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Let $A$ be a non-zero abelian variety defined over a number field $F$. Let $v$ be a finite place of $F$, and let $f_v(A)$ be the usual conductor exponent of $A$ at $v$ (defined e.g. on p.500 of the article of Serre and Tate on good reduction of abelian varieties).

Questions (a): Let $B$ be a non-zero abelian variety over $F$. Is it true that $$f_v(A\times_F B)=f_v(A)+f_v(B)?$$ If not, what is the relation of $f_v(A\times_F B)$ to $f_v(A)$ and $f_v(B)$?

Suppose $A=E$ has dimension one (i.e. $A$ is an elliptic curve). Then $f_v(E)=1$ if and only if $E$ has semi-stable bad reduction at $v$.

Question (b): For arbitrary $A$ as above, can one say something similar (probably involving $dim(A)$) about $f_v$ in the case when $A$ has semi-stable reduction at $v$?

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up vote 4 down vote accepted

The answer to (a) is yes. The conductor is given by the representation of an inertia group $I_v$ in the Tate module. As $T_\ell(A\times B)=T_\ell(A)\times T_\ell(B)$, the additivity is easy to see from definition (Serre: Facteurs locaux des fonctions zêta des variétés algébriques, §2. The definition you cite is the same, but only when there is potentially good reduction).

For (b): let $t$ (resp. $u$) be the toric (resp. unipotent) rank of $A$ at $v$ (this means that the fiber at $v$ of the identity component of the Néron model of $A$ is extension an abelian variety by the product of a torus of dimension $t$ and a unipotent group of dimension $u$), then $$f_v(A)=t+2u+\delta_v$$ where $\delta_v$ is the Swan conductor. If $f_v=0$, then $A$ has good reduction. If $f_v=1$, then $u=0$ and by definition $A$ has semi-stable bad reduction at $v$ (equal to the extension of an abelian variety of dimension $\dim A -1$ by a torus of dimension $1$).

If $f_v(A)\ge 2$, this doesn't imply semi-stable reduction in general. You can have $u=1$, $t=\delta_v=0$ and $f_v(A)=2=\dim A$: e.g. if $A$ is the Jacobian of the curve $y^2=(x^3+1)(x^3+p)$, $p>3$, over $\mathbb Q$.

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What an amazing answer. Thanks a lot QiL'8!!!! –  Umit Demir Jul 13 '13 at 7:02
    
One additional question: Suppose $A$ has semi-stable bad reduction at $v$. Then $u=0$. Is it true that $\delta_v=0$, i.e. $f_v=t$? If yes, is it further true that $t=g$ (or at least $t\leq g$)? –  Umit Demir Aug 5 '13 at 14:36
    
Yes, if $u=0$, then $\delta_v=0$. The toric rank $t$ can take any value from $0$ to $g$. For any type of reduction we have $g=t+u+a$ where $a$ is the dimension of the abelian variety which occurs in the fiber at $v$ of the Néron model. –  user36560 Aug 5 '13 at 22:41
    
Thanks a lot. I realized that $g=t+u+a$ follows directly from the definitions. Is it easy to see that $\delta_v=0$ if $u=0$? –  Umit Demir Aug 8 '13 at 15:23
    
Not so easy. See Serre, "facteurs locaux des fonctions zêta...", page 19-06. –  user36560 Aug 8 '13 at 16:44
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