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If we are given an algebraic number field L, and $ \alpha $ is an element of L whose field trace over Q is zero and whose field norm over Q belongs to Z, then does $ \alpha $ necessarily belong to the integral closure of Z in L?
The Hilbert's theorem 90 states the necessary and sufficient condition for elements to have the trace zero or the norm one, however, if we are given such kind of elements can we know more about them? And the original question is from the quadratic number fields.

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As soon as the degree of the minimal polynomial is $> 2$, norm and trace just determine two of the at least three coefficients of the minimal polynomial. So you can easily pick an element whose minimal polynomial has a non-integral coefficient, as in the examples of Zev and JSE. In particular: integral norm and trace in general only force elements to be integral if you are in quadratic number fields. – felix Jan 26 '11 at 9:36
up vote 15 down vote accepted

The answer is no. Let $\alpha\in\mathbb{C}$ be a root of $x^3+tx+1$ for $t\in\mathbb{Q}\setminus\mathbb{Z}$, and let $L=\mathbb{Q}(\alpha)$, so that $N_\mathbb{Q}^L(\alpha)=1$ and $tr_{\mathbb{Q}}^L(\alpha)=0$. The element $\alpha$ cannot be integral over $\mathbb{Z}$, because its minimal polynomial over $\mathbb{Q}$ is not integral.

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This answer is pretty inspiring, thanks. – awllower Feb 3 '11 at 2:06

No. Consider a root of x^3 - (1/100000000)x + 1.

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