network profile
| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 6 months |
| seen | Dec 3 '12 at 17:59 | |
| stats | profile views | 4 |
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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? |
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Dec 3 |
asked | Singular Matrix and Characteristic Polynomial |
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Nov 28 |
comment |
Prime and Irreducible ideal (d) or ideal (x). Since p and x are associates then (p)=(x). |
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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?? |
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Nov 28 |
comment |
Prime and Irreducible Ok thanks so much. |
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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? |
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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? |
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Nov 28 |
asked | Prime and Irreducible |
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Nov 28 |
comment |
Associates of Integral Domain Really? Thanks so much Sir. :) |
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Nov 28 |
asked | Associates of Integral Domain |
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Nov 19 |
asked | Regular open/closed set |
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Nov 19 |
awarded | Student |
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Nov 19 |
awarded | Editor |
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Nov 19 |
revised |
Dense Set and Topology deleted 1 characters in body |
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Nov 19 |
asked | Dense Set and Topology |
