1
$\begingroup$

$\newcommand{\Hom}{{\rm Hom}} \newcommand{\Gm}{{{\mathbb G}_{m,{\Bbb C}}}} \newcommand{\X}{{\sf X}} $ I am looking for a reference for the following lemma (for which I know a proof):

Lemma. Let $\varphi\colon S\to T$ be a surjective homomorphism of algebraic $\Bbb C$-tori with finite kernel $A=\ker\varphi$. Write $$\X^*(A):=\Hom(A,\Gm)$$ for the character group of $A$. Write $\X_*(T):=\Hom(\Gm\,,T)$ for the cocharacter group of $T$, and set $$M={\rm coker}\big[\varphi_*\colon \X_*(S)\to \X_*(T)\big].$$ Then there is a canonical nondegenerate pairing of finite abelian groups $$\X^*(A)\times M\to {\mathbb Q}/{\mathbb Z}.$$

Improve this question
$\endgroup$
16
  • 1
    $\begingroup$ @LSpice: $A=\mu_n$, ${\sf X}^*(A)={\Bbb Z}/n{\Bbb Z}$, $M=\frac1n{\Bbb Z}/{\Bbb Z}$, $$\langle \chi, x\rangle =\chi\cdot x.$$ $\endgroup$
    – Mikhail Borovoi
    Commented Jan 29, 2023 at 15:47
  • 1
    $\begingroup$ @LSpice: I regard $A$ as a finite commutative group scheme. $\endgroup$
    – Mikhail Borovoi
    Commented Jan 29, 2023 at 16:11
  • 1
    $\begingroup$ @LSpice: "Did I get the right map?" Yes, I think so. $\endgroup$
    – Mikhail Borovoi
    Commented Jan 29, 2023 at 16:16
  • 1
    $\begingroup$ @LSpice: "Did I get the right map?" Not exactly. If $A=\mu_2\times \mu_8$, you must take $n=8$, which will kill $\mu_2$. One should work with free abelian groups (the character and cocharacter groups of $S$ and $T$). $\endgroup$
    – Mikhail Borovoi
    Commented Jan 29, 2023 at 16:22
  • 1
    $\begingroup$ @LSpice: Yes, your map is correct, I did not notice that you actually compute in lattices. $\endgroup$
    – Mikhail Borovoi
    Commented Jan 29, 2023 at 16:45

0

Reset to default

You must log in to answer this question.