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I am reading Silverman's book the arithmetic of elliptic curves. On page 20, the author says:

Let $C_{1}/K$ and $C_{2}/K$ be curves and let $\phi:C_{1}\to C_{2}$ be a nonconstant rational map defined over $K$. Then composition with $\phi$ induces an injection of function fields fixing $K$, $$ \phi^{*}:K(C_{2})\to K(C_{1}), \quad \phi^{*}f=f\circ \phi. $$

My question: why do we have $f\circ \phi\in K(C_{1})$?

Some background:

(1). I think here the curves should be smooth, because then we can apply proposition 2.1 to get $\phi$ is a morphism, and then by theorem 2.3, we get $\phi$ is surjective. But I'm not sure if we have that. At the beginning of the chapter, the author only says "We generally deal with curves that are smooth."

(2). From example 2.2, we have that for $C/K$ a smooth curve, there is a bijection of these two sets: $$ K(C)\cup \{\infty\}\longleftrightarrow \{\textrm{rational maps } C\to \mathbb{P}^{1} \textrm{ defined over } K\}, $$ where $\infty$ corresponds to $[1,0]$. The author says "we will often implicitly identify these two sets".

Based on the above information, I tried the following:

If I assume they are smooth, then $\phi:C_{1}\to C_{2}$ is a map and $f:C_{2}\to \mathbb{P}^{1}$ is a map. Then $f\circ \phi:C_{1}\to \mathbb{P}^{1}$ is a map. I want to show it is a rational map. Let $C_{1}\subset \mathbb{P}^{m}$ and $C_{2}\subset \mathbb{P}^{n}$. Let $\phi=[\phi_{0},\cdots,\phi_{n}]$ and $f=[f_{0},f_{1}]$. Then some $\phi_{i}\notin I(C_{1})$ and some $f_{j}\notin I(C_{2})$. I'm trying to prove $[f_{0}(\phi_{0},\cdots,\phi_{n}),f_{1}(\phi_{0},\cdots,\phi_{n})]$ is a rational map and it equals $f\circ \phi$. I'm checking the conditions for it to be a rational map. But I cannot verify that $f_{0}(\phi_{0},\cdots,\phi_{n})$ or $f_{1}(\phi_{0},\cdots,\phi_{n})$ is not in $I(C_{1})$.

Thank you for your help!

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  • $\begingroup$ A map $\phi:C_1\to C_2$ is a rational map means it is a morphism on a dense open subset of $C_1$. Similarly, $f\in K(C)$ for a curve means you have a morphism $f:U\to\mathbb{P}^1$ for a dense open subset of $C$. Can you finish the argument now? $\endgroup$
    – Mohan
    Mar 22 '18 at 1:43
  • $\begingroup$ @Mohan I used Silverman's definitions which is given below. I'm not familiar with your definition. math.stackexchange.com/questions/2651096/… $\endgroup$
    – Delong
    Mar 22 '18 at 1:53
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I included below the way I would've approached the exercise - it doesn't require the identification

$$ K(C)\cup \{\infty\}\longleftrightarrow \{\textrm{rational maps } C\to \mathbb{P}^{1} \textrm{ defined over } K\}. $$


Let's treat $K(C_1)$ as the set of $g/h$ where:

  1. $g,h$ are homogeneous polynomials of the same degree,
  2. $h \not \in I(C_1)$,
  3. we identify $g_1/h_1$ with $g_2/h_2$ if $g_1h_2 - g_2h_1 \in I(C_1)$.

(This is the contents of Remark 2.9 - I rewrote for those who might not have access to the text).

Write $f=G/H$ where $G,H$ are homogeneous polynomials of the same degree and $H\not \in I(C_2)$. Then there exists some point $Q$ in $C_2$ such that $H(Q)$ is nonzero. The surjectivity of $\phi$ guarantees a $P\in C_1$ such that $\phi(P) = Q$. Choose a sequence of representative (homogeneous) polynomials $[\phi_0,\ldots,\phi_n]$ of $\phi$ which is well-defined at $P$. Then we claim that $f \circ \phi = G(\phi_0,\ldots,\phi_n)/H(\phi_0,\ldots, \phi_n)$ lies in $K(C_1)$ by the definition above.

Condition 1 is clear. Condition 2 follows because $H(\phi_0,\ldots,\phi_n)(P) \ne 0$. There is nothing to check for condition 3, so we are done.

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  • $\begingroup$ What is $H\circ \phi$? If you define $f\circ \phi = G(\phi_0,\ldots, \phi_n)/H(\phi_0,\ldots,\phi_n)$, then the goal is to show $H(\phi_0,\ldots,\phi_n)\notin I(C_{1})$ not $H\circ \phi\notin I(C_{1})$. $\endgroup$
    – Delong
    Mar 29 '18 at 20:34
  • $\begingroup$ @Delong $H\circ \phi$ is shorthand for $H(\phi_0,\ldots,\phi_n)$ $\endgroup$
    – abc
    Mar 30 '18 at 1:15
  • $\begingroup$ I know $\phi$ is surjective, but if you fix one representation of $\phi$, then you may not be able to get all the points in $C_{2}$. $\endgroup$
    – Delong
    Mar 30 '18 at 1:50
  • $\begingroup$ @Delong, of course, good point! I tried to rectify the argument. See above. Let me know if you believe that we can: "Choose a sequence of representative (homogeneous) polynomials $[\phi_0,\ldots,\phi_n]$ of $\phi$ which is well-defined at $P$" - I haven't had a chance to properly justify this to myself. $\endgroup$
    – abc
    Mar 30 '18 at 14:38
  • $\begingroup$ I can believe that. Then the definition of $f\circ \phi$ would be: choose $f=G/H$, then $H\notin I(C_{2})$, so $H(Q)\neq 0$ for some $Q\in C_{2}$. Since $\phi$ is surjective, $\phi(P)=Q$ for some $P\in C_{1}$. Choose a representative such that $[\phi_{0},\cdots,\phi_{n}](P)=\phi(P)$. Then define $f\circ\phi=G(\phi_{0},\cdots,\phi_{n})/H(\phi_{0},\cdots,\phi_{n})$. $\endgroup$
    – Delong
    Mar 30 '18 at 19:10

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