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How did Gauss conjecture there were nine Heegner numbers?
How does topology enter Number theory and how we can grasp its essence?Irreducibility of cyclotomic polynomials via schemesShow that $mathbb Q(sqrt p) notsimeqmathbb Q(sqrt q)$Extending the set of algebraic numbersIntuition in studying splitting and ramification of prime idealsUsing quadratic reciprocity to motivate higher reciprocity laws?Euler prime generating polynomial and Heegner numbersIs Algebraic Number Theory still an active research field?Number of ways to sum two Egyptian fractions and satisfy a given inequality.How did Gauss find the units of the cubic field $Q[n^1/3]$?
$begingroup$
Coming from someone not very knowledgable in algebraic number theory it seems odd. At the time they didn't have the computing power to determine whether very high values (>>163) were Heegner numbers; so, why even assume there was a finite amount rather than infinite (let alone exactly the nine there are)?
algebraic-number-theory
asked yesterday
Joshua FarrellJoshua Farrell
654219
$endgroup$
add a comment |
$begingroup$
Coming from someone not very knowledgable in algebraic number theory it seems odd. At the time they didn't have the computing power to determine whether very high values (>>163) were Heegner numbers; so, why even assume there was a finite amount rather than infinite (let alone exactly the nine there are)?
algebraic-number-theory
asked yesterday
Joshua FarrellJoshua Farrell
654219
$endgroup$
add a comment |
$begingroup$
Coming from someone not very knowledgable in algebraic number theory it seems odd. At the time they didn't have the computing power to determine whether very high values (>>163) were Heegner numbers; so, why even assume there was a finite amount rather than infinite (let alone exactly the nine there are)?
algebraic-number-theory
asked yesterday
Joshua FarrellJoshua Farrell
654219
$endgroup$
Coming from someone not very knowledgable in algebraic number theory it seems odd. At the time they didn't have the computing power to determine whether very high values (>>163) were Heegner numbers; so, why even assume there was a finite amount rather than infinite (let alone exactly the nine there are)?
algebraic-number-theory
algebraic-number-theory
asked yesterday
Joshua FarrellJoshua Farrell
654219
asked yesterday
Joshua FarrellJoshua Farrell
654219
asked yesterday
Joshua FarrellJoshua Farrell
654219
asked yesterday
Joshua FarrellJoshua Farrell
654219
asked yesterday
Joshua FarrellJoshua Farrell
654219
654219
add a comment |
add a comment |
2 Answers
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Gauss had plenty of computing power. He calculated class numbers up to 2000, and found they got scarcer as he climbed higher, with none at all after 163. That seemed enough to conjecture there weren't any more.
Gauss was working with quadratic forms, rather than quadratic fields, and the bigger the discriminant, the easier it was to find inequivalent forms, so it stood to reason that eventually there would be no discriminants with just one class of forms.
answered yesterday
Gerry MyersonGerry Myerson
147k8149302
$endgroup$
$begingroup$
So other than just simply computing a lot there was no strong evidence to suggest there couldn't be another?
$endgroup$
– Joshua Farrell
yesterday
add a comment |
$begingroup$
The first answer is that he did not. Gauss worked with binary quadratic forms with even middle coefficient (the determinant of a form $Ax^2 + 2Bxy + Cy^2$ is $B^2 - AC$), so some of his class numbers actually are ring class numbers modulo $2$ (equal to $3h$, where $h$ is the usual class number). In the Disquisitiones (art. 303) he gives a list, of which I give an extract here:
$$ beginarrayc
h & rm determinant \ hline
1 & 1, 2, 3, 4, 7 \
3 & 11, 19, 23, 27, 31, 43, 67, 163
endarray $$
It is of course easy to transfer this to our usual class numbers,
which is where the nine value up to $163$ are coming from.
In Gauss's case, proving that there are only finitely many determinants
with Gauss class number $1$ is actually quite easy; the difficult
part is showing that those with Gauss class number $3$ are finite.
Gauss also observed that there seem to be only finitely many determinants with given small class numbers.
In addition to computational evidence Gauss also knew that the class number of his forms gros asymptotically as
$$ gamma sqrtD - delta, $$
where $gamma = 2pi/7e$ and $delta = 2/pi^2$ (art. 302).
answered yesterday
franz lemmermeyerfranz lemmermeyer
7,19422047
$endgroup$
$begingroup$
I was hoping you would come to this question!
