Recent content by pantboio

I have the following assignment:consider the map $$|\cdot|:\mathbb{Z}[i]\longrightarrow \mathbb{N},\qquad |a+ib|:=a^2+b^2$$1) Prove that $|\alpha|<|\beta|$ iff $|\alpha|\leq |\beta|-1$ and $|\alpha|<1$ iff $\alpha=0$2) Let $\alpha,\beta\in\mathbb{Z}[i],\beta\neq 0$. Prove that the map...

Almost all is clear now for me, thanks. According to your suggestion, i take the prime $$p=11$$, which is not of the form $$A^2+13\cdot B^2$$, but has a multiple, namely its square $$121$$ such that $$121=2^2+13\cdot 3^2$$ so that the ideal $$(11,2+3\sqrt{-13})$$ is not principal, thanks to 2)...

Let $$R=\mathbb{Z}[\sqrt{-13}]$$, let $$p$$ be a prime in $$\mathbb{N}$$, $$p\neq 2,13$$. Suppose that $$p$$ divides an integer of the form $$a^2+13b^2$$, with $$a,b$$ integers and coprime. Let $$P=(p,a+b\sqrt{-13})$$ be the ideal generated in $$R$$ by $$p$$ and $$a+b\sqrt{-13}$$ and let...

Re: integral closure in finite extension fields There is a theorem which says: let $R$ be a domain, le $K$ be the fraction field of $R$, and finally let $L$ be a finite extension ok $K$. Take an element $\alpha$ in $L$. Then $\alpha$ is algebraic over $K$ and call $m_{\alpha}(X)$ its minimal...

Let $K=\mathbb{Q}[\omega]$ where $\omega^2+\omega+1=0$ and let $R$ be the polynomial ring $K[x]$. Let $L$ be the field $K(x)[y]$ where $y$ satisfies $y^3=1+x^2$.Which is the integral closure of $R$ in $L$, why?

Two facts we saw $$\displaystyle\sum_{k}\frac1{|z_k|^{2+\epsilon}}\<\infty\ \forall\epsilon, \ \textrm{and}\ \displaystyle\sum_{k}\frac1{|z_k|^2}=\infty$$ precisely mean that $b=2$, where $b$ is the convergence exponent of the sequence of zeros of $F$, which is defined as...

Thank you very much. I used a theorem which tells that if $\{f_j\}$ is a sequence of functions, analytic somewhere, say in a domain $\Omega$, such that the series $\displaystyle\sum_{j=1}^{+\infty}|f_j|$ converges uniformly on compact subsets of $\Omega$, then the infinite product...

Consider the product $$\displaystyle\prod_{n=1}^{+\infty}(1-e^{-2\pi n}e^{2\pi iz})$$ I've proven that this product converges uniformly on compact subsets of complex plane since the serie $\sum_{n=0}^{+\infty}|\frac{e^{2\pi iz}}{e^{2\pi n}}|$ does. Now I'm interested to zeros of $F$, the...

using your trick i got what follows $$f(z)=\frac{ e^{ \frac{1}{z-1} }}{e^z-1}=\frac{ \sum\frac{1}{n!(z-1)^n} }{e-1+e(z-1)+\frac{e}{2}(z-1)^2+\ldots}=\frac{ \sum\frac{1}{n!(z-1)^n}}{(e-1)(1-h)}$$ where $h=\frac{e}{1-e}(z-1)+\frac{e}{2(1-e)}(z-1)^2+\ldots$ So what we have is...

unfortunately, i don't know how to divide one series by another...

$e^{z}-1=e^{z-1+1}-1=e\cdot e^{z-1}-1=e\cdot\sum\frac{(z-1)^n}{n!} -1$

Consider the function $$f(z)=\frac{e^{\frac{1}{z-1}}}{e^z -1}$$ $z_0=1$ is an essential singularity, hence $$f(z)=\displaystyle\sum_{-\infty}^{+\infty}a_n(z-1)^n$$ near to $z_0=1$ and i want to find $a_{-1}$. I can write $$f(z)=\frac{\sum\frac{1}{n!(z-1)^n}}{e\cdot...

so the residue is the same for all poles, maybe because of $\cos$ periodicity

How can i compute $Res(f,z_k)$ where $$f(z)=\frac{z-1}{1+cos\pi z}$$ and $z_k=2k+1, k\neq 0$?

Take $a_j=|z_j|$, where $\{z_j\}$ is the sequence of zeros of an entire function. We suppose $|z_j|\rightarrow +\infty$. Then we introduce two numbers:$b$ which is the exponent of convergence of $\{z_j\}$, defined as the $\inf B$ where $B$ is the set of all $\lambda>0$ such that the series...