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Jul
22
comment Subgroups of Prüfer Group
A Prufer Group only involves one prime. Do you intend to consider $\bigoplus_{p}\mathbb{Z}[p^{\infty}]$? Although yes it is trival, just note that $ord(1/p^n)$ has arbitrarly high order for large $n$.
Jul
22
comment Subgroups of Prüfer Group
Yes, in general it is, but what did we literally just prove?
Jul
22
comment Subgroups of Prüfer Group
Any finite set of elements will lie in the subgroup generated by the largest inverse power of $p$ in their denominator, and the subgroup of such elements in finite in $\mathbb{Z}[p^{\infty}]$. Now, if $\mathbb{Z}[p^{\infty}]$ was finitely generated, wouldn't that imply it was finite?
Jul
22
comment Are there necessary and sufficient conditions on $A$ and $B$ such that each row of $AB$ has a nonzero entry?
In terms of condition on the rank, the only possible condition is that the rank are maximal, and equal to the dimension of the image space. The matrices $A=diag\{1, \dots ,1, 0,\dots 0\}$ and $B=diag\{0, \dots ,0, 1,\dots 1\}$. In terms of the second question, $|B^tA^te_i|^2>0$ is an equivalent condition.
Jul
21
comment $\mathbb{R}P^n$ is orientable iff $n$ is odd, without homology, without differential geometry
I should also note that your Homology characterization is quite wrong. The only requirement you need is $H^{dim(M)}(M)=\mathbb{Z}$, not the vanishing of the lower ones. $H^1(S^1\times S^1)=\mathbb{Z}^2$ after all.
Jul
21
comment $\mathbb{R}P^n$ is orientable iff $n$ is odd, without homology, without differential geometry
Well, what is your definition of orientation character?
Jul
21
comment Proof of Wilson's Theorem using concept of group.
@joriki Wow, I might be up a bit too late. Thanks for catching that!
Jul
21
revised Proof of Wilson's Theorem using concept of group.
deleted 23 characters in body
Jul
21
answered Proof of Wilson's Theorem using concept of group.
Jul
21
comment For which varieties is the natural map from the Chow ring to integral cohomology an injection?
This condition probably won't be of much help to you, but there is a factorization $CH^*\to \mathbb{Z}\otimes_{MU^*(*)} MU^*\to H^*$, and the latter map is known to not be a injection or surjection in even the case of complex analytic manifolds. This is proven here, and in section 7 there is given an example of a smooth complex projective variety where the map is not an injection, so the conditions would have to be pretty strong.
Jul
20
comment Prove $\mathbb Q( \sqrt2)$ has only two orderings
Uniquenss of orderings on $\mathbb{Q}$ is also your friend.
Jul
19
comment integration of $\int_0^{2\pi} cos^{2n}(t)dt$
I disagree. This is actually a decent approach, as demanstrated by the simplicity of the answer.
Jul
18
comment Can two non-abelian groups have an abelian product or coproduct?
A product of groups is abelian if and only if all of there components are. This is since two elements commute if and if all of there components commute.
Jul
17
comment Centre of matrix ring over skew field
$e^ie_j=e_{ij}$.
Jul
17
answered Verify that $\binom{n+1}{4} = \frac{\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)}{3}$ for $n \geq 4$
Jul
17
answered Centre of matrix ring over skew field
Jul
17
answered Continuity of distance function
Jul
17
comment Enlarging set where “weighted” MacLaurin series of $\frac{1}{1 - x}$ equals $\frac{1}{1 - x}$
The Mac-Laurin series of $1/(1-x)$ evaluated at $y$ will have radius of convergence $|1-y|$, For example. $\sum_i b^{-i}x^i$ converges to $1/(b-x)$ while $|x|<b$. So $\sum_i b^{-i}(x+(b-1))^i$ converges to $1/(1-x)$ in a larger radius.
Jul
17
answered Extension of rings decreasing Krull dimension
Jul
17
comment Limit of continuous convex functions.
If $f(x)=x^{2n}$, $x>0$ and $f(x)=x$ for $x<0$, this has non-continous limit and is convex by the second derivative test.