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1728
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location United Kingdom
age 23
visits member for 2 years, 9 months
seen Jul 22 at 14:39

Currently studying for an Msc in Mathematics


Apr
8
asked functors on Zero-Object in $_RMod$-category
Apr
1
revised Order and element set of the group with presentation $\langle a,b \;|\; a^9=1, b^3=a^3, [a,b]=a^3\rangle$
formatting latex
Apr
1
comment Order and element set of the group with presentation $\langle a,b \;|\; a^9=1, b^3=a^3, [a,b]=a^3\rangle$
Why is the title of this question elements of a p group?
Apr
1
suggested suggested edit on Order and element set of the group with presentation $\langle a,b \;|\; a^9=1, b^3=a^3, [a,b]=a^3\rangle$
Mar
29
accepted Global Dimension 1, right annihilator
Mar
29
comment Global Dimension 1, right annihilator
@MannyReyes yes thanks
Mar
29
revised Global Dimension 1, right annihilator
edited body
Mar
29
asked Global Dimension 1, right annihilator
Mar
26
comment Sets is not Equivalent to $sets^{op}$
Ok sorry I misunderstood, so I need to show that that is an invariant under equivalence then right?
Mar
26
comment Sets is not Equivalent to $sets^{op}$
Ok but in order to do this I would need to show that the number of morphisms is an equivalence invariant right?
Mar
26
comment Sets is not Equivalent to $sets^{op}$
Thanks but why does that follow? Why is the number of invertible maps some sort of invaraint under equivalence?
Mar
26
asked Sets is not Equivalent to $sets^{op}$
Mar
25
asked Why is a perfect group called a perfect group
Mar
22
awarded  Revival
Mar
22
comment Noetherian Ring where all primes are maximal
Thanks. I have not studied the Hopkins-Levitzki theorem but I will have a look. I have posted my solution below using only what I know , which from a quick look is how it is done in Atiyah-Macdonald, for completeness. I have still excepted your answer though. Thanks!
Mar
22
answered Noetherian Ring where all primes are maximal
Mar
22
accepted Showing that $R$ is an ID with $0$ and $P$ only prime ideals.
Mar
20
asked Showing that $R$ is an ID with $0$ and $P$ only prime ideals.
Mar
17
answered Abstract algebra: Proof that if $\mathit{H}$ is a subgroup of index 2 in a finite group G, then gH=Hg for all g $\in$ G
Mar
16
comment Sylow subgroups of soluble groups
mathoverflow.net/questions/24081/… may be of some interest?