# Sanchez

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

 Nov16 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? Nov16 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 $kX = -(k+1)X$. Nov14 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 Nov14 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 :) Nov14 comment How find this maximum $\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{1+c^2}$ $a+b+c=3$.${}{}$ Nov14 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. Nov14 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. Nov13 comment Trouble computing a sum of Dirichlet characters. $\sum_{\chi} \chi(a)$ can be simplified. Do you know that? Nov12 comment Primes that divide a number or $2(2n(n+1)+1) = (2n+1)^2 + 1$ Nov12 comment Primes that divide a number Do you know about quadratic residues? Nov12 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. Nov12 comment Exterior power respects $G$-action Good job :) ${}{}$ Nov11 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 $\psi(w^*)( ? )$? Nov11 comment Exterior power respects $G$-action Can you write down exactly what you get for $g \psi(w^*)$, in the form of $(w \to P(-,-))$? Nov11 comment Exterior power respects $G$-action Yes. It shouldn't be hard to check. Nov11 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. Nov11 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. Nov10 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. Nov10 comment Is the direct limit of Noetherian rings necessarily Noetherian? For your general problem, try to look at $A_m = \mathbb{C}[x_1,\cdots,x_m]$ Nov9 comment Is $A + A^{-1}$ always invertible? Good point. Thanks!