$O$ is dedekind domain and $K=Frac(O)$. Suppose $a\in L$ where $L/K$ is finite separable extension. If $Tr(a), Norm(a)\in O$, is $a\in O_L$?

Question

$O$ is dedekind domain and $K=Frac(O)$. Suppose $a\in L$ where $L/K$ is finite separable extension.

$\textbf{Q:}$ If $Tr(a), Norm(a)\in O$, is $a\in O_L$ where $O_L$ is integral closure of $O$ in $L$? I knew this holds for $K$ complete/$O$ complete as polynomial ring valuation is induced by the valuation on irreducible polynomial. If $f=\sum_ia_ix^i$ with $f$ irreducible, then $v(f)=\min\{v(a_0),v(a_n)\}$. It seems that the statement is true as I can use local statement for each valuation. Completion is flat. So I do not change any inherent information as $Tr$ and $Norm$ decompose into local sums or local products.

Answer 1

This is false, consider a root $\alpha$ of an irreducible cubic polynomial over $\mathbb{Q}$ of the form $$ x^3 + ax^2 + bx + c $$

Then the norm of $\alpha$ is $-c$ and the trace is $-a$, but $\alpha$ is an algebraic integer if and only if $b$ is an integer.