1
$\begingroup$

I do not quite understand the sentence in the Wikipedia article:

http://en.wikipedia.org/wiki/Weil_pairing

Section "Formulation" line 3:

"... for given points $P,Q \in E(K)[n]$, where $E(K)[n]=\{T \in E(K) \mid n \cdot T = O \}$ and $\mu_n = \{x\in K \mid x^n =1 \} $, by means of [[Kummer theory]]. "

The sentence seems to me un-understandable. Can one comment ?

More mathematical question: what Kummer theory has to do with Weil pairing ?

Improve this question
$\endgroup$
0

3 Answers 3

Reset to default
4
$\begingroup$

Let $G$ be a commutative algebraic group defined over a (number) field $K$. Write $G[n]$ for the kernel of the map $G(\overline{K})\to G(\overline{K})$ defined by $x\mapsto nx$. Let $K_n=K(G[n])$ be the field generated by $G[n]$. In this setting, Kummer theory is the theory of the extensions $K_n(y)$, where $y\in G(\overline{K})$ is a point satisfying $ny\in G(K)$. In one of the definitions of the Weil pairing on an elliptic curve, one pulls back the point $P\in E[n]$ by the multiplication-by-$n$ map and uses the resulting divisor to create a function, which is then evaluated (more or less) at the other $n$-torsion point. This procedure may be viewed as occurring in the Kummer extension $K([n]^{-1}(P),E[n])$. As Charles said, the heuristic remark in the Wikipedia may be helpful to some people, although maybe there's a way to phrase it so as to make it more helpful to more people.

Improve this answer
$\endgroup$
1
  • $\begingroup$ @Joe thank You for the answer. May I ask for the clarification of the following issue - the "relation Kummer<-> Weil" is just on the level that some objects are involved are similar or there is deeper relation (I guess it is) ? I would expect that class field theory for Kummer extensions has something to do with Weil pairing - I mean that probably Weil pairing is related to some symbol (tame, Hilbert???) is there something like that ? PS I like heuristics very much, just I like when it is clearly written - "this is heuristics, comes as it is"... $\endgroup$
    – Alexander Chervov
    Commented Jan 26, 2012 at 8:07
2
$\begingroup$

It is perhaps better to say that $\mu_n(\overline{K})$ admits a Galois action described by Kummer theory, and that the Weil pairing is Galois equivariant. This does not require the assumption in the article that $K$ contains $n$th roots of unity.

Improve this answer
$\endgroup$
2
  • $\begingroup$ @Scott Thank You again ! That sounds quite clarifying and reasonable - any natural construction like Weil pairing should be Galois equivariant. And it seems from what is written below the Weil pairing is so... $\endgroup$
    – Alexander Chervov
    Commented Jan 25, 2012 at 12:53
  • $\begingroup$ Galois equivariance - it seems we do not need Kummer theory - it is due to everything is defined algebraically, is it correct ? So not quite clear why Kummer has to do with this. $\endgroup$
    – Alexander Chervov
    Commented Jan 25, 2012 at 13:08
2
$\begingroup$

The reason Kummer theory is involved is that the Galois covering of an elliptic curve E created by multiplication by n, assuming that n is prime to the characteristic of the base field K, has Galois group that is the product of two copies of a cyclic group of order n. Under the same assumption on the characteristic, when K is algebraically closed, there are n roots of unity of order n in K. Therefore when you look at the extension of function fields corresponding to the isogeny "multiply by n on E", Kummer theory can be applied. We know the Galois group is built up from two cyclic groups, so the extension of function fields is built up, by extracting two n-th roots of functions. That is what the remark says. It shouldn't be too hard to work out the details: the typical proofs for the pairing actually construct functions and translate them.

Improve this answer
$\endgroup$
5
  • $\begingroup$ @Charles Thank You very much for your answer, but I am somehow lost with logic. Is possible to say something like: "to prove proposition ... we need ..." or "to consturct map ... we need... " ? $\endgroup$
    – Alexander Chervov
    Commented Jan 25, 2012 at 13:22
  • $\begingroup$ As far as I understand from what is written on Wiki to define Weil pairing - you do not need to know anything about Kummer. Correct ? Galois equivariance - follows that it is defined algebraically. Correct ? May be Kummer allows to give alternative construction ? Or prove some properties (what properties) ? Or there is just relation - that we see roots of unity in both subjs? $\endgroup$
    – Alexander Chervov
    Commented Jan 25, 2012 at 13:44
  • $\begingroup$ The root of w(P,Q) is defined as the ratio of two functions. What you actually need to know here is the existence theorem for a function with a given divisor on the curve: the Abel-Jacobi map in this case, but the basic result is in any book on classical elliptic functions, i.e. the divisors of functions have the same number of zeroes and poles, and when you "sum up" the divisor on the curve you get 0. Write down a divisor determined by P and Q, note that it is the same when you translate it by P, and so the function changes by a constant multiple. But please write on the Talk page! $\endgroup$
    – Charles Matthews
    Commented Jan 25, 2012 at 13:45
  • $\begingroup$ @Charles do you mean Wiki Talk page for Weil pairing ? Any way - what you write is true, but I do not see any relation with previous discussion and it does not answer the question what Kummer has to do with Weil ? $\endgroup$
    – Alexander Chervov
    Commented Jan 25, 2012 at 18:50
  • $\begingroup$ Yes, I do mean the Talk page on Wikipedia (please do not call it "Wiki"), since the point you raise is expository, and if the article needs to be improved, that is what should happen. It is also typical of Wikipedia that heuristic remarks get taken out because one person doesn't find them helpful, when maybe ten others with a different background would find them helpful. The root of unity is of form σ(F)/F where F is in the function field and σ is in the Galois group. That is exactly how roots of unity occur in Kummer theory. $\endgroup$
    – Charles Matthews
    Commented Jan 25, 2012 at 19:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .