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asked Factor rings of polynomial rings
Sep
12
comment “Bad” primitive root $\bmod p^{2}$
Thank you very much for the clarification. I understand that, since the Artin conjecture for primitive roots is not resolved, my question is not resolved either.
Sep
11
revised “Bad” primitive root $\bmod p^{2}$
added 44 characters in body; edited title
Sep
11
comment “Bad” primitive root $\bmod p^{2}$
Thomas Andrews, KCd: Mea culpa. Yes I should have said "bad primitive root", and should have written "$m \bmod p$ is a primitive root".
Sep
11
comment “Bad” primitive root $\bmod p^{2}$
It's the terminology used in blms.oxfordjournals.org/content/6/1/42.full.pdf.
Sep
11
asked “Bad” primitive root $\bmod p^{2}$
Aug
9
comment Good books and lecture notes about category theory.
@JW: It's published. See maths.ed.ac.uk/~tl/bct. And it will be available free (online) in January, 2016.
Jul
14
awarded  Revival
Jul
2
awarded  Curious
May
31
asked Numbers of $m$-simplices in the barycentric subdivision of an $n$-simplex ($m \leq n$).
Mar
8
comment $A'B=I \rightarrow B'B \geq (A'A)^{-1}$
@Dreamer: Do you mean that they are not square matrices? Then vous avez raison. Sorry for my confusion.
Mar
8
comment $A'B=I \rightarrow B'B \geq (A'A)^{-1}$
@Dreamer: Not at all. Thomas uses the fact that $A^{\prime}$ is invertible (note that $A^{\prime} B = I$.)
Mar
7
comment Is a simply connected set connected?
I don't like to start a meta-discussion here, but why did someone give me a down vote? The first statement is a reminder of the definition, and the second statement is an easily exercise.
Mar
7
comment Is a simply connected set connected?
> the union of two open disjoint discs are simply connected< It is a disjoint union of simply connected sets, but since it is not path-connected, it cannot be simply connected by definition.
Mar
6
comment Simply connected and homotopic
>I see lots of proving left to do.< I quite agree. Besides, as I rewrote, we need many postulates to bypass pathology. So, this problem is not so easy as it appears.
Mar
6
revised Simply connected and homotopic
added 7 characters in body
Mar
6
revised Simply connected and homotopic
Reflection incorporated.
Mar
6
answered Is a simply connected set connected?
Mar
6
comment Simply connected and homotopic
>the sausage shape (assuming it includes the boundary) is not contractible< Do you mean the sausage to be $E$? Well $E$ is supposed to be a domain so must be open.
Mar
6
comment Simply connected and homotopic
On second thought if you meant, by `$C$ is homotopic to a point', that $C$ is shrinkable to a point of $E$ inside $E$, then that homotopy can be used to define a contraction of $E$ to the point inside $E$, thus $E$ is simply connected.