hmIII

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seen Feb 23 at 20:45
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Feb
23
 revised Why is the cyclic decomposition of a primary torsion module not unique?
added 7 characters in body; edited tags; edited title
Feb
23
 awarded  Yearling
Feb
23
 asked Why is the cyclic decomposition of a primary torsion module not unique?
Feb
23
 comment If $A\equiv 1\pmod{3}$, then $4p=A^2+27B^2$ uniquely determines $A$.
Thanks for the detailed answer, Hurkyl.
Feb
23
 accepted If $A\equiv 1\pmod{3}$, then $4p=A^2+27B^2$ uniquely determines $A$.
Feb
5
 asked If $A\equiv 1\pmod{3}$, then $4p=A^2+27B^2$ uniquely determines $A$.
Feb
5
 awarded  Cleanup
Feb
5
 revised Uniqueness up to sign of $L$ and $M$ in $p=\frac{1}{4}(L^2+27M^2)$.
rolled back to a previous revision
Feb
5
 comment Uniqueness up to sign of $L$ and $M$ in $p=\frac{1}{4}(L^2+27M^2)$.
sorry, I changed the question. I'll roll it back to prevent confusion, and post the new question separately.
Feb
5
 revised Uniqueness up to sign of $L$ and $M$ in $p=\frac{1}{4}(L^2+27M^2)$.
added 310 characters in body; edited tags; edited title
Feb
5
 comment Uniqueness up to sign of $L$ and $M$ in $p=\frac{1}{4}(L^2+27M^2)$.
@WillJagy Here in the article on cubic reciprocity in the subsection of $p\equiv 1\pmod{3}$. I think I might change the question as I've been digging around since I posted it.
Feb
5
 asked Uniqueness up to sign of $L$ and $M$ in $p=\frac{1}{4}(L^2+27M^2)$.
Jan
29
 accepted Existence of irreducible polynomial of arbitrary degree over finite field without use of primitive element theorem?
Jan
27
 asked Existence of irreducible polynomial of arbitrary degree over finite field without use of primitive element theorem?
Jan
27
 comment Determining the sign of the Gauss sum under the change of variable $x\mapsto 1+u$.
Very helpful, thanks!
Jan
27
 accepted Determining the sign of the Gauss sum under the change of variable $x\mapsto 1+u$.
Jan
17
 revised Determining the sign of the Gauss sum under the change of variable $x\mapsto 1+u$.
added 80 characters in body
Jan
17
 asked Determining the sign of the Gauss sum under the change of variable $x\mapsto 1+u$.
Jan
10
 accepted Why does $(2/p)=\prod_{k=1}^{(p-1)/2}2\cos\left(\frac{2\pi k}{p}\right)$?
Jan
10
 comment Why does $(2/p)=\prod_{k=1}^{(p-1)/2}2\cos\left(\frac{2\pi k}{p}\right)$?
Yes, thanks for the proof!
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