Mixing
$n$th power residue conjecture
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 '18 at 1:40
J. Linne
846315
add a comment |
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 '18 at 1:40
J. Linne
846315
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
Nov 20 '18 at 23:51
@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
Nov 21 '18 at 7:32
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
Nov 21 '18 at 11:24
@KCd In theory, yes your assertion is correct, but here we only care about the solvability of $a=x^n pmod p$, and there is only two "values". Solvable $(1)$ or non-solvable $(-1)$ or a particular root of unity.
– J. Linne
Nov 24 '18 at 1:46
add a comment |
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 '18 at 1:40
J. Linne
846315
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 '18 at 1:40
J. Linne
846315
asked Nov 20 '18 at 1:40
J. Linne
846315
edited Nov 21 '18 at 7:29
asked Nov 20 '18 at 1:40
J. Linne
846315
asked Nov 20 '18 at 1:40
J. Linne
846315
asked Nov 20 '18 at 1:40
J. Linne
846315
846315
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
Nov 20 '18 at 23:51
@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
Nov 21 '18 at 7:32
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
Nov 21 '18 at 11:24
@KCd In theory, yes your assertion is correct, but here we only care about the solvability of $a=x^n pmod p$, and there is only two "values". Solvable $(1)$ or non-solvable $(-1)$ or a particular root of unity.
– J. Linne
Nov 24 '18 at 1:46
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
Nov 20 '18 at 23:51
@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
Nov 21 '18 at 7:32
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
Nov 21 '18 at 11:24
@KCd In theory, yes your assertion is correct, but here we only care about the solvability of $a=x^n pmod p$, and there is only two "values". Solvable $(1)$ or non-solvable $(-1)$ or a particular root of unity.
– J. Linne
Nov 24 '18 at 1:46
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
Nov 20 '18 at 23:51
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
Nov 20 '18 at 23:51
@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
Nov 21 '18 at 7:32
@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
Nov 21 '18 at 7:32
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
Nov 21 '18 at 11:24
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
Nov 21 '18 at 11:24
@KCd In theory, yes your assertion is correct, but here we only care about the solvability of $a=x^n pmod p$, and there is only two "values". Solvable $(1)$ or non-solvable $(-1)$ or a particular root of unity.
– J. Linne
Nov 24 '18 at 1:46
@KCd In theory, yes your assertion is correct, but here we only care about the solvability of $a=x^n pmod p$, and there is only two "values". Solvable $(1)$ or non-solvable $(-1)$ or a particular root of unity.
– J. Linne
Nov 24 '18 at 1:46
add a comment |
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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
Nov 20 '18 at 23:51
@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
Nov 21 '18 at 7:32
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
Nov 21 '18 at 11:24
@KCd In theory, yes your assertion is correct, but here we only care about the solvability of $a=x^n pmod p$, and there is only two "values". Solvable $(1)$ or non-solvable $(-1)$ or a particular root of unity.
– J. Linne
Nov 24 '18 at 1:46