$n$th power residue conjecture
up vote
1
down vote
favorite
Let $left(dfrac{a}{p}right)_n$ be the $n$th power residue symbol such that when $a$ is an $n$th power residue $pmod p$, $left(dfrac{a}{p}right)_n=1$ otherwise $left(dfrac{a}{p}right)_n=zeta$. Let $f_n(x)$ be a defining polynomial for the $n$th cyclotomic field $K=mathbb{Q}(e^{2pi i/n})$. Suppose that there exists some integer $m$ such that $f_n(m)=0 pmod {n^2}$, and there are primes $p$ and $q$ such that $p=f_n(z)$ and $q=f_n(z+a)$. Then $left(dfrac{a}{p}right)_n = left(dfrac{a}{q}right)_n$. When $n=3$, this conjecture agrees with cubic reciprocity theorems. The same is true for $n=4$.
Here is an example of the conjecture for $n=5$:
$f_5(x)=x^4+5x^3+10x^2+25$ and $f_5(0)=0pmod {5^2}$
$a=3$, $p=f_5(1)=41$ and $q=f_5(1+a)=f_5(4)=761$ are both prime, thus
$left(dfrac{3}{41}right)_5 = left(dfrac{3}{761}right)_5$.
Since $3$ is a quintic residue $pmod {41}$, then it is also a quintic residue of $761$. In general, if $p=f_5(3a+1)$ is prime, then $3$ is a quintic residue $pmod p$.
elementary-number-theory cubic-reciprocity
asked Nov 20 at 1:40
J. Linne
829315
add a comment |
up vote
1
down vote
favorite
Let $left(dfrac{a}{p}right)_n$ be the $n$th power residue symbol such that when $a$ is an $n$th power residue $pmod p$, $left(dfrac{a}{p}right)_n=1$ otherwise $left(dfrac{a}{p}right)_n=zeta$. Let $f_n(x)$ be a defining polynomial for the $n$th cyclotomic field $K=mathbb{Q}(e^{2pi i/n})$. Suppose that there exists some integer $m$ such that $f_n(m)=0 pmod {n^2}$, and there are primes $p$ and $q$ such that $p=f_n(z)$ and $q=f_n(z+a)$. Then $left(dfrac{a}{p}right)_n = left(dfrac{a}{q}right)_n$. When $n=3$, this conjecture agrees with cubic reciprocity theorems. The same is true for $n=4$.
Here is an example of the conjecture for $n=5$:
$f_5(x)=x^4+5x^3+10x^2+25$ and $f_5(0)=0pmod {5^2}$
$a=3$, $p=f_5(1)=41$ and $q=f_5(1+a)=f_5(4)=761$ are both prime, thus
$left(dfrac{3}{41}right)_5 = left(dfrac{3}{761}right)_5$.
Since $3$ is a quintic residue $pmod {41}$, then it is also a quintic residue of $761$. In general, if $p=f_5(3a+1)$ is prime, then $3$ is a quintic residue $pmod p$.
elementary-number-theory cubic-reciprocity
asked Nov 20 at 1:40
J. Linne
829315
2
Your $n$-th power residue symbol is not standard when $n > 2$: its value at numbers that are not $n$th powers mod $p$ should not be $-1$, but some $n$th root of unity if you want the symbol to be multiplicative in any reasonable way. Also, using ordinary prime numbers rather than prime ideals in $mathbf Z[e^{2pi i/n}]$ seems unusual, e.g., cubic reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/3}]$ and quartic (biquadratic) reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/4}] = mathbf Z[i]$. Your conjecture has a gap: what type of number is $z$?
– KCd
2 days ago
@KCd changing -1 to a root of unity is only nessesary when dealing with cosets of the groups of $n$th power residues but I changed it anyways. As for $z$, like $x, a,$ and $n$ is an integer.
– J. Linne
yesterday
For $n > 2$, an $n$th power residue symbol does not just have two nonzero values, "1 and $zeta$". Its nonzero values run over all $n$th roots of unity, in order to be multiplicative in $a$, so your description of $(frac{a}{p})_n$ is unusual (esp. since $p$ need not be prime in $mathbf Z[e^{2pi i/n}]$, e.g., $p = 41$ when $n = 5$. Where are you getting your description of power residue symbols from?
