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Let $X$ be a regular, projective flat scheme over $Spec(\mathbb{Z})$, of relative dimension $d$, and look at the $L$-function $L(X, s)$, that is, the Hasse-Weil zeta function completed by the Gamma factors.

Conjecturally, it satisfies a functional equation $$ L(X, s)=\varepsilon(X) A(X)^{-s} L(X, d+1-s) $$ where $\varepsilon(X)$ and $A(X)$ (the so called conductor) are real numbers which can be defined unconditionally.

1) How is the conductor defined?

2) It is a theorem or essentially part of the definition that $X$ has everywhere good reduction if and only if $|A(X)|=1$?

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1) The conductor measures ramification in the Galois representations of $X$. It is the product (or alternating product?) of the Artin conductors of all these Galois representations.

2) No. Very much not so. $A(X)$ only measures bad reduction in the Galois representation of $X$, or it is a property only of the $L$-function. But different varieties can share the same Galois representation and $L$-function, and some of the varieties can have good reduction and some not. The simplest example is genus $0$ curves. These all have the same $L$-fucntion, and so the same conductor, $1$, but for every finite set of primes there is a genus $0$ curve with bad reduction at those priems.

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  • $\begingroup$ Is the conductor a motivic invariant? $\endgroup$
    – Qiaochu Yuan
    Commented Jan 25, 2014 at 0:23
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    $\begingroup$ I think so. Also, the $L$-function is a motivic invariant, and the equation in the question enables you to derive the conductor from the $L$-function, so it had better be. $\endgroup$
    – Will Sawin
    Commented Jan 25, 2014 at 0:52
  • $\begingroup$ Thanks for your answer Will. So the ramification of the Galois representation does not say nothing at all about the bad reduction of the variety? $\endgroup$
    – conduc
    Commented Jan 25, 2014 at 9:32
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    $\begingroup$ If the variety has good reduction then the Galois representation is unramified. The converse holds in some special cases (e.g. for abelian varieties), but it does not hold in general. $\endgroup$
    – Daniel Loughran
    Commented Jan 25, 2014 at 10:33
  • $\begingroup$ @conduc See Chandan Singh Dalawat's notes arxiv.org/abs/math/0605326 for many examples. There are also several other relevant questions/answers on MO that you can find on these matters. For instance, mathoverflow.net/questions/133840/… $\endgroup$
    – Ariyan Javanpeykar
    Commented Jan 27, 2014 at 10:04

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