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This question is from Silverman's 'the arithmetic of elliptic curves',$p134$.

Let $K$ be a field of characteristic $p > 0$, let $E_1/ K$ and $ E_2/K$ be elliptic curves, and let $\phi : E_1 \to E_2$ be a nonzero isogeny defined over $K$. Further, let $f: \hat{E_1} \to \hat{E_2} $ be the homomorphism of formal groups induced by $\phi$.

My question:

How does an isogeny $\phi$ on elliptic curves induces a homomorphism of corresponding formal groups?

I guess $f(T)＝\phi(T)$,but I cannot check this is actually homomorphism.

My question is, I would like to know the confirming process

$\phi(F_1(x,y))＝F_2(\phi(x),\phi(y))$,where $F_1$ and $F_2$ are formal group law of elliptic curve $E_1$, $E_2$.

Poitou-Tate
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  • Thank you so much reuns for answering my questions. But your explanation is always a little hard to understand for me. This is just because my fault. You are clever and I admire you. But I often can understand other's answers, so that is just enough to me. If possible, please leave me alone from now and quit down voting other my questions. – Poitou-Tate Apr 14 '21 at 12:57
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    Here you are stuck at the coordinate ring, function field, and why it embeds in some ring of formal series, ie. you don't understand what means "$1/y_j$ is in $k[[x_j/y_j]]$". Note that it is explained in the first few pages of Silverman's chapter on the formal group law. The explanations are too hard because you are asking about non-trivial topics without clarifying the very first steps at which you are stuck. – reuns Apr 14 '21 at 13:24
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    With $f$ the isogeny and $G_1,G_2$ the group law (both some rational-algebraic functions) in the $x,y$ coordinates then $f(G_1(x,y)) = G_2(f(x),f(y))$. Then $\phi, F_1,F_2$ are just the same functions, in a different coordinates, and embedded in a ring of formal series. So it doesn't make sense to bother with formulas, just to understand this point. – reuns Apr 14 '21 at 14:09
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    @reuns (and OP) It is usually counterproductive to focus on who possibly downvoted what. Please don't let that sour your interactions. If you're interested in discussing solutions to the problem then do so in a respectful way. I have deleted some comments that were rude/unkind. – Pedro Apr 14 '21 at 15:26

1 Answers1

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If you have an elliptic curve $E_j/k:y_j^2=x_j^3+a_j x+b_j$ then

  • the group law is given by a rational function in 4 variables $x_j,y_j,x_j',y_j'$ which can be rewritten as a rational function in $x_j/y_j,1/y_j,x_j'/y_j',1/y_j'$.

  • $1/y_j$ is in $k[[x_j/y_j]]$

  • whence the group law is given by a formal series $F_j\in k[[x_j/y_j, x_j'/y_j']]$,

the power series expansion of the algebraic-rational function giving $x_j(A+B)/y_j(A+B)$ from $x_j(A)/y_j(A)$ and $x_j(B)/y_j(B)$. This is the formal group law. When $k=\Bbb{C}$ this power series is an analytic function.

  • Your isogeny $f:E_1\to E_2$ is given by a pair of rational functions in $x_1,y_1$, can be rewritten as a pair of rational functions in the $x_1/y_1,1/y_1$ coordinates. Replacing $1/y_1$ by its formal series $\in k[[x_1/y_1]$ this pair of rational functions is given by one formal series $\phi(x_1/y_1)\in k[[x_1/y_1]]$, the power series expansion of the algebraic-rational function giving the $x_2/y_2$ coordinate from $x_1/y_1$.

  • $T\to \phi(T)$ is your homomorphism of formal group law.

    We have just made a change of coordinate and embedded the rational-algebraic functions defining the group law and the isogeny in some larger ring of formal series. The group law and the isogeny commute, so do their formal series representation.

reuns
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  • Sorry, but what is $A$ and $B$? And what about when the field has character 2 or 3? And can't I check f（T）＝Φ（T）・・・① directly by calculation? I cannot see in which process you check ① ? – Poitou-Tate Apr 13 '21 at 18:00
  • Two points of the curve that you want to add. When $k=\Bbb{C}$ the formal series are analytic functions, it is formal only because we are rewriting rational/algebraic functions as formal series, but it stays the same rational/algebraic functions giving the group law and the isogeny. – reuns Apr 13 '21 at 18:11
  • Sorry, but did you comment my third question in my first comment? Sorry to bother, but I still cannot see how you checked ① is a hom. – Poitou-Tate Apr 13 '21 at 18:25
  • Which question? – reuns Apr 13 '21 at 18:26
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    Could you tell me how can I confirm $\phi(F_1(x,y))＝F_2(\phi(x),\phi(y))$,where $F_1$ and $F_2$ are formal group law of elliptic curve $E_1$, $E_2$? This is my titled question. I specified this in my question. – Poitou-Tate Apr 13 '21 at 19:37
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    It follows from how $F_1,F_2,\phi$ are constructed... Again this is just a change of coordinates and replacing the rational and algebraic functions by their formal series. The (rational/algebraic functions defining) the isogeny and the group law commute, thus so do their formal series representation. – reuns Apr 13 '21 at 20:11