# Sanchez

less info
reputation
1928
bio website location age member for 1 year, 11 months seen 4 hours ago profile views 934

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

# 928 Actions

 Nov16 comment $\mathcal O_K/(p) \cong \mathbb F_p[T]/(T^n)$? not exactly - $p = w^n u$ for some unit $u$. I'm not sure why you need that though. Nov16 comment $\mathcal O_K/(p) \cong \mathbb F_p[T]/(T^n)$? $\mathcal{O}_K/(\omega^n)$ should look like polynomials in $\omega$ wih degree less than $n$. Can you then write down an explicit isomorphism with $\mathbb{F}_p[T]/(T^n)$? Nov16 comment Equicontinuity and Uniform Boundedness In the one-variable case, you can use mean value theorem to proceed. Can you see how that generalizes? 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 $kX = -(k+1)X$. Nov16 answered Why can't you square both sides of an equation? Nov16 comment asymptotic behavior of primitive characters ... and you only need to do summation by part to get back to $\sum_{p \leq x} \chi(p)$ Nov16 comment asymptotic behavior of primitive characters $\sum_{p \leq x} \chi(p) \log p$ is related to $\sum_{p \leq x} \chi(p)$ by summation by part. (Do you see why? If not, I suggest you look it up in standard texts on how to do the transformation) $$\sum_{p \leq x} \chi(p) \log p + \sum_{p^k \leq x, k \ge 2} \chi(p) \log p = \sum_{n \leq x} \chi(n) \Lambda(n)$$ The second quantity on the left is already bounded, in my earlier comment. This bound is smaller than the bound from Prime Number Theorem you get on the right. This means that the first quantity on the left, can be bounded by your Prime-Number-Theorem Bound, Nov16 comment asymptotic behavior of primitive characters Your character is non-trivial, so there's no $x$ term (i.e. $\delta_{\chi} = 0$). Siegel zero bound would tell you that the second term is bounded by $\frac{x}{(\log x)^A}$. The $\sqrt{q}$ in your statement comes from your big $O$ term. As for summation by part, what trouble do you have? Nov16 comment asymptotic behavior of primitive characters Once you use the bounds I mentioned, you can bound $\sum_{p \leq x} \chi(p) \log p$, then a summation by part argument would suffice. Nov15 comment asymptotic behavior of primitive characters Well, you only need to bound $\sum_{p^k \leq x, k \ge 2} \chi(p^k)\Lambda(p^k)$. Since $|\chi(p^k)| \leq 1$, you can bound it by $\sum_{p^k \leq x, k \ge 2} \Lambda(p^k)$. But then all the $p$ involved is less than $\sqrt{x}$. So this is bounded by $\sum_{p \leq \sqrt{x}} \log p \frac{\log x}{\log p}$, trivially bounded by $\sqrt{x} \log x$. Nov15 comment asymptotic behavior of primitive characters "But I have problems, as restricting the sum without the function of von Mangoldt and consider the main part within the error term." What do you mean by this? Are you looking for a derivation from a bound of $\sum_{n \leq x} \chi(n) \Lambda(n)$ to what you've written down? 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 revised How find this inequality minimum $\sum_{cyc}\sqrt{a^2+b^2+ab-2a-b+1}$ edited body 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 $2(2n(n+1)+1) = (2n+1)^2 + 1$