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Question 1:

If $E/K$ is an elliptic curve defined over a number field $K$, then the Weil Pairing gives me that $K(\mu_n) \subseteq K(E[n](\bar{K}))$. If i identify $Gal(K(\mu_n)/K)$ with a subgroup of $(\mathbb{Z}/n\mathbb{Z})^*$ i gain a map from $Gal(K(E[n](\bar{K}))/K) \to (\mathbb{Z}/n\mathbb{Z})^*$ doing via the surjection $Gal(K(E[n](\bar{K}))/K) \to Gal(K(\mu_n)/K)$. This is the determinant map $Gal(K(E[n](\bar{K}))/K) \to Aut(\Lambda^2 E[n])$ right?

Question 2 If $A/K$ is an abelian variety over a number field $K$ does the Weil pairing gives me again an embedding $K(\mu_n) \subseteq K(A[n](\bar{K}))$ (as far as i understand the above mentioned embedding is garanteed by the non degenerancy of the form, but i have read that in this general case this last thing may fail) or are there examples where this inclusion fails? If it fails does it fails only for finitely many n(given K)? Is it possible to bound in term of the degree of K over Q, and the dimension of A the set of integers where it fails?

Question 3 Given the setting of the first two lines of Question 2, does the picture of the last lines of Question 1 generalize in this setting, for the n that gives positive answer to question 2? That is, if i take the determinant of an element $ g \in Gal(K(A[n](\bar{K}))/K)$(once thought as an automorphism of A[n]) do i get the same thing as restricting to $K(\mu_n)$ and then embedding to $(\mathbb{Z}/n\mathbb{Z})^*$?

I'm sorry for the possibly trivial question, but i'm a newbie in this field.

Thanks in advance

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For abelian varieties, there is no a priori defined Weil pairing of $A[n]$ with itself.

Rather, there is a dual abelian variety $\widehat{A},$ and a perfect pairing $A[n] \times \widehat{A}[n] \to \mu_n.$

A polarization on $A$ is an isogeny $A \to \widehat{A}$, satisfying some additional conditions. (See wikipedia, or these notes of Brian Conrad.) Given a polarization, we now get a pairing $A[n] \times A[n] \to \mu_n,$ by utilizing the induced morphism $A[n] \to \widehat{A}[n].$

If the kernel of the polarization has order prime to $n$, then this induced morphism is an isomorphism, and so the pairing on $A[n]$ is perfect. (Since this kernel has finite order, the pairing will be perfect for all $n$ not divisibly by a finite list of primes.)

If the isogeny $A \to \widehat{A}$ is actually an isomorphism, we say we have a principal polarization; in this case the induced pairing on $A[n]$ is perfect for all $n$.

Jacobians of curves always come equipped with a canonical principal polarization. (This generalizes the fact that elliptic curves are canonically principally polarized, since they are isomorphic to their own Jacobians.) But not all varieties admit a principal polarization (even if we work over $\mathbb C$). (See e.g. this post on MO.)

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