Show that the natural density is $1/2$.
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Let $$A_b= leftlbrace{p in mathbb{P}| left(frac{b}{p}right)=1 } rightrbrace $$ and $$ nu(A_b)=limlimits_{x to infty} frac{#lbrace {p in A_b|ple x}rbrace}{pi(x)}$$ the natural density.
For $b in mathbb{Z}$ square-free, prove that $$nu(A_b)=frac{1}{2}. $$
I do not have really an idea, but I know that $$ left(frac{b}{p}right)$$ is a Dirichlet-character modulo $4|b|$.
Can someone give me a hint ? Thanks for helping .
number-theory limits prime-numbers analytic-number-theory
edited Nov 15 at 15:57
Zvi
3,230223
asked Nov 15 at 14:26
Matillo
196
|
show 2 more comments
up vote
0
down vote
favorite
Let $$A_b= leftlbrace{p in mathbb{P}| left(frac{b}{p}right)=1 } rightrbrace $$ and $$ nu(A_b)=limlimits_{x to infty} frac{#lbrace {p in A_b|ple x}rbrace}{pi(x)}$$ the natural density.
For $b in mathbb{Z}$ square-free, prove that $$nu(A_b)=frac{1}{2}. $$
I do not have really an idea, but I know that $$ left(frac{b}{p}right)$$ is a Dirichlet-character modulo $4|b|$.
Can someone give me a hint ? Thanks for helping .
number-theory limits prime-numbers analytic-number-theory
edited Nov 15 at 15:57
Zvi
3,230223
asked Nov 15 at 14:26
Matillo
196
Does quadratic reciprocity help?
– Richard Martin
Nov 15 at 15:16
not really , do I have to use the Dirichlet prime number theorem ?
– Matillo
Nov 15 at 15:24
Well I was thinking that mod $b$, $p$ is half the time a quadratic residue and half the time not. But then again, I'm not sure how to prove that either. So my next guess is something on $L$ functions ...
– Richard Martin
Nov 15 at 15:26
Yes , but I am trying to understand why $$ left(frac{p}{p_{n_j}}right) $$ is +1 for 1/2 of nonzero residues mod p_j . I know this is +1 if p is a quadratic residue mod p_nj .
– Matillo
Nov 16 at 16:23
@Matillo the quadratic residues mod $p_{n_j}$ are $1^2,2^2,dots, (frac{p_{n_j}-1}{2})^2$.
– mathworker21
Nov 17 at 3:09
|
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $$A_b= leftlbrace{p in mathbb{P}| left(frac{b}{p}right)=1 } rightrbrace $$ and $$ nu(A_b)=limlimits_{x to infty} frac{#lbrace {p in A_b|ple x}rbrace}{pi(x)}$$ the natural density.
For $b in mathbb{Z}$ square-free, prove that $$nu(A_b)=frac{1}{2}. $$
I do not have really an idea, but I know that $$ left(frac{b}{p}right)$$ is a Dirichlet-character modulo $4|b|$.
Can someone give me a hint ? Thanks for helping .
number-theory limits prime-numbers analytic-number-theory
edited Nov 15 at 15:57
Zvi
3,230223
asked Nov 15 at 14:26
Matillo
196
Let $$A_b= leftlbrace{p in mathbb{P}| left(frac{b}{p}right)=1 } rightrbrace $$ and $$ nu(A_b)=limlimits_{x to infty} frac{#lbrace {p in A_b|ple x}rbrace}{pi(x)}$$ the natural density.
For $b in mathbb{Z}$ square-free, prove that $$nu(A_b)=frac{1}{2}. $$
I do not have really an idea, but I know that $$ left(frac{b}{p}right)$$ is a Dirichlet-character modulo $4|b|$.
Can someone give me a hint ? Thanks for helping .
number-theory limits prime-numbers analytic-number-theory
number-theory limits prime-numbers analytic-number-theory
edited Nov 15 at 15:57
Zvi
3,230223
asked Nov 15 at 14:26
Matillo
196
edited Nov 15 at 15:57
Zvi
3,230223
asked Nov 15 at 14:26
Matillo
196
edited Nov 15 at 15:57
Zvi
3,230223
edited Nov 15 at 15:57
Zvi
3,230223
edited Nov 15 at 15:57
Zvi
3,230223
3,230223
asked Nov 15 at 14:26
Matillo
196
asked Nov 15 at 14:26
Matillo
196
asked Nov 15 at 14:26
Matillo
196
196
Does quadratic reciprocity help?
– Richard Martin
Nov 15 at 15:16
not really , do I have to use the Dirichlet prime number theorem ?
– Matillo
Nov 15 at 15:24
Well I was thinking that mod $b$, $p$ is half the time a quadratic residue and half the time not. But then again, I'm not sure how to prove that either. So my next guess is something on $L$ functions ...
