Factorisation of fundamental discriminants
$begingroup$
$D$ is a fundamental discriminant if $Dequiv 1pmod 4$ and $D$ is square-free or $D=4m,$ where $mequiv 2,3 pmod 4$ and $m$ is square-free.
On wikipedia I found the following characterisation: fundamental discriminants can be characterized by their factorisation into positive and negative prime powers. If we define the set $$S={-8,-4,8,-3,-5,-7,-11,13,dots}$$ where the prime numbers $equiv 1pmod4$ are positive and those $equiv 3pmod 4$ are negative. Then, $Dneq 1$ is a fundamental discriminant if, and only if, it is the product of pairwise relatively prime elements of S. However, unfortunately there is no reference given. Or is this just a trivial observation because in one case in order to have $Dequiv 1pmod 4$ either the primes $equiv 3pmod 4$ have to come in even mutlipliciticity which is a contradiction against the square-freeness. So we have to take the negative of these primes. Same argument for the primes $equiv 1 pmod 4$ here taking the positive ones. Using a similiar argument One could argue that $-8,-4,8$ are the only possible even integers which satisfie $D=4m$ where $mequiv 2,3 pmod 4$ and $m$ squarefree.
abstract-algebra discriminant
asked Dec 2 '18 at 22:53
MasterexeMasterexe
54
$endgroup$
add a comment |
$begingroup$
$D$ is a fundamental discriminant if $Dequiv 1pmod 4$ and $D$ is square-free or $D=4m,$ where $mequiv 2,3 pmod 4$ and $m$ is square-free.
On wikipedia I found the following characterisation: fundamental discriminants can be characterized by their factorisation into positive and negative prime powers. If we define the set $$S={-8,-4,8,-3,-5,-7,-11,13,dots}$$ where the prime numbers $equiv 1pmod4$ are positive and those $equiv 3pmod 4$ are negative. Then, $Dneq 1$ is a fundamental discriminant if, and only if, it is the product of pairwise relatively prime elements of S. However, unfortunately there is no reference given. Or is this just a trivial observation because in one case in order to have $Dequiv 1pmod 4$ either the primes $equiv 3pmod 4$ have to come in even mutlipliciticity which is a contradiction against the square-freeness. So we have to take the negative of these primes. Same argument for the primes $equiv 1 pmod 4$ here taking the positive ones. Using a similiar argument One could argue that $-8,-4,8$ are the only possible even integers which satisfie $D=4m$ where $mequiv 2,3 pmod 4$ and $m$ squarefree.
abstract-algebra discriminant
asked Dec 2 '18 at 22:53
MasterexeMasterexe
54
$endgroup$
add a comment |
$begingroup$
$D$ is a fundamental discriminant if $Dequiv 1pmod 4$ and $D$ is square-free or $D=4m,$ where $mequiv 2,3 pmod 4$ and $m$ is square-free.
On wikipedia I found the following characterisation: fundamental discriminants can be characterized by their factorisation into positive and negative prime powers. If we define the set $$S={-8,-4,8,-3,-5,-7,-11,13,dots}$$ where the prime numbers $equiv 1pmod4$ are positive and those $equiv 3pmod 4$ are negative. Then, $Dneq 1$ is a fundamental discriminant if, and only if, it is the product of pairwise relatively prime elements of S. However, unfortunately there is no reference given. Or is this just a trivial observation because in one case in order to have $Dequiv 1pmod 4$ either the primes $equiv 3pmod 4$ have to come in even mutlipliciticity which is a contradiction against the square-freeness. So we have to take the negative of these primes. Same argument for the primes $equiv 1 pmod 4$ here taking the positive ones. Using a similiar argument One could argue that $-8,-4,8$ are the only possible even integers which satisfie $D=4m$ where $mequiv 2,3 pmod 4$ and $m$ squarefree.
abstract-algebra discriminant
asked Dec 2 '18 at 22:53
MasterexeMasterexe
54
$endgroup$
$D$ is a fundamental discriminant if $Dequiv 1pmod 4$ and $D$ is square-free or $D=4m,$ where $mequiv 2,3 pmod 4$ and $m$ is square-free.
On wikipedia I found the following characterisation: fundamental discriminants can be characterized by their factorisation into positive and negative prime powers. If we define the set $$S={-8,-4,8,-3,-5,-7,-11,13,dots}$$ where the prime numbers $equiv 1pmod4$ are positive and those $equiv 3pmod 4$ are negative. Then, $Dneq 1$ is a fundamental discriminant if, and only if, it is the product of pairwise relatively prime elements of S. However, unfortunately there is no reference given. Or is this just a trivial observation because in one case in order to have $Dequiv 1pmod 4$ either the primes $equiv 3pmod 4$ have to come in even mutlipliciticity which is a contradiction against the square-freeness. So we have to take the negative of these primes. Same argument for the primes $equiv 1 pmod 4$ here taking the positive ones. Using a similiar argument One could argue that $-8,-4,8$ are the only possible even integers which satisfie $D=4m$ where $mequiv 2,3 pmod 4$ and $m$ squarefree.
abstract-algebra discriminant
abstract-algebra discriminant
asked Dec 2 '18 at 22:53
MasterexeMasterexe
54
asked Dec 2 '18 at 22:53
MasterexeMasterexe
54
asked Dec 2 '18 at 22:53
MasterexeMasterexe
54
asked Dec 2 '18 at 22:53
MasterexeMasterexe
54
asked Dec 2 '18 at 22:53
MasterexeMasterexe
54
54
add a comment |
add a comment |
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