Uniform boundedness of the number of number fields having fixed discriminantDensity of monogenic number fields?Orders in number fieldsConstructing quintic number fields with certain splitting behaviourNumber of polynomials whose Galois group is a subgroup of the alternating groupA class number in a family of number fieldsRecognizing the Galois group from the field discriminantDoes there always exist a non-rational algebraic integer in a number field whose discriminant divides its norm?Number fields ordered by discriminantField of minimum discriminant of a given degreeHow small may the discriminant of an $S_d$-field be?
Uniform boundedness of the number of number fields having fixed discriminant
Density of monogenic number fields?Orders in number fieldsConstructing quintic number fields with certain splitting behaviourNumber of polynomials whose Galois group is a subgroup of the alternating groupA class number in a family of number fieldsRecognizing the Galois group from the field discriminantDoes there always exist a non-rational algebraic integer in a number field whose discriminant divides its norm?Number fields ordered by discriminantField of minimum discriminant of a given degreeHow small may the discriminant of an $S_d$-field be?
$begingroup$
Let $d$ be an integer. It is a well-known theorem, attributed to Hermite and Minkowski, which asserts that the number of number fields $K$, allowed to have any degree over $mathbbQ$, having discriminant $Delta_K = d$ is finite.
Let $S(d)$ be the number of isomorphism classes of number fields of discriminant $d$. The Hermite-Minkowski theorem is the assertion that $S(d) < infty$ for all $d in mathbbZ$. Is $S(d)$ uniformly bounded? That is, does there exist a positive integer $N$ such that $S(d) leq N$ for all $d in mathbbZ$?
We can refine the question and instead only count (isomorphism classes of) number fields of fixed degree over $mathbbQ$. Let $S_n(d)$ be the number of isomorphism classes of number fields of discriminant equal to $d$ and degree equal to $n$. Does there exist a number $N(n)$ such that $S_n(d) leq N(n)$ for all $d in mathbbZ$?
Observe that the answer is yes for $n = 2$. Indeed we find that $S_2(d) leq 1$ for all $d in mathbbZ$, with equality if and only if $d$ is a fundamental discriminant.
nt.number-theory algebraic-number-theory
asked May 2 at 17:05
Stanley Yao XiaoStanley Yao Xiao
8,72642787
$endgroup$
add a comment |
$begingroup$
Let $d$ be an integer. It is a well-known theorem, attributed to Hermite and Minkowski, which asserts that the number of number fields $K$, allowed to have any degree over $mathbbQ$, having discriminant $Delta_K = d$ is finite.
Let $S(d)$ be the number of isomorphism classes of number fields of discriminant $d$. The Hermite-Minkowski theorem is the assertion that $S(d) < infty$ for all $d in mathbbZ$. Is $S(d)$ uniformly bounded? That is, does there exist a positive integer $N$ such that $S(d) leq N$ for all $d in mathbbZ$?
We can refine the question and instead only count (isomorphism classes of) number fields of fixed degree over $mathbbQ$. Let $S_n(d)$ be the number of isomorphism classes of number fields of discriminant equal to $d$ and degree equal to $n$. Does there exist a number $N(n)$ such that $S_n(d) leq N(n)$ for all $d in mathbbZ$?
Observe that the answer is yes for $n = 2$. Indeed we find that $S_2(d) leq 1$ for all $d in mathbbZ$, with equality if and only if $d$ is a fundamental discriminant.
nt.number-theory algebraic-number-theory
asked May 2 at 17:05
Stanley Yao XiaoStanley Yao Xiao
8,72642787
$endgroup$
$begingroup$
For the first question, the answer seems to be no: algebra.at/multi.htm For the second, if this was known then we would get a linear bound on the number of number fields of fixed degree $n$ and discriminant $<X$ which I think is a conjecture even for low values of $n$.
$endgroup$
– François Brunault
May 2 at 17:46
add a comment |
$begingroup$
Let $d$ be an integer. It is a well-known theorem, attributed to Hermite and Minkowski, which asserts that the number of number fields $K$, allowed to have any degree over $mathbbQ$, having discriminant $Delta_K = d$ is finite.
Let $S(d)$ be the number of isomorphism classes of number fields of discriminant $d$. The Hermite-Minkowski theorem is the assertion that $S(d) < infty$ for all $d in mathbbZ$. Is $S(d)$ uniformly bounded? That is, does there exist a positive integer $N$ such that $S(d) leq N$ for all $d in mathbbZ$?
We can refine the question and instead only count (isomorphism classes of) number fields of fixed degree over $mathbbQ$. Let $S_n(d)$ be the number of isomorphism classes of number fields of discriminant equal to $d$ and degree equal to $n$. Does there exist a number $N(n)$ such that $S_n(d) leq N(n)$ for all $d in mathbbZ$?
Observe that the answer is yes for $n = 2$. Indeed we find that $S_2(d) leq 1$ for all $d in mathbbZ$, with equality if and only if $d$ is a fundamental discriminant.
nt.number-theory algebraic-number-theory
asked May 2 at 17:05
Stanley Yao XiaoStanley Yao Xiao
8,72642787
$endgroup$
Let $d$ be an integer. It is a well-known theorem, attributed to Hermite and Minkowski, which asserts that the number of number fields $K$, allowed to have any degree over $mathbbQ$, having discriminant $Delta_K = d$ is finite.
Let $S(d)$ be the number of isomorphism classes of number fields of discriminant $d$. The Hermite-Minkowski theorem is the assertion that $S(d) < infty$ for all $d in mathbbZ$. Is $S(d)$ uniformly bounded? That is, does there exist a positive integer $N$ such that $S(d) leq N$ for all $d in mathbbZ$?