– Richard Martin
Nov 15 at 15:26
Yes , but I am trying to understand why $$ left(frac{p}{p_{n_j}}right) $$ is +1 for 1/2 of nonzero residues mod p_j . I know this is +1 if p is a quadratic residue mod p_nj .
– Matillo
Nov 16 at 16:23
@Matillo the quadratic residues mod $p_{n_j}$ are $1^2,2^2,dots, (frac{p_{n_j}-1}{2})^2$.
– mathworker21
Nov 17 at 3:09
|
show 2 more comments
Does quadratic reciprocity help?
– Richard Martin
Nov 15 at 15:16
not really , do I have to use the Dirichlet prime number theorem ?
– Matillo
Nov 15 at 15:24
Well I was thinking that mod $b$, $p$ is half the time a quadratic residue and half the time not. But then again, I'm not sure how to prove that either. So my next guess is something on $L$ functions ...
– Richard Martin
Nov 15 at 15:26
Yes , but I am trying to understand why $$ left(frac{p}{p_{n_j}}right) $$ is +1 for 1/2 of nonzero residues mod p_j . I know this is +1 if p is a quadratic residue mod p_nj .
– Matillo
Nov 16 at 16:23
@Matillo the quadratic residues mod $p_{n_j}$ are $1^2,2^2,dots, (frac{p_{n_j}-1}{2})^2$.
– mathworker21
Nov 17 at 3:09
Does quadratic reciprocity help?
– Richard Martin
Nov 15 at 15:16
Does quadratic reciprocity help?
– Richard Martin
Nov 15 at 15:16
not really , do I have to use the Dirichlet prime number theorem ?
– Matillo
Nov 15 at 15:24
not really , do I have to use the Dirichlet prime number theorem ?
– Matillo
Nov 15 at 15:24
Well I was thinking that mod $b$, $p$ is half the time a quadratic residue and half the time not. But then again, I'm not sure how to prove that either. So my next guess is something on $L$ functions ...
– Richard Martin
Nov 15 at 15:26
Well I was thinking that mod $b$, $p$ is half the time a quadratic residue and half the time not. But then again, I'm not sure how to prove that either. So my next guess is something on $L$ functions ...
– Richard Martin
Nov 15 at 15:26
Yes , but I am trying to understand why $$ left(frac{p}{p_{n_j}}right) $$ is +1 for 1/2 of nonzero residues mod p_j . I know this is +1 if p is a quadratic residue mod p_nj .
– Matillo
Nov 16 at 16:23
Yes , but I am trying to understand why $$ left(frac{p}{p_{n_j}}right) $$ is +1 for 1/2 of nonzero residues mod p_j . I know this is +1 if p is a quadratic residue mod p_nj .
– Matillo
Nov 16 at 16:23
@Matillo the quadratic residues mod $p_{n_j}$ are $1^2,2^2,dots, (frac{p_{n_j}-1}{2})^2$.
– mathworker21
Nov 17 at 3:09
@Matillo the quadratic residues mod $p_{n_j}$ are $1^2,2^2,dots, (frac{p_{n_j}-1}{2})^2$.
– mathworker21
Nov 17 at 3:09
|
show 2 more comments
1 Answer
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Write $b =p_{n_1}dots p_{n_k}$ a product of distinct primes. First assume each $p_{n_i}$ is odd. Then for $p not = p_{n_1},dots,p_{n_k}$, by quadratic reciprocity, $$(frac{b}{p}) = (frac{p_{n_1}}{p})dots(frac{p_{n_k}}{p}) = (frac{p}{p_{n_1}})dots (frac{p}{p_{n_k}})(-1)^{epsilon_p}$$ where $epsilon_p$ depends (at most) on the residue of $p$ mod $4$. By Dirichlet's theorem on primes in arithmetic progression, $p$ is equidistributed mod $4p_{n_1}dots p_{n_k}$ and thus, since $(frac{p}{p_{n_j}})$ is $+1$ for $frac{1}{2}$ of the nonzero residues mod $p_j$ and $-1$ for the others, we conclude that $(frac{b}{p})$ is $+1$ half of the time and $-1$ half of the time.
If some $p_{n_j} = 2$... (try to finish this case).
answered Nov 15 at 16:09
mathworker21
7,9831827
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Write $b =p_{n_1}dots p_{n_k}$ a product of distinct primes. First assume each $p_{n_i}$ is odd. Then for $p not = p_{n_1},dots,p_{n_k}$, by quadratic reciprocity, $$(frac{b}{p}) = (frac{p_{n_1}}{p})dots(frac{p_{n_k}}{p}) = (frac{p}{p_{n_1}})dots (frac{p}{p_{n_k}})(-1)^{epsilon_p}$$ where $epsilon_p$ depends (at most) on the residue of $p$ mod $4$. By Dirichlet's theorem on primes in arithmetic progression, $p$ is equidistributed mod $4p_{n_1}dots p_{n_k}$ and thus, since $(frac{p}{p_{n_j}})$ is $+1$ for $frac{1}{2}$ of the nonzero residues mod $p_j$ and $-1$ for the others, we conclude that $(frac{b}{p})$ is $+1$ half of the time and $-1$ half of the time.
