Quadratic reciprocity and Cox
Cox proves the following statement using quadratic reciprocity:
If $Dequiv 0, 1mod 4$ is a nonzero integer, then there is a unique homomorphism $chi:(mathbb{Z}/Dmathbb{Z})^*to{pm 1}$ such that $chi([p]) = (D/p)$ for odd primes $p$ not dividing $D$.
Exercise 1.13 then asks to show that the above statement implies quadratic reciprocity, showing that the two are equivalent. The hint given is as follows:
If $p$, $q$ are distinct odd primes, let $q^* = (-1)^{(q-1)/2}q$; then $q^*equiv 1mod{4}$, so that $(q^*/cdot)$ gives a homomorphism from $(mathbb{Z}/qmathbb{Z})^*to{pm 1}$. $(cdot/q)$ is also a homomorphism between these groups. Show that these two homomorphisms are the same.
The second homomorphism, $(./q)$, is obviously nontrivial. If we can show that the first is nontrivial, then the two must be equal, since they must both equal $-1$ on some generator of $(mathbb{Z}/qmathbb{Z})^*$. But I can't see why (without using quadratic reciprocity or heavier machinery) the first map must be nontrivial.
So why is that true?
number-theory
asked Nov 28 at 22:23
rogerl
17.4k22746
|
show 4 more comments
Cox proves the following statement using quadratic reciprocity:
If $Dequiv 0, 1mod 4$ is a nonzero integer, then there is a unique homomorphism $chi:(mathbb{Z}/Dmathbb{Z})^*to{pm 1}$ such that $chi([p]) = (D/p)$ for odd primes $p$ not dividing $D$.
Exercise 1.13 then asks to show that the above statement implies quadratic reciprocity, showing that the two are equivalent. The hint given is as follows:
If $p$, $q$ are distinct odd primes, let $q^* = (-1)^{(q-1)/2}q$; then $q^*equiv 1mod{4}$, so that $(q^*/cdot)$ gives a homomorphism from $(mathbb{Z}/qmathbb{Z})^*to{pm 1}$. $(cdot/q)$ is also a homomorphism between these groups. Show that these two homomorphisms are the same.
The second homomorphism, $(./q)$, is obviously nontrivial. If we can show that the first is nontrivial, then the two must be equal, since they must both equal $-1$ on some generator of $(mathbb{Z}/qmathbb{Z})^*$. But I can't see why (without using quadratic reciprocity or heavier machinery) the first map must be nontrivial.
So why is that true?
number-theory
asked Nov 28 at 22:23
rogerl
17.4k22746
As it is your question is ill-posed. At first it is not at all obvious that $(frac{D}{p})$ is periodic in $p$, that's the key of quadratic reciprocity, and it follows from expressing $pm sqrt{D}$ as a product of quadratic Gauss sums $sum_{n=1}^m e^{2i pi n^2/m}$.
– reuns
Nov 29 at 5:01
@reuns In the text, the shaded statement is proved using quadratic reciprocity. This exercise is an attempt to show that the shaded statement is in fact equivalent to quadratic reciprocity. Although I understand your point, I don't think the question is ill-posed, since the shaded statement is assumed true in the exercise statement.
– rogerl
Nov 29 at 13:01
It is. Do you assume that for $q equiv 1 bmod 4$ and $p$ prime $f(p) = (frac{q}{p})$ then $f(p_1) = f(p_2)$ when $q | p_1-p_2$ ? That $f(p_1)f(p_2) = f(p_3)$ when $q | p_1p_2-p_3$ ?
– reuns
Nov 29 at 14:10
Yes. Clearly part of the assumption is that the given map is well-defined.
– rogerl
Nov 29 at 14:31
Another way of asking my question is: is there a simple way to see that if $q$ is an odd prime, then there is some odd prime $p$ such that $(q/p)=-1$?
– rogerl
Nov 29 at 14:33
|
show 4 more comments
Cox proves the following statement using quadratic reciprocity:
If $Dequiv 0, 1mod 4$ is a nonzero integer, then there is a unique homomorphism $chi:(mathbb{Z}/Dmathbb{Z})^*to{pm 1}$ such that $chi([p]) = (D/p)$ for odd primes $p$ not dividing $D$.
