ghet

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seen Dec 3 '12 at 17:59

Dec
3
comment Singular Matrix and Characteristic Polynomial
I already proved that A is singular iff 0 is the eigenvalue of A. What is the next step?
Dec
3
asked Singular Matrix and Characteristic Polynomial
Nov
28
comment Prime and Irreducible
ideal (d) or ideal (x). Since p and x are associates then (p)=(x).
Nov
28
comment Prime and Irreducible
Ok. So p and x are associates then (p)=(x). Since p is nonzero then (p) is not {0} and (x) is not {0} so x is not zero. Since p is not unit (p) is not D and (x) is not D and x is not a unit. It remains to show that if x|ab then x|a or x|b. How? Any more tips??
Nov
28
comment Prime and Irreducible
Ok thanks so much.
Nov
28
comment Prime and Irreducible
I used a theorem that says p and x are associates if and only if (p)=(x). Does it follow that (x) is a prime ideal since (p) is prime ideal? Then does it imply that x is prime?
Nov
28
comment Prime and Irreducible
So p is prime if and only if (p) is a prime ideal. And p and x are associates if and only if (p)=(x) So (x) is a prime ideal (????) that implies x is also prime. Does it make sense?
Nov
28
asked Prime and Irreducible
Nov
28
comment Associates of Integral Domain
Really? Thanks so much Sir. :)
Nov
28
asked Associates of Integral Domain
Nov
19
asked Regular open/closed set
Nov
19
awarded  Student
Nov
19
awarded  Editor
Nov
19
revised Dense Set and Topology
deleted 1 characters in body
Nov
19
asked Dense Set and Topology