$endgroup$
– Gerry Myerson
yesterday
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Gauss had plenty of computing power. He calculated class numbers up to 2000, and found they got scarcer as he climbed higher, with none at all after 163. That seemed enough to conjecture there weren't any more.
Gauss was working with quadratic forms, rather than quadratic fields, and the bigger the discriminant, the easier it was to find inequivalent forms, so it stood to reason that eventually there would be no discriminants with just one class of forms.
answered yesterday
Gerry MyersonGerry Myerson
147k8149302
$endgroup$
$begingroup$
So other than just simply computing a lot there was no strong evidence to suggest there couldn't be another?
$endgroup$
– Joshua Farrell
yesterday
add a comment |
$begingroup$
Gauss had plenty of computing power. He calculated class numbers up to 2000, and found they got scarcer as he climbed higher, with none at all after 163. That seemed enough to conjecture there weren't any more.
Gauss was working with quadratic forms, rather than quadratic fields, and the bigger the discriminant, the easier it was to find inequivalent forms, so it stood to reason that eventually there would be no discriminants with just one class of forms.
answered yesterday
Gerry MyersonGerry Myerson
147k8149302
$endgroup$
$begingroup$
So other than just simply computing a lot there was no strong evidence to suggest there couldn't be another?
$endgroup$
– Joshua Farrell
yesterday
add a comment |
$begingroup$
Gauss had plenty of computing power. He calculated class numbers up to 2000, and found they got scarcer as he climbed higher, with none at all after 163. That seemed enough to conjecture there weren't any more.
Gauss was working with quadratic forms, rather than quadratic fields, and the bigger the discriminant, the easier it was to find inequivalent forms, so it stood to reason that eventually there would be no discriminants with just one class of forms.
answered yesterday
Gerry MyersonGerry Myerson
147k8149302
$endgroup$
Gauss had plenty of computing power. He calculated class numbers up to 2000, and found they got scarcer as he climbed higher, with none at all after 163. That seemed enough to conjecture there weren't any more.
Gauss was working with quadratic forms, rather than quadratic fields, and the bigger the discriminant, the easier it was to find inequivalent forms, so it stood to reason that eventually there would be no discriminants with just one class of forms.
answered yesterday
Gerry MyersonGerry Myerson
147k8149302
answered yesterday
Gerry MyersonGerry Myerson
147k8149302
answered yesterday
Gerry MyersonGerry Myerson
147k8149302
answered yesterday
Gerry MyersonGerry Myerson
147k8149302
147k8149302
$begingroup$
So other than just simply computing a lot there was no strong evidence to suggest there couldn't be another?
$endgroup$
– Joshua Farrell
yesterday
add a comment |
$begingroup$
So other than just simply computing a lot there was no strong evidence to suggest there couldn't be another?
$endgroup$
– Joshua Farrell
yesterday
$begingroup$
So other than just simply computing a lot there was no strong evidence to suggest there couldn't be another?
$endgroup$
– Joshua Farrell
yesterday
$begingroup$
So other than just simply computing a lot there was no strong evidence to suggest there couldn't be another?
$endgroup$
– Joshua Farrell
yesterday
add a comment |
$begingroup$
The first answer is that he did not. Gauss worked with binary quadratic forms with even middle coefficient (the determinant of a form $Ax^2 + 2Bxy + Cy^2$ is $B^2 - AC$), so some of his class numbers actually are ring class numbers modulo $2$ (equal to $3h$, where $h$ is the usual class number). In the Disquisitiones (art. 303) he gives a list, of which I give an extract here:
$$ beginarrayc
h & rm determinant \ hline
1 & 1, 2, 3, 4, 7 \
3 & 11, 19, 23, 27, 31, 43, 67, 163
endarray $$
It is of course easy to transfer this to our usual class numbers,
which is where the nine value up to $163$ are coming from.
In Gauss's case, proving that there are only finitely many determinants
with Gauss class number $1$ is actually quite easy; the difficult
part is showing that those with Gauss class number $3$ are finite.
Gauss also observed that there seem to be only finitely many determinants with given small class numbers.
In addition to computational evidence Gauss also knew that the class number of his forms gros asymptotically as
$$ gamma sqrtD - delta, $$
where $gamma = 2pi/7e$ and $delta = 2/pi^2$ (art. 302).
answered yesterday
franz lemmermeyerfranz lemmermeyer
7,19422047
$endgroup$
$begingroup$
I was hoping you would come to this question!
