Let $\alpha \in \mathbb{C}$ be an algebraic number and $\alpha = \alpha_1, \alpha_2,...,\alpha_n$ its conjugates and $N(\alpha) = \prod_i \alpha_i$ its norm. Is it true that $|\alpha| < 1 \Rightarrow |N(\alpha)| < 1$ and if so why? (|| denotes the complex absolut value) Thanks.
norm of an algebraic number with abs value smaller than 1
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3Let $\alpha=2-\sqrt{2}$. – 2011-05-02
3 Answers
Of course $P_c(z) = z^3-2cz^2-c$ has three roots, let us call them $a$,$b$ and $\bar b$. The roots $b$ and $\bar b$ lie within the unit circle but the root $a$ escapes along the real axis to infinity as the parameter $c$ increase from $1$ to $\infty$. Note that these are algebraic integers with norm $c$, two of which have "absolute value" below 1 (and getting smaller) and one of which has as big absolute value close to $2c$.
The geometric interpretation of algebraic norm (in some number field) is quite unrelated to the metric of the complex plane. You actually form a higher dimensional lattice spanned by an integral basis and the norm of an algebraic number, expressed in terms of the basis, is the distance of that point to the origin.
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2$2c$ is the $\it trace$, not the norm. The norm is $c$. – 2011-05-03
No. For example, let $\alpha=\frac{1-\sqrt{5}}{2}$. Then $|\alpha|<1$ but $N(\alpha)=1$.
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0@Qiaochu: Exactly. For a second I was thinking that we needed reciