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- Thread starter pantboio
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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 $K[X][Y]$...)