– KCd
yesterday
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $left(dfrac{a}{p}right)_n$ be the $n$th power residue symbol such that when $a$ is an $n$th power residue $pmod p$, $left(dfrac{a}{p}right)_n=1$ otherwise $left(dfrac{a}{p}right)_n=zeta$. Let $f_n(x)$ be a defining polynomial for the $n$th cyclotomic field $K=mathbb{Q}(e^{2pi i/n})$. Suppose that there exists some integer $m$ such that $f_n(m)=0 pmod {n^2}$, and there are primes $p$ and $q$ such that $p=f_n(z)$ and $q=f_n(z+a)$. Then $left(dfrac{a}{p}right)_n = left(dfrac{a}{q}right)_n$. When $n=3$, this conjecture agrees with cubic reciprocity theorems. The same is true for $n=4$.
Here is an example of the conjecture for $n=5$:
$f_5(x)=x^4+5x^3+10x^2+25$ and $f_5(0)=0pmod {5^2}$
$a=3$, $p=f_5(1)=41$ and $q=f_5(1+a)=f_5(4)=761$ are both prime, thus
$left(dfrac{3}{41}right)_5 = left(dfrac{3}{761}right)_5$.
Since $3$ is a quintic residue $pmod {41}$, then it is also a quintic residue of $761$. In general, if $p=f_5(3a+1)$ is prime, then $3$ is a quintic residue $pmod p$.
elementary-number-theory cubic-reciprocity
asked Nov 20 at 1:40
J. Linne
829315
Let $left(dfrac{a}{p}right)_n$ be the $n$th power residue symbol such that when $a$ is an $n$th power residue $pmod p$, $left(dfrac{a}{p}right)_n=1$ otherwise $left(dfrac{a}{p}right)_n=zeta$. Let $f_n(x)$ be a defining polynomial for the $n$th cyclotomic field $K=mathbb{Q}(e^{2pi i/n})$. Suppose that there exists some integer $m$ such that $f_n(m)=0 pmod {n^2}$, and there are primes $p$ and $q$ such that $p=f_n(z)$ and $q=f_n(z+a)$. Then $left(dfrac{a}{p}right)_n = left(dfrac{a}{q}right)_n$. When $n=3$, this conjecture agrees with cubic reciprocity theorems. The same is true for $n=4$.
Here is an example of the conjecture for $n=5$:
$f_5(x)=x^4+5x^3+10x^2+25$ and $f_5(0)=0pmod {5^2}$
$a=3$, $p=f_5(1)=41$ and $q=f_5(1+a)=f_5(4)=761$ are both prime, thus
$left(dfrac{3}{41}right)_5 = left(dfrac{3}{761}right)_5$.
Since $3$ is a quintic residue $pmod {41}$, then it is also a quintic residue of $761$. In general, if $p=f_5(3a+1)$ is prime, then $3$ is a quintic residue $pmod p$.
elementary-number-theory cubic-reciprocity
elementary-number-theory cubic-reciprocity
asked Nov 20 at 1:40
J. Linne
829315
asked Nov 20 at 1:40
J. Linne
829315
edited yesterday
asked Nov 20 at 1:40
J. Linne
829315
asked Nov 20 at 1:40
J. Linne
829315
asked Nov 20 at 1:40
J. Linne
829315
829315
2
Your $n$-th power residue symbol is not standard when $n > 2$: its value at numbers that are not $n$th powers mod $p$ should not be $-1$, but some $n$th root of unity if you want the symbol to be multiplicative in any reasonable way. Also, using ordinary prime numbers rather than prime ideals in $mathbf Z[e^{2pi i/n}]$ seems unusual, e.g., cubic reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/3}]$ and quartic (biquadratic) reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/4}] = mathbf Z[i]$. Your conjecture has a gap: what type of number is $z$?
– KCd
2 days ago
@KCd changing -1 to a root of unity is only nessesary when dealing with cosets of the groups of $n$th power residues but I changed it anyways. As for $z$, like $x, a,$ and $n$ is an integer.
– J. Linne
yesterday
For $n > 2$, an $n$th power residue symbol does not just have two nonzero values, "1 and $zeta$". Its nonzero values run over all $n$th roots of unity, in order to be multiplicative in $a$, so your description of $(frac{a}{p})_n$ is unusual (esp. since $p$ need not be prime in $mathbf Z[e^{2pi i/n}]$, e.g., $p = 41$ when $n = 5$. Where are you getting your description of power residue symbols from?
– KCd
yesterday
add a comment |
2
Your $n$-th power residue symbol is not standard when $n > 2$: its value at numbers that are not $n$th powers mod $p$ should not be $-1$, but some $n$th root of unity if you want the symbol to be multiplicative in any reasonable way. Also, using ordinary prime numbers rather than prime ideals in $mathbf Z[e^{2pi i/n}]$ seems unusual, e.g., cubic reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/3}]$ and quartic (biquadratic) reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/4}] = mathbf Z[i]$. Your conjecture has a gap: what type of number is $z$?
