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Let $z=e^{i\theta}$ be a complex number of modulus $1$. Trivially if $z$ is a root of unity then $z$ is also an algebraic number, but the converse is known to be false : $z$ can be algebraic without being a root of unity (consider, e.g. $\frac{3+4i}{5}$).

Is there a simple procedure to produce for each $n$, an algebraic number $z_n$ in this situation, with degree $n$ over ${\mathbb Q}$ ?

There are several related questions on this theme on MSE, for example here.

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Ewan Delanoy
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  • Since multiplication rotates the number around the origin isn't it sufficient to just take any complex number with modulus one such that the argument of it is not a "divisor" of $\pi$? – chubakueno Jun 13 '14 at 04:46
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    @chubakueno you mean rational multiple of $\pi$, and no: that will generally give you transcendental numbers, not algebraic ones. – anon Jun 13 '14 at 04:47
  • @seaturtles true! So the question is asking for algebraic values of $y=\sin x$ (and therefore also $\cos x$) such that $x$ is not a rational multiple of $\pi$ but $y$ is algebraic...tricky. – chubakueno Jun 13 '14 at 04:50
  • So to clarify: you want $\theta$ so that $e^{i\theta}$ is algebraic of degree $n$ but not a root of unity. Is that correct? – David Jun 13 '14 at 04:52
  • I think that's pretty clear. Although Ewan said we want the actual algebraic number itself, not just its argument - while specifying one is equivalent to specifying the other (it may or may not be useful to look at arguments). – anon Jun 13 '14 at 04:54
  • Because the complex conjugates will share the minimal polynomial I suspect that you need $n$ to be even. – Jyrki Lahtonen Jun 13 '14 at 04:59
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    @Jyrki is right, although I am not sure about ruling out the minimal polynomial having real roots. Suppose $f(T)$ is $x$'s minimal polynomial of degree $n$. As $\bar{x}=x^{-1}$ is a root, we know $T^nf(1/T)$ shares $x$ as a root and has the same degree, so $f(T)=c,T^nf(1/T)$. Evaluating at $T=1$ gives $c=1$ so $f$'s roots (none of which are $\pm1$ or $0$) must come in reciprocal pairs. – anon Jun 13 '14 at 05:13
  • The title question is substantially easier than the body question. To answer the title question you just divide most non-real algebraic numbers by their complex conjugates. – Qiaochu Yuan Jun 13 '14 at 07:48

1 Answers1

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The comment by Sea Turtles shows that we need to assume that $n$ is even, say $n=2k$. I proffer a general recipe that works for all $k\ge3$. To add something extra to the recipe in Qiaochu Yuan's comment I add the property that $z_n$ should be an algebraic integer to the wish list.

Let $\alpha=\root k\of 2 -1$. The minimal polynomial of $\alpha$ is $(x+1)^k-2$ which is irreducible by Eisenstein, so $[\Bbb{Q}(\alpha):\Bbb{Q}]=k$. We have clearly $0<\alpha<1$, so if $\beta=\sqrt{1-\alpha^2}$, then $i\beta$ is purely imaginary, and $|\alpha+i\beta|=1$.

I claim that $z_n=\alpha+i\beta$ works.

Because $i\beta$ and $\alpha$ are algebraic integers, so is $z_n$. We also have $[\Bbb{Q}(\alpha,i\beta):\Bbb{Q}(\alpha)]=2$. Furthermore, $$ z_n+z_n^{-1}=2\alpha,\qquad\text{and}\qquad z_n-z_n^{-1}=2i\beta, $$ so $\Bbb{Q}(z_n)=\Bbb{Q}(\alpha,i\beta)$. Thus $[\Bbb{Q}(z_n):\Bbb{Q}]=2k=n$.

It remains to show that $z_n$ is not a root of unity. Assume contrariwise that this would be the case. By the well known Galois theory of cyclotomic fields this implies that the extension $\Bbb{Q}(z_n)/\Bbb{Q}$ is abelian and Galois. By basic Galois theory the same would hod for the subextension $\Bbb{Q}(\alpha)/\Bbb{Q}$. But $\Bbb{Q}(\alpha)=\Bbb{Q}(\root k\of 2)$ is not normal, because the irreducible polynomial $p(x)=x^k-2$ has some but not all of its roots in there.

Jyrki Lahtonen
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