※ 引述《WasabiSushi (田馥甄Hebe)》之銘言：
: Let R be a integral closed integral domain with its fraction field F. Let K be a finite separable extension field of F, and let A be the integral closure of R in K.
: It is well known the trace map Tr: K -> F is non-trivial and hence is
: surjective because of separable extension. If we restrict Tr to A, then it is
: obvious that Tr maps A into R because R is integral closed.
: QUESTION: Is the restriction Tr: A -> R also surjective?
No, you can take ANY extension of Q, the field of rational numbers.
Tr must not be surjective according to Minkowski's theorem.
https://en.wikipedia.org/wiki/Discriminant_of_an_algebraic_number_field#Basic_results
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