– KCd
2 days ago
@KCd changing -1 to a root of unity is only nessesary when dealing with cosets of the groups of $n$th power residues but I changed it anyways. As for $z$, like $x, a,$ and $n$ is an integer.
– J. Linne
yesterday
For $n > 2$, an $n$th power residue symbol does not just have two nonzero values, "1 and $zeta$". Its nonzero values run over all $n$th roots of unity, in order to be multiplicative in $a$, so your description of $(frac{a}{p})_n$ is unusual (esp. since $p$ need not be prime in $mathbf Z[e^{2pi i/n}]$, e.g., $p = 41$ when $n = 5$. Where are you getting your description of power residue symbols from?
– KCd
yesterday
2
2
Your $n$-th power residue symbol is not standard when $n > 2$: its value at numbers that are not $n$th powers mod $p$ should not be $-1$, but some $n$th root of unity if you want the symbol to be multiplicative in any reasonable way. Also, using ordinary prime numbers rather than prime ideals in $mathbf Z[e^{2pi i/n}]$ seems unusual, e.g., cubic reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/3}]$ and quartic (biquadratic) reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/4}] = mathbf Z[i]$. Your conjecture has a gap: what type of number is $z$?
– KCd
2 days ago
Your $n$-th power residue symbol is not standard when $n > 2$: its value at numbers that are not $n$th powers mod $p$ should not be $-1$, but some $n$th root of unity if you want the symbol to be multiplicative in any reasonable way. Also, using ordinary prime numbers rather than prime ideals in $mathbf Z[e^{2pi i/n}]$ seems unusual, e.g., cubic reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/3}]$ and quartic (biquadratic) reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/4}] = mathbf Z[i]$. Your conjecture has a gap: what type of number is $z$?
– KCd
2 days ago
@KCd changing -1 to a root of unity is only nessesary when dealing with cosets of the groups of $n$th power residues but I changed it anyways. As for $z$, like $x, a,$ and $n$ is an integer.
– J. Linne
yesterday
@KCd changing -1 to a root of unity is only nessesary when dealing with cosets of the groups of $n$th power residues but I changed it anyways. As for $z$, like $x, a,$ and $n$ is an integer.
– J. Linne
yesterday
For $n > 2$, an $n$th power residue symbol does not just have two nonzero values, "1 and $zeta$". Its nonzero values run over all $n$th roots of unity, in order to be multiplicative in $a$, so your description of $(frac{a}{p})_n$ is unusual (esp. since $p$ need not be prime in $mathbf Z[e^{2pi i/n}]$, e.g., $p = 41$ when $n = 5$. Where are you getting your description of power residue symbols from?
– KCd
yesterday
For $n > 2$, an $n$th power residue symbol does not just have two nonzero values, "1 and $zeta$". Its nonzero values run over all $n$th roots of unity, in order to be multiplicative in $a$, so your description of $(frac{a}{p})_n$ is unusual (esp. since $p$ need not be prime in $mathbf Z[e^{2pi i/n}]$, e.g., $p = 41$ when $n = 5$. Where are you getting your description of power residue symbols from?
– KCd
yesterday
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005811%2fnth-power-residue-conjecture%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
Your $n$-th power residue symbol is not standard when $n > 2$: its value at numbers that are not $n$th powers mod $p$ should not be $-1$, but some $n$th root of unity if you want the symbol to be multiplicative in any reasonable way. Also, using ordinary prime numbers rather than prime ideals in $mathbf Z[e^{2pi i/n}]$ seems unusual, e.g., cubic reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/3}]$ and quartic (biquadratic) reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/4}] = mathbf Z[i]$. Your conjecture has a gap: what type of number is $z$?
– KCd
2 days ago
@KCd changing -1 to a root of unity is only nessesary when dealing with cosets of the groups of $n$th power residues but I changed it anyways. As for $z$, like $x, a,$ and $n$ is an integer.
– J. Linne
yesterday
For $n > 2$, an $n$th power residue symbol does not just have two nonzero values, "1 and $zeta$". Its nonzero values run over all $n$th roots of unity, in order to be multiplicative in $a$, so your description of $(frac{a}{p})_n$ is unusual (esp. since $p$ need not be prime in $mathbf Z[e^{2pi i/n}]$, e.g., $p = 41$ when $n = 5$. Where are you getting your description of power residue symbols from?
– KCd
yesterday