Exercise 1.13 then asks to show that the above statement implies quadratic reciprocity, showing that the two are equivalent. The hint given is as follows:
If $p$, $q$ are distinct odd primes, let $q^* = (-1)^{(q-1)/2}q$; then $q^*equiv 1mod{4}$, so that $(q^*/cdot)$ gives a homomorphism from $(mathbb{Z}/qmathbb{Z})^*to{pm 1}$. $(cdot/q)$ is also a homomorphism between these groups. Show that these two homomorphisms are the same.
The second homomorphism, $(./q)$, is obviously nontrivial. If we can show that the first is nontrivial, then the two must be equal, since they must both equal $-1$ on some generator of $(mathbb{Z}/qmathbb{Z})^*$. But I can't see why (without using quadratic reciprocity or heavier machinery) the first map must be nontrivial.
So why is that true?
number-theory
asked Nov 28 at 22:23
rogerl
17.4k22746
Cox proves the following statement using quadratic reciprocity:
If $Dequiv 0, 1mod 4$ is a nonzero integer, then there is a unique homomorphism $chi:(mathbb{Z}/Dmathbb{Z})^*to{pm 1}$ such that $chi([p]) = (D/p)$ for odd primes $p$ not dividing $D$.
Exercise 1.13 then asks to show that the above statement implies quadratic reciprocity, showing that the two are equivalent. The hint given is as follows:
If $p$, $q$ are distinct odd primes, let $q^* = (-1)^{(q-1)/2}q$; then $q^*equiv 1mod{4}$, so that $(q^*/cdot)$ gives a homomorphism from $(mathbb{Z}/qmathbb{Z})^*to{pm 1}$. $(cdot/q)$ is also a homomorphism between these groups. Show that these two homomorphisms are the same.
The second homomorphism, $(./q)$, is obviously nontrivial. If we can show that the first is nontrivial, then the two must be equal, since they must both equal $-1$ on some generator of $(mathbb{Z}/qmathbb{Z})^*$. But I can't see why (without using quadratic reciprocity or heavier machinery) the first map must be nontrivial.
So why is that true?
number-theory
number-theory
asked Nov 28 at 22:23
rogerl
17.4k22746
asked Nov 28 at 22:23
rogerl
17.4k22746
edited Nov 29 at 15:07
asked Nov 28 at 22:23
rogerl
17.4k22746
asked Nov 28 at 22:23
rogerl
17.4k22746
asked Nov 28 at 22:23
rogerl
17.4k22746
17.4k22746
As it is your question is ill-posed. At first it is not at all obvious that $(frac{D}{p})$ is periodic in $p$, that's the key of quadratic reciprocity, and it follows from expressing $pm sqrt{D}$ as a product of quadratic Gauss sums $sum_{n=1}^m e^{2i pi n^2/m}$.
– reuns
Nov 29 at 5:01
@reuns In the text, the shaded statement is proved using quadratic reciprocity. This exercise is an attempt to show that the shaded statement is in fact equivalent to quadratic reciprocity. Although I understand your point, I don't think the question is ill-posed, since the shaded statement is assumed true in the exercise statement.
– rogerl
Nov 29 at 13:01
It is. Do you assume that for $q equiv 1 bmod 4$ and $p$ prime $f(p) = (frac{q}{p})$ then $f(p_1) = f(p_2)$ when $q | p_1-p_2$ ? That $f(p_1)f(p_2) = f(p_3)$ when $q | p_1p_2-p_3$ ?
– reuns
Nov 29 at 14:10
Yes. Clearly part of the assumption is that the given map is well-defined.
– rogerl
Nov 29 at 14:31
Another way of asking my question is: is there a simple way to see that if $q$ is an odd prime, then there is some odd prime $p$ such that $(q/p)=-1$?
– rogerl
Nov 29 at 14:33
|
show 4 more comments
As it is your question is ill-posed. At first it is not at all obvious that $(frac{D}{p})$ is periodic in $p$, that's the key of quadratic reciprocity, and it follows from expressing $pm sqrt{D}$ as a product of quadratic Gauss sums $sum_{n=1}^m e^{2i pi n^2/m}$.
– reuns
Nov 29 at 5:01
@reuns In the text, the shaded statement is proved using quadratic reciprocity. This exercise is an attempt to show that the shaded statement is in fact equivalent to quadratic reciprocity. Although I understand your point, I don't think the question is ill-posed, since the shaded statement is assumed true in the exercise statement.
– rogerl
Nov 29 at 13:01
It is. Do you assume that for $q equiv 1 bmod 4$ and $p$ prime $f(p) = (frac{q}{p})$ then $f(p_1) = f(p_2)$ when $q | p_1-p_2$ ? That $f(p_1)f(p_2) = f(p_3)$ when $q | p_1p_2-p_3$ ?