We can refine the question and instead only count (isomorphism classes of) number fields of fixed degree over $mathbbQ$. Let $S_n(d)$ be the number of isomorphism classes of number fields of discriminant equal to $d$ and degree equal to $n$. Does there exist a number $N(n)$ such that $S_n(d) leq N(n)$ for all $d in mathbbZ$?
Observe that the answer is yes for $n = 2$. Indeed we find that $S_2(d) leq 1$ for all $d in mathbbZ$, with equality if and only if $d$ is a fundamental discriminant.
nt.number-theory algebraic-number-theory
nt.number-theory algebraic-number-theory
asked May 2 at 17:05
Stanley Yao XiaoStanley Yao Xiao
8,72642787
asked May 2 at 17:05
Stanley Yao XiaoStanley Yao Xiao
8,72642787
asked May 2 at 17:05
Stanley Yao XiaoStanley Yao Xiao
8,72642787
asked May 2 at 17:05
Stanley Yao XiaoStanley Yao Xiao
8,72642787
asked May 2 at 17:05
Stanley Yao XiaoStanley Yao Xiao
8,72642787
8,72642787
$begingroup$
For the first question, the answer seems to be no: algebra.at/multi.htm For the second, if this was known then we would get a linear bound on the number of number fields of fixed degree $n$ and discriminant $<X$ which I think is a conjecture even for low values of $n$.
$endgroup$
– François Brunault
May 2 at 17:46
add a comment |
$begingroup$
For the first question, the answer seems to be no: algebra.at/multi.htm For the second, if this was known then we would get a linear bound on the number of number fields of fixed degree $n$ and discriminant $<X$ which I think is a conjecture even for low values of $n$.
$endgroup$
– François Brunault
May 2 at 17:46
$begingroup$
For the first question, the answer seems to be no: algebra.at/multi.htm For the second, if this was known then we would get a linear bound on the number of number fields of fixed degree $n$ and discriminant $<X$ which I think is a conjecture even for low values of $n$.
$endgroup$
– François Brunault
May 2 at 17:46
$begingroup$
For the first question, the answer seems to be no: algebra.at/multi.htm For the second, if this was known then we would get a linear bound on the number of number fields of fixed degree $n$ and discriminant $<X$ which I think is a conjecture even for low values of $n$.
$endgroup$
– François Brunault
May 2 at 17:46
add a comment |
1 Answer
1
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votes
$begingroup$
The answers to both questions is no. For instance for $n=3$, if $p_1$,... $p_k$
are primes congruent to $1$ modulo $3$ there exist $(3^k-1)/2$ cyclic cubic fields
of discriminant equal to $(p_1...p_k)^2$ by elementary class field theory.
answered May 2 at 19:47
Henri CohenHenri Cohen
3,467525
$endgroup$
$begingroup$
Great, thanks for the example!
$endgroup$
– Stanley Yao Xiao
May 3 at 6:54
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The answers to both questions is no. For instance for $n=3$, if $p_1$,... $p_k$
are primes congruent to $1$ modulo $3$ there exist $(3^k-1)/2$ cyclic cubic fields
of discriminant equal to $(p_1...p_k)^2$ by elementary class field theory.
answered May 2 at 19:47
Henri CohenHenri Cohen
3,467525
$endgroup$
$begingroup$
Great, thanks for the example!
$endgroup$
– Stanley Yao Xiao
May 3 at 6:54
add a comment |
$begingroup$
The answers to both questions is no. For instance for $n=3$, if $p_1$,... $p_k$
are primes congruent to $1$ modulo $3$ there exist $(3^k-1)/2$ cyclic cubic fields
of discriminant equal to $(p_1...p_k)^2$ by elementary class field theory.
answered May 2 at 19:47
Henri CohenHenri Cohen
3,467525
$endgroup$
$begingroup$
Great, thanks for the example!
$endgroup$
– Stanley Yao Xiao
May 3 at 6:54
add a comment |
$begingroup$
The answers to both questions is no. For instance for $n=3$, if $p_1$,... $p_k$
are primes congruent to $1$ modulo $3$ there exist $(3^k-1)/2$ cyclic cubic fields
of discriminant equal to $(p_1...p_k)^2$ by elementary class field theory.
answered May 2 at 19:47
Henri CohenHenri Cohen
3,467525
$endgroup$
The answers to both questions is no. For instance for $n=3$, if $p_1$,... $p_k$
are primes congruent to $1$ modulo $3$ there exist $(3^k-1)/2$ cyclic cubic fields
of discriminant equal to $(p_1...p_k)^2$ by elementary class field theory.
answered May 2 at 19:47
Henri CohenHenri Cohen
3,467525
answered May 2 at 19:47
Henri CohenHenri Cohen
3,467525
answered May 2 at 19:47
Henri CohenHenri Cohen
3,467525
answered May 2 at 19:47
Henri CohenHenri Cohen
3,467525
3,467525
$begingroup$
Great, thanks for the example!
$endgroup$
– Stanley Yao Xiao
May 3 at 6:54
add a comment |
$begingroup$
Great, thanks for the example!
$endgroup$
– Stanley Yao Xiao
May 3 at 6:54
$begingroup$
Great, thanks for the example!
$endgroup$
– Stanley Yao Xiao
May 3 at 6:54
$begingroup$
Great, thanks for the example!
$endgroup$
– Stanley Yao Xiao
May 3 at 6:54
add a comment |
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For the first question, the answer seems to be no: algebra.at/multi.htm For the second, if this was known then we would get a linear bound on the number of number fields of fixed degree $n$ and discriminant $<X$ which I think is a conjecture even for low values of $n$.
$endgroup$
– François Brunault
May 2 at 17:46