If some $p_{n_j} = 2$... (try to finish this case).
answered Nov 15 at 16:09
mathworker21
7,9831827
add a comment |
up vote
1
down vote
Write $b =p_{n_1}dots p_{n_k}$ a product of distinct primes. First assume each $p_{n_i}$ is odd. Then for $p not = p_{n_1},dots,p_{n_k}$, by quadratic reciprocity, $$(frac{b}{p}) = (frac{p_{n_1}}{p})dots(frac{p_{n_k}}{p}) = (frac{p}{p_{n_1}})dots (frac{p}{p_{n_k}})(-1)^{epsilon_p}$$ where $epsilon_p$ depends (at most) on the residue of $p$ mod $4$. By Dirichlet's theorem on primes in arithmetic progression, $p$ is equidistributed mod $4p_{n_1}dots p_{n_k}$ and thus, since $(frac{p}{p_{n_j}})$ is $+1$ for $frac{1}{2}$ of the nonzero residues mod $p_j$ and $-1$ for the others, we conclude that $(frac{b}{p})$ is $+1$ half of the time and $-1$ half of the time.
If some $p_{n_j} = 2$... (try to finish this case).
answered Nov 15 at 16:09
mathworker21
7,9831827
add a comment |
up vote
1
down vote
up vote
1
down vote
Write $b =p_{n_1}dots p_{n_k}$ a product of distinct primes. First assume each $p_{n_i}$ is odd. Then for $p not = p_{n_1},dots,p_{n_k}$, by quadratic reciprocity, $$(frac{b}{p}) = (frac{p_{n_1}}{p})dots(frac{p_{n_k}}{p}) = (frac{p}{p_{n_1}})dots (frac{p}{p_{n_k}})(-1)^{epsilon_p}$$ where $epsilon_p$ depends (at most) on the residue of $p$ mod $4$. By Dirichlet's theorem on primes in arithmetic progression, $p$ is equidistributed mod $4p_{n_1}dots p_{n_k}$ and thus, since $(frac{p}{p_{n_j}})$ is $+1$ for $frac{1}{2}$ of the nonzero residues mod $p_j$ and $-1$ for the others, we conclude that $(frac{b}{p})$ is $+1$ half of the time and $-1$ half of the time.
If some $p_{n_j} = 2$... (try to finish this case).
answered Nov 15 at 16:09
mathworker21
7,9831827
Write $b =p_{n_1}dots p_{n_k}$ a product of distinct primes. First assume each $p_{n_i}$ is odd. Then for $p not = p_{n_1},dots,p_{n_k}$, by quadratic reciprocity, $$(frac{b}{p}) = (frac{p_{n_1}}{p})dots(frac{p_{n_k}}{p}) = (frac{p}{p_{n_1}})dots (frac{p}{p_{n_k}})(-1)^{epsilon_p}$$ where $epsilon_p$ depends (at most) on the residue of $p$ mod $4$. By Dirichlet's theorem on primes in arithmetic progression, $p$ is equidistributed mod $4p_{n_1}dots p_{n_k}$ and thus, since $(frac{p}{p_{n_j}})$ is $+1$ for $frac{1}{2}$ of the nonzero residues mod $p_j$ and $-1$ for the others, we conclude that $(frac{b}{p})$ is $+1$ half of the time and $-1$ half of the time.
If some $p_{n_j} = 2$... (try to finish this case).
answered Nov 15 at 16:09
mathworker21
7,9831827
answered Nov 15 at 16:09
mathworker21
7,9831827
answered Nov 15 at 16:09
mathworker21
7,9831827
answered Nov 15 at 16:09
mathworker21
7,9831827
7,9831827
add a comment |
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Does quadratic reciprocity help?
– Richard Martin
Nov 15 at 15:16
not really , do I have to use the Dirichlet prime number theorem ?
– Matillo
Nov 15 at 15:24
Well I was thinking that mod $b$, $p$ is half the time a quadratic residue and half the time not. But then again, I'm not sure how to prove that either. So my next guess is something on $L$ functions ...
– Richard Martin
Nov 15 at 15:26
Yes , but I am trying to understand why $$ left(frac{p}{p_{n_j}}right) $$ is +1 for 1/2 of nonzero residues mod p_j . I know this is +1 if p is a quadratic residue mod p_nj .
– Matillo
Nov 16 at 16:23
@Matillo the quadratic residues mod $p_{n_j}$ are $1^2,2^2,dots, (frac{p_{n_j}-1}{2})^2$.
– mathworker21
Nov 17 at 3:09