– reuns
Nov 29 at 14:10
Yes. Clearly part of the assumption is that the given map is well-defined.
– rogerl
Nov 29 at 14:31
Another way of asking my question is: is there a simple way to see that if $q$ is an odd prime, then there is some odd prime $p$ such that $(q/p)=-1$?
– rogerl
Nov 29 at 14:33
As it is your question is ill-posed. At first it is not at all obvious that $(frac{D}{p})$ is periodic in $p$, that's the key of quadratic reciprocity, and it follows from expressing $pm sqrt{D}$ as a product of quadratic Gauss sums $sum_{n=1}^m e^{2i pi n^2/m}$.
– reuns
Nov 29 at 5:01
As it is your question is ill-posed. At first it is not at all obvious that $(frac{D}{p})$ is periodic in $p$, that's the key of quadratic reciprocity, and it follows from expressing $pm sqrt{D}$ as a product of quadratic Gauss sums $sum_{n=1}^m e^{2i pi n^2/m}$.
– reuns
Nov 29 at 5:01
@reuns In the text, the shaded statement is proved using quadratic reciprocity. This exercise is an attempt to show that the shaded statement is in fact equivalent to quadratic reciprocity. Although I understand your point, I don't think the question is ill-posed, since the shaded statement is assumed true in the exercise statement.
– rogerl
Nov 29 at 13:01
@reuns In the text, the shaded statement is proved using quadratic reciprocity. This exercise is an attempt to show that the shaded statement is in fact equivalent to quadratic reciprocity. Although I understand your point, I don't think the question is ill-posed, since the shaded statement is assumed true in the exercise statement.
– rogerl
Nov 29 at 13:01
It is. Do you assume that for $q equiv 1 bmod 4$ and $p$ prime $f(p) = (frac{q}{p})$ then $f(p_1) = f(p_2)$ when $q | p_1-p_2$ ? That $f(p_1)f(p_2) = f(p_3)$ when $q | p_1p_2-p_3$ ?
– reuns
Nov 29 at 14:10
It is. Do you assume that for $q equiv 1 bmod 4$ and $p$ prime $f(p) = (frac{q}{p})$ then $f(p_1) = f(p_2)$ when $q | p_1-p_2$ ? That $f(p_1)f(p_2) = f(p_3)$ when $q | p_1p_2-p_3$ ?
– reuns
Nov 29 at 14:10
Yes. Clearly part of the assumption is that the given map is well-defined.
– rogerl
Nov 29 at 14:31
Yes. Clearly part of the assumption is that the given map is well-defined.
– rogerl
Nov 29 at 14:31
Another way of asking my question is: is there a simple way to see that if $q$ is an odd prime, then there is some odd prime $p$ such that $(q/p)=-1$?
– rogerl
Nov 29 at 14:33
Another way of asking my question is: is there a simple way to see that if $q$ is an odd prime, then there is some odd prime $p$ such that $(q/p)=-1$?
– rogerl
Nov 29 at 14:33
|
show 4 more comments
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As it is your question is ill-posed. At first it is not at all obvious that $(frac{D}{p})$ is periodic in $p$, that's the key of quadratic reciprocity, and it follows from expressing $pm sqrt{D}$ as a product of quadratic Gauss sums $sum_{n=1}^m e^{2i pi n^2/m}$.
– reuns
Nov 29 at 5:01
@reuns In the text, the shaded statement is proved using quadratic reciprocity. This exercise is an attempt to show that the shaded statement is in fact equivalent to quadratic reciprocity. Although I understand your point, I don't think the question is ill-posed, since the shaded statement is assumed true in the exercise statement.
– rogerl
Nov 29 at 13:01
It is. Do you assume that for $q equiv 1 bmod 4$ and $p$ prime $f(p) = (frac{q}{p})$ then $f(p_1) = f(p_2)$ when $q | p_1-p_2$ ? That $f(p_1)f(p_2) = f(p_3)$ when $q | p_1p_2-p_3$ ?
– reuns
Nov 29 at 14:10
Yes. Clearly part of the assumption is that the given map is well-defined.
– rogerl
Nov 29 at 14:31
Another way of asking my question is: is there a simple way to see that if $q$ is an odd prime, then there is some odd prime $p$ such that $(q/p)=-1$?
– rogerl
Nov 29 at 14:33