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I really appreciate it when you take time to answer my questions. Thanks!


Nov
16
comment $ \mathcal O_K/(p) \cong \mathbb F_p[T]/(T^n)$?
Let your uniformizer at $K$ be $\omega$, and $(p) = (\omega)^n$. Everything element in $\mathcal{O_K}$ is then a power series in $\omega$. The coefficients of the power series lie in $\mathcal{O_K}/(\omega)$. You then mod $\mathcal{O_K}$ by $(\omega^n)$ - does it ring a bell now?
Nov
16
comment Order of subgroup on elliptic curve over $Z_p$
If you use infinity as the origin, then $(x,y)$ and $(x,y')$ should be inverse to each other, meaning that $kX = -(k+1)X$.
Nov
14
comment What happens if one multiplies two elements belonging to two different groups?
By definition the group operation is defined only for two elements in the same group. So you can't multiply two elements from different groups, unless you can somehow put them together in a bigger group, e.g. the direct product of them. In your context, it maybe the case that $\mathbb{Z}/p^2\mathbb{Z} \cong \mathbb{Z}_p/p^2 \mathbb{Z}_p$ is regarded as a $\mathbb{Z}_p$-module
Nov
14
comment How find this maximum $\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{1+c^2}$
Not yet actually, you need to fix $c \ge 1$, $a \leq 1$ and so forth, but otherwise it's correct :)
Nov
14
comment How find this maximum $\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{1+c^2}$
$a+b+c=3$.${}{}$
Nov
14
comment How find this maximum $\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{1+c^2}$
$f(x) = \frac{1}{x^2+1}$ has only one inflection point in $[0,3]$. By playing with convexity it's easy to deduce that two of the variables are equal, then reduce it to a one-variable inequality.
Nov
14
comment Inequality. $\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3$
I don't think your assertion is true, although I don't have a particular counterexample in my mind. However, note that this inequality is cyclic, not symmetric.
Nov
13
comment Trouble computing a sum of Dirichlet characters.
$\sum_{\chi} \chi(a)$ can be simplified. Do you know that?
Nov
12
comment Primes that divide a number
or $2(2n(n+1)+1) = (2n+1)^2 + 1$
Nov
12
comment Primes that divide a number
Do you know about quadratic residues?
Nov
12
comment How find this inequality minimum $\sum_{i=1}^{n}a^2_{i}-2\sum_{i=1}^{n-1}a_{i}a_{i+1}$
The main issue to me is rather: if $A \ge B$, with a certain equality case $X$, you can't say that $A - 2B$ must attain minimum when $X$ happens, without more explanation.
Nov
12
comment Exterior power respects $G$-action
Good job :) ${}{}$
Nov
11
comment Exterior power respects $G$-action
No. $P(w, w^*) \in k$, where there is no $G$-action. If you figure out what $g \cdot (w \to P(w,w^*))$ is supposed to be, you are probably done. Hint: $\psi(w^*)$ lies in $(\wedge^k V)^*$. When I do the evaluation $(g \cdot \psi(w^*)) (w)$, it should be equal to $\psi(w^*)( ? )$?
Nov
11
comment Exterior power respects $G$-action
Can you write down exactly what you get for $g \psi(w^*)$, in the form of $(w \to P(-,-))$?
Nov
11
comment Exterior power respects $G$-action
Yes. It shouldn't be hard to check.
Nov
11
comment Exterior power respects $G$-action
It's probably conceptually cleaner to prove that $P(gw, gw^*) = P(w,w^*)$. This would follow from $(gv, gv^*) = (v,v^*)$, by the definition of dual representation.
Nov
11
comment Finding norm of a functional
You are on the right path. Do you know when Holder's inequality attains its equality? That may help.
Nov
10
comment That submodule generated by one element leads to submodule being finitely generated
because $R$ is Noetherian, then your $K$ is a Noetherian $R$-module, meaning that any $R$-submodule is finitely generated.
Nov
10
comment Is the direct limit of Noetherian rings necessarily Noetherian?
For your general problem, try to look at $A_m = \mathbb{C}[x_1,\cdots,x_m]$
Nov
9
comment Is $A + A^{-1}$ always invertible?
Good point. Thanks!