Mixing
Euler factors of L-function at bad primes
Clash Royale CLAN TAG#URR8PPP
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This is of course a very-well known problem, but still let me ask the questions my way. Let $L(s)$ be a "motivic" $L$-function, whatever that means: in particular, it has an Euler product (including at "bad" primes), and a (possibly conjectural) functional equation of standard type with an arithmetic conductor $N$ occurring as $N^s/2$ in the factor at infinity. Let $p$ be a prime, denote by $L_p(s)$ the denominator of the Euler factor at $p$, by $d_p$ the degree of $L_p(s)$ in $p^-s$, and by $v_p(N)$ the valuation of $N$ at $p$. Finally let $d$ be the degree of the $L$-function, equal both to the number of $Gamma_mathbb R(s)$ factors and to the generic degree in $p^-s$ of $L_p(s)$.
I define a prime to be good if it does not divide $N$ (note that if $L$ comes from a variety with bad reduction at $p$, it may still be a good prime in this sense). For the following statements, please tell me if they are true, or only sometimes true, sometimes false (and in that case possibly some extra info):
1) If $p$ is good then $d_p=d$.
2) If $d_p=d$ then $p$ is good.
3) $v_pge d-d_p$
4) I know that one can have $v_p>d-d_p$, but can one give a reasonably simple criterion for equality ?
nt.number-theory l-functions
asked 59 mins ago
Henri Cohen
2,779420
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up vote
3
down vote
favorite
This is of course a very-well known problem, but still let me ask the questions my way. Let $L(s)$ be a "motivic" $L$-function, whatever that means: in particular, it has an Euler product (including at "bad" primes), and a (possibly conjectural) functional equation of standard type with an arithmetic conductor $N$ occurring as $N^s/2$ in the factor at infinity. Let $p$ be a prime, denote by $L_p(s)$ the denominator of the Euler factor at $p$, by $d_p$ the degree of $L_p(s)$ in $p^-s$, and by $v_p(N)$ the valuation of $N$ at $p$. Finally let $d$ be the degree of the $L$-function, equal both to the number of $Gamma_mathbb R(s)$ factors and to the generic degree in $p^-s$ of $L_p(s)$.
I define a prime to be good if it does not divide $N$ (note that if $L$ comes from a variety with bad reduction at $p$, it may still be a good prime in this sense). For the following statements, please tell me if they are true, or only sometimes true, sometimes false (and in that case possibly some extra info):
1) If $p$ is good then $d_p=d$.
2) If $d_p=d$ then $p$ is good.
3) $v_pge d-d_p$
4) I know that one can have $v_p>d-d_p$, but can one give a reasonably simple criterion for equality ?
nt.number-theory l-functions
asked 59 mins ago
Henri Cohen
2,779420
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
This is of course a very-well known problem, but still let me ask the questions my way. Let $L(s)$ be a "motivic" $L$-function, whatever that means: in particular, it has an Euler product (including at "bad" primes), and a (possibly conjectural) functional equation of standard type with an arithmetic conductor $N$ occurring as $N^s/2$ in the factor at infinity. Let $p$ be a prime, denote by $L_p(s)$ the denominator of the Euler factor at $p$, by $d_p$ the degree of $L_p(s)$ in $p^-s$, and by $v_p(N)$ the valuation of $N$ at $p$. Finally let $d$ be the degree of the $L$-function, equal both to the number of $Gamma_mathbb R(s)$ factors and to the generic degree in $p^-s$ of $L_p(s)$.
I define a prime to be good if it does not divide $N$ (note that if $L$ comes from a variety with bad reduction at $p$, it may still be a good prime in this sense). For the following statements, please tell me if they are true, or only sometimes true, sometimes false (and in that case possibly some extra info):
1) If $p$ is good then $d_p=d$.
2) If $d_p=d$ then $p$ is good.
3) $v_pge d-d_p$
4) I know that one can have $v_p>d-d_p$, but can one give a reasonably simple criterion for equality ?
nt.number-theory l-functions
asked 59 mins ago
Henri Cohen
2,779420
This is of course a very-well known problem, but still let me ask the questions my way. Let $L(s)$ be a "motivic" $L$-function, whatever that means: in particular, it has an Euler product (including at "bad" primes), and a (possibly conjectural) functional equation of standard type with an arithmetic conductor $N$ occurring as $N^s/2$ in the factor at infinity. Let $p$ be a prime, denote by $L_p(s)$ the denominator of the Euler factor at $p$, by $d_p$ the degree of $L_p(s)$ in $p^-s$, and by $v_p(N)$ the valuation of $N$ at $p$. Finally let $d$ be the degree of the $L$-function, equal both to the number of $Gamma_mathbb R(s)$ factors and to the generic degree in $p^-s$ of $L_p(s)$.
I define a prime to be good if it does not divide $N$ (note that if $L$ comes from a variety with bad reduction at $p$, it may still be a good prime in this sense). For the following statements, please tell me if they are true, or only sometimes true, sometimes false (and in that case possibly some extra info):
1) If $p$ is good then $d_p=d$.
2) If $d_p=d$ then $p$ is good.
3) $v_pge d-d_p$
4) I know that one can have $v_p>d-d_p$, but can one give a reasonably simple criterion for equality ?
nt.number-theory l-functions
nt.number-theory l-functions
asked 59 mins ago
Henri Cohen
2,779420
asked 59 mins ago
Henri Cohen
2,779420
asked 59 mins ago
Henri Cohen
2,779420
asked 59 mins ago
Henri Cohen
2,779420
asked 59 mins ago
Henri Cohen
2,779420
2,779420
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
1) True.
2) True.
3) True.
