On algebraic integers.

algebraic-number-theorynumber theory

I've question on algebraic integers. $z$ is a complex number and $n$ is a natural number, such that, $z^n=\pm 1$. So, $z-1/z$ is an algebraic integer.

I. If $r$ is a rational number, is $(z-1/z)^{r}$ still an algebraic integer?
Or more general, if $a$ is an alg. int. then $a^r$ is also alg. int.?

II. If $a-1/a$ is an alg. int., is $a$ an alg. int.?

Thanks

Edit. I've been reading about some more simple transcendental proofs, and I came across a few similar statements. But I just had these questions, because of this I can't find the answers so easily online.

Best Answer

I. Yes if $r$ is positive. Not necessarily if $r$ is negative. If $a_1,\ldots,a_{n-1}$ are algebraic integers, then any root of $x^n + a_{n-1}x^{n-1}+\cdots + a_1x + a_0$ is also an algebraic integer. Writing $r=p/q$ with $p$ and $q$ (positive) integers, it is clear that $(z-1/z)^p$ is an algebraic integer, and $(z-1/z)^r$ is a root of $x^q - (z-1/z)^p$, hence an algebraic integer itself. If $r$ is negative, then the answer may be negative: e.g., $z=i$, $z-(1/z) = i-(-i) = 2i$, and taking $r=-1$ yields $(2i)^{-1} = -\frac{i}{2}$, which is not an algebraic integer.

II. As mr_e_man noted, if $a-(1/a)$ is an algebraic integer, then $a$ is a root of $x^2 - (a-1/a)x -1$ which is a monic polynomial with algebraic integer coefficients, hence $a$ is an algebraic integer.

In both cases, the key is

Theorem. If $f(x)$ is a monic polynomial with algebraic integer coefficients, and $a$ is a root of $f$, then $a$ is an algebraic integer.

So that’s the theorem you want to prove.