Integral closure in finite extension fields

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SUMMARY

The integral closure of the polynomial ring $R = K[x]$ in the field $L = K(x)[y]$, where $y$ satisfies $y^3 = 1 + x^2$, is determined by the algebraic properties of elements in $L$. According to the theorem discussed, if $R$ is a domain and $K$ is its fraction field, then any element $\alpha$ in $L$ is algebraic over $K$. The coefficients of the minimal polynomial $m_{\alpha}(X)$ of $\alpha$ over $K(X)$ lie in the integral closure of $R$ in $K(X)$, which leads to the conclusion that the integral closure of $K[x]$ in $L$ is $K[X][Y]$.

PREREQUISITES
  • Understanding of integral closure in ring theory
  • Familiarity with finite extension fields
  • Knowledge of minimal polynomials and their properties
  • Basic concepts of algebraic elements over fields
NEXT STEPS
  • Study the properties of integral closures in polynomial rings
  • Learn about finite extensions and their implications in field theory
  • Explore the computation of minimal polynomials in algebraic extensions
  • Investigate the structure of $K[X][Y]$ and its applications in algebra
USEFUL FOR

Mathematicians, algebraists, and students studying field theory and ring theory, particularly those interested in integral closures and finite extensions.

pantboio
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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?
 
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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 $K[X][Y]$...)
 

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