4) They are equal in the case of "tame ramification", e.g. if the degree of $K(mu_p)$ over $K$, $K$ the coefficient field of the motive, is greater than $d$.
One just has to recall that for $V$ the $ell$-adic Galois representation associated to the motive, $dim V$ is the degree of the $L$-function, the local $L$-factor is the characteristic polynomial of $operatornameFrob_p$ on the invariants of the inertia group at $p$, and a prime is good if and only if the inertia group acts nontrivially. Furthermore $v_p$ is the Artin conductor which consists of the codimension of the inertia invariant subspace (i.e. $d-d_p$) plus the Swan conductor.
The Swan conductor vanishes if and only if the wild inertia group acts trivially. The last claim depends on something that I think is still conjectural, depending on the theory of motives, which is that the characteristic polynomials of elements of the Weil group are defined over $K$. From this, it follows immediately, as these characteristic polynomials have degree $d$, and for elements of teh wild inertia group, roots $p$-power roots of unity.
answered 38 mins ago
Will Sawin
65k6130272
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
1) True.
2) True.
3) True.
4) They are equal in the case of "tame ramification", e.g. if the degree of $K(mu_p)$ over $K$, $K$ the coefficient field of the motive, is greater than $d$.
One just has to recall that for $V$ the $ell$-adic Galois representation associated to the motive, $dim V$ is the degree of the $L$-function, the local $L$-factor is the characteristic polynomial of $operatornameFrob_p$ on the invariants of the inertia group at $p$, and a prime is good if and only if the inertia group acts nontrivially. Furthermore $v_p$ is the Artin conductor which consists of the codimension of the inertia invariant subspace (i.e. $d-d_p$) plus the Swan conductor.
The Swan conductor vanishes if and only if the wild inertia group acts trivially. The last claim depends on something that I think is still conjectural, depending on the theory of motives, which is that the characteristic polynomials of elements of the Weil group are defined over $K$. From this, it follows immediately, as these characteristic polynomials have degree $d$, and for elements of teh wild inertia group, roots $p$-power roots of unity.
answered 38 mins ago
Will Sawin
65k6130272
add a comment |Â
up vote
4
down vote
1) True.
2) True.
3) True.
4) They are equal in the case of "tame ramification", e.g. if the degree of $K(mu_p)$ over $K$, $K$ the coefficient field of the motive, is greater than $d$.
One just has to recall that for $V$ the $ell$-adic Galois representation associated to the motive, $dim V$ is the degree of the $L$-function, the local $L$-factor is the characteristic polynomial of $operatornameFrob_p$ on the invariants of the inertia group at $p$, and a prime is good if and only if the inertia group acts nontrivially. Furthermore $v_p$ is the Artin conductor which consists of the codimension of the inertia invariant subspace (i.e. $d-d_p$) plus the Swan conductor.
The Swan conductor vanishes if and only if the wild inertia group acts trivially. The last claim depends on something that I think is still conjectural, depending on the theory of motives, which is that the characteristic polynomials of elements of the Weil group are defined over $K$. From this, it follows immediately, as these characteristic polynomials have degree $d$, and for elements of teh wild inertia group, roots $p$-power roots of unity.
answered 38 mins ago
Will Sawin
65k6130272
add a comment |Â
up vote
4
down vote
up vote
4
down vote
1) True.
2) True.
3) True.
4) They are equal in the case of "tame ramification", e.g. if the degree of $K(mu_p)$ over $K$, $K$ the coefficient field of the motive, is greater than $d$.
One just has to recall that for $V$ the $ell$-adic Galois representation associated to the motive, $dim V$ is the degree of the $L$-function, the local $L$-factor is the characteristic polynomial of $operatornameFrob_p$ on the invariants of the inertia group at $p$, and a prime is good if and only if the inertia group acts nontrivially. Furthermore $v_p$ is the Artin conductor which consists of the codimension of the inertia invariant subspace (i.e. $d-d_p$) plus the Swan conductor.
The Swan conductor vanishes if and only if the wild inertia group acts trivially. The last claim depends on something that I think is still conjectural, depending on the theory of motives, which is that the characteristic polynomials of elements of the Weil group are defined over $K$. From this, it follows immediately, as these characteristic polynomials have degree $d$, and for elements of teh wild inertia group, roots $p$-power roots of unity.
answered 38 mins ago
Will Sawin
65k6130272
1) True.
2) True.
3) True.
4) They are equal in the case of "tame ramification", e.g. if the degree of $K(mu_p)$ over $K$, $K$ the coefficient field of the motive, is greater than $d$.
One just has to recall that for $V$ the $ell$-adic Galois representation associated to the motive, $dim V$ is the degree of the $L$-function, the local $L$-factor is the characteristic polynomial of $operatornameFrob_p$ on the invariants of the inertia group at $p$, and a prime is good if and only if the inertia group acts nontrivially. Furthermore $v_p$ is the Artin conductor which consists of the codimension of the inertia invariant subspace (i.e. $d-d_p$) plus the Swan conductor.
The Swan conductor vanishes if and only if the wild inertia group acts trivially. The last claim depends on something that I think is still conjectural, depending on the theory of motives, which is that the characteristic polynomials of elements of the Weil group are defined over $K$. From this, it follows immediately, as these characteristic polynomials have degree $d$, and for elements of teh wild inertia group, roots $p$-power roots of unity.
answered 38 mins ago
Will Sawin
65k6130272
answered 38 mins ago
Will Sawin
65k6130272
answered 38 mins ago
Will Sawin
65k6130272
answered 38 mins ago
Will Sawin
65k6130272
65k6130272
add a comment |Â
add a comment |Â
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