MHB Integral closure in finite extension fields

pantboio
Messages
43
Reaction score
0
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?
 
Physics news on Phys.org
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 polynomial over $K$. Then the coefficients of $m_{\alpha}(X)$ lie in the integral closure of $R$ in $K$, hence in $R$ if we assume $R$ to be integrally closed.

My strategy to solve the problem i posted was:
1) take an element $\alpha$ in $L$ and express it in the most general form you can;
2) compute the minimal polynomial of $\alpha$ over $K(X)$.
3) now suppose $\alpha$ integral over $K[X]$, hence the coefficients of its minimal polynomial lie in the integral closure of $K[X]$ in $K(X)$, which is $K[X]$ itself.
4) deduce from 3) that the writing for $\alpha$ in 1) implies $\alpha\in K[X][Y]$ (my guess is that the integral closure of $K[x]$ in $L$ is $K[X][Y]$...)
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top