prefer if the solution uses only the algebraic geometry discussed so far in Silverman.

This question has been asked before at exercise 2.10 in silverman AEC, but it received no responses.

algebraic-geometry algebraic-curves

share|cite|improve this question

asked Mar 22 at 10:15

Devang AgarwalDevang Agarwal

182

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $P in C_1,Q=phi(P)in C_2, f(X) in k(C_1), g(X) = N_k(C_1)/phi^* k(C_2)(f(X))$ you want to show $g(P) = prod_Rin phi^-1(Q) f(R)^e(R)$ where $e(R)$ is the ramification index defined by $e(R)=ord_R h(phi(X))$ for $h(Y) in k(C_2)$ with a simple zero at $Q$. If $k(C_1)/phi^* k(C_2)$ is a normal extension this is easy because $N_k(C_1)/phi^* k(C_2)(f(X)) = (prod_j=1^n f(psi_j(X)))^d$ where $f(X) mapsto f(psi_j(X))$ are the automorphisms $in Aut(k(C_1)/phi^* k(C_2))$ and $n = [k(C_1):phi^* k(C_2)]_s$ and $d =[k(C_1):phi^*k(C_2)]_i=frac[k(C_1):phi^* k(C_2)]n$.
  $endgroup$
  – reuns
  Mar 23 at 11:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The $psi_j$ are some birational maps $C_1 to C_1$ such that $phi( psi_j(X))=phi(X)$, equality of rational maps so there are a few problematic points at the poles of the $psi_j$. Now if $k(C_1)/phi^* k(C_2)$ is not a normal extension (for example in the elliptic curve/isogeny case) then the $psi_j$ are replaced by the embeddings from $k(C_1)$ to $L$ the normal closure of $k(C_1)/phi^* k(C_2)$ and I'm not sure how to deal with it.
  $endgroup$
  – reuns
  Mar 23 at 11:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Even in the case when $k(C_1)/phi^*k(C_2)$ is a normal extension, I don't see why the ramification indices should appear in the product expression for the norm.
  $endgroup$
  – Devang Agarwal
  Mar 23 at 21:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Let $u(X)$ with a unique zero of order $l$ at $P$ then $N_k(C_1)/phi^* k(C_2)(u(X)) = (prod_j u(psi_j(X))^d=v(X)=w(phi(X))$ where the zeros of $v(X)$ are at the $R in phi^-1(phi(P)$ so $w(Y)$ has a unique zero of order $m$ at $phi(P)$. If necessary replace $u(X)$ by $1/(1/u(X)+c)$ so there is no cancellation in the zeros/poles of the $u(psi_j(X))$ so that $l = m$ obtaining $e(R) /d$ is the number of $psi_j$ sending $P$ to $R$
  $endgroup$
  – reuns
  Mar 24 at 13:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! I don't really see what to do in case of a non-normal extension either. Assuming a result from Hartshorne, that given any finitely generated field of trancendence degree one over $k$ is a function field of some curve, using that on the normal closure of the extension $k(C_1)/phi^*k(C_2)$ and that $N_K/M = N_L/Mcirc N_K/L$, I found that $f(phi^*D)^d = ((phi_*f)(D))^d$, where $d = [K:k(C_1)]_s$ and $K$ is the normal closure of $k(C_1)/phi^*k(C_2)$, but I am not sure how to get rid of the exponent.
  $endgroup$
  – Devang Agarwal
  Mar 24 at 19:29
  
  

  
  
  
  

add a comment | 

$begingroup$

For a curve $C$, $fin barK(C)^*$ and a divisor $D = sum n_P(P) in textDiv(C)$ such that $D$ and $textdiv(f)$ have disjoint supports (support of a divisor is the set of points with non-zero coefficients in its sum), we define :
$$f(D) = prod_Pin Cf(P)^n_P$$

Let $phi:C_1rightarrow C_2$ be a non-constant map of smooth curves. Then show that $f(phi^*D) = (phi_*f)(D)$ for all $fin barK(C_1)^*$ and all $Din textDiv(C_2)$ whenever both sides are defined. Here, $phi^*:barK(C_2)rightarrow barK(C_1)$ is the induced map on function fields and $phi_*:barK(C_1)rightarrowbarK(C_2)$ is the map defined as $$phi_* = (phi^*)^-1circ N_barK(C_1)/phi^*barK(C_2)$$
where $N$ denotes the usual norm of field extensions. This is question 2.10(a) from Silverman's Arithmetic of Elliptic Curves. So far, I have managed to reduce it to the case where $phi$ is separable, but I don't understand how to proceed further, since I don't really have a good understanding of what the norm map does in this context. I would prefer if the solution uses only the algebraic geometry discussed so far in Silverman.

This question has been asked before at exercise 2.10 in silverman AEC, but it received no responses.

algebraic-geometry algebraic-curves

share|cite|improve this question

asked Mar 22 at 10:15

Devang AgarwalDevang Agarwal

182

$endgroup$

share|cite|improve this question

asked Mar 22 at 10:15

Devang AgarwalDevang Agarwal

182

share|cite|improve this question

asked Mar 22 at 10:15

Devang AgarwalDevang Agarwal

182

asked Mar 22 at 10:15

Devang AgarwalDevang Agarwal

182

asked Mar 22 at 10:15

Devang AgarwalDevang Agarwal

182