# Question about algebraic integers and roots of unit

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

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

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 $$\Bbb R$$.