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). There are left only...
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 \overline{P}=(p,a-b\sqrt{-13}).
1)Prove that...
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?
Consider the ring $\mathbb{Q}[X]$ of polynomials in $X$ with coefficients in the field of rational numbers. Consider the quotient field $\mathbb{Q}(X)$ and let $K$ be the finite extension of $\mathbb{Q}(X)$ given by $K:=\mathbb{Q}(X)[Y]$, where $Y^2-X=0$.Let $O_{K}$ be the integral closure of...
Let $X\subset \mathbb{A}^n$ be an affine variety, let $I(X)=\{f\in k[X_1,\ldots,X_n]:f(P)=0,\ \forall P \in X\}$. We consider the ring
$$A=k[a_1,\ldots,a_n]=\frac{k[X_1,\ldots,X_n]}{I(X)}$$
where $a_i=X_i \mod I(X)$.Noether normalization says that there are algebraically indipendent linear forms...
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...
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...