$endgroup$
– Gerry Myerson
yesterday
add a comment |
$begingroup$
The first answer is that he did not. Gauss worked with binary quadratic forms with even middle coefficient (the determinant of a form $Ax^2 + 2Bxy + Cy^2$ is $B^2 - AC$), so some of his class numbers actually are ring class numbers modulo $2$ (equal to $3h$, where $h$ is the usual class number). In the Disquisitiones (art. 303) he gives a list, of which I give an extract here:
$$ beginarrayc
h & rm determinant \ hline
1 & 1, 2, 3, 4, 7 \
3 & 11, 19, 23, 27, 31, 43, 67, 163
endarray $$
It is of course easy to transfer this to our usual class numbers,
which is where the nine value up to $163$ are coming from.
In Gauss's case, proving that there are only finitely many determinants
with Gauss class number $1$ is actually quite easy; the difficult
part is showing that those with Gauss class number $3$ are finite.
Gauss also observed that there seem to be only finitely many determinants with given small class numbers.
In addition to computational evidence Gauss also knew that the class number of his forms gros asymptotically as
$$ gamma sqrtD - delta, $$
where $gamma = 2pi/7e$ and $delta = 2/pi^2$ (art. 302).
answered yesterday
franz lemmermeyerfranz lemmermeyer
7,19422047
$endgroup$
$begingroup$
I was hoping you would come to this question!
$endgroup$
– Gerry Myerson
yesterday
add a comment |
$begingroup$
The first answer is that he did not. Gauss worked with binary quadratic forms with even middle coefficient (the determinant of a form $Ax^2 + 2Bxy + Cy^2$ is $B^2 - AC$), so some of his class numbers actually are ring class numbers modulo $2$ (equal to $3h$, where $h$ is the usual class number). In the Disquisitiones (art. 303) he gives a list, of which I give an extract here:
$$ beginarrayc
h & rm determinant \ hline
1 & 1, 2, 3, 4, 7 \
3 & 11, 19, 23, 27, 31, 43, 67, 163
endarray $$
It is of course easy to transfer this to our usual class numbers,
which is where the nine value up to $163$ are coming from.
In Gauss's case, proving that there are only finitely many determinants
with Gauss class number $1$ is actually quite easy; the difficult
part is showing that those with Gauss class number $3$ are finite.
Gauss also observed that there seem to be only finitely many determinants with given small class numbers.
In addition to computational evidence Gauss also knew that the class number of his forms gros asymptotically as
$$ gamma sqrtD - delta, $$
where $gamma = 2pi/7e$ and $delta = 2/pi^2$ (art. 302).
answered yesterday
franz lemmermeyerfranz lemmermeyer
7,19422047
$endgroup$
The first answer is that he did not. Gauss worked with binary quadratic forms with even middle coefficient (the determinant of a form $Ax^2 + 2Bxy + Cy^2$ is $B^2 - AC$), so some of his class numbers actually are ring class numbers modulo $2$ (equal to $3h$, where $h$ is the usual class number). In the Disquisitiones (art. 303) he gives a list, of which I give an extract here:
$$ beginarrayc
h & rm determinant \ hline
1 & 1, 2, 3, 4, 7 \
3 & 11, 19, 23, 27, 31, 43, 67, 163
endarray $$
It is of course easy to transfer this to our usual class numbers,
which is where the nine value up to $163$ are coming from.
In Gauss's case, proving that there are only finitely many determinants
with Gauss class number $1$ is actually quite easy; the difficult
part is showing that those with Gauss class number $3$ are finite.
Gauss also observed that there seem to be only finitely many determinants with given small class numbers.
In addition to computational evidence Gauss also knew that the class number of his forms gros asymptotically as
$$ gamma sqrtD - delta, $$
where $gamma = 2pi/7e$ and $delta = 2/pi^2$ (art. 302).
answered yesterday
franz lemmermeyerfranz lemmermeyer
7,19422047
edited 16 hours ago
answered yesterday
franz lemmermeyerfranz lemmermeyer
7,19422047
answered yesterday
franz lemmermeyerfranz lemmermeyer
7,19422047
answered yesterday
franz lemmermeyerfranz lemmermeyer
7,19422047
7,19422047
$begingroup$
I was hoping you would come to this question!
$endgroup$
– Gerry Myerson
yesterday
add a comment |
$begingroup$
I was hoping you would come to this question!
$endgroup$
– Gerry Myerson
yesterday
$begingroup$
I was hoping you would come to this question!
$endgroup$
– Gerry Myerson
yesterday
$begingroup$
I was hoping you would come to this question!
$endgroup$
– Gerry Myerson
yesterday
add a comment |
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