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Let $y$ be an algebraic integer in a finite field extension $K:Q$. Hence y is an element in the maximal Z-order $O_K$. The question is if $y$ and all its conjugates have absolute value 1, then y in a root of unit.

Here is my proof: Since $O_K$ is Z-lattice, it only have finite many elements in a finite area. Let $S$ denote the unit circle. We have $K \bigcap O_K$ is finite. Since $y^n$ also have absolute value 1 and it is still in $O_K$, we get $y^i=y^j$ for some different i and j. Hence, we get $y^{i-j}=1$ which means y is a root of unit.

In the proof, i didn't use the property that all its conjugates have absolute value 1.

But I heared that without this property, y not have to be a root of unit. (Are all algebraic integers with absolute value 1 roots of unity?)

Can anyone help me find the error in my proof. Thanks

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  • $\begingroup$ I guess the part the intersection is finite is false. But i can't imagine that. $\endgroup$
    – CJJ
    Mar 6, 2022 at 19:20

1 Answer 1

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Note that you need to consider all embedings $K \to \Bbb C$ (up to complex conjugation) to obtain a lattice (how exactly you do this is in every algebraic number theory book that treats Minkowski theory). So if you want $y$ to lie in the unit circle under that embedding, all conjugates need to have absolute value $1$.

For example consider $K=\Bbb Q(\sqrt{2})$, then $\mathcal O_K=\Bbb Z[\sqrt{2}]$ is not a lattice if you just embed it via one real embedding into $\Bbb R$.

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  • $\begingroup$ Oh! Z[sqrt2] is dense in R? Yes take (sqrt2 -1)^n it converges to 0 $\endgroup$
    – CJJ
    Mar 6, 2022 at 19:53
  • $\begingroup$ @CJJ exactly! a one dimensional R-vector space can't have a rank two lattice, of course, but you can also show directly that it is dense. $\endgroup$ Mar 6, 2022 at 19:54
  • $\begingroup$ thanks you so much! $\endgroup$
    – CJJ
    Mar 6, 2022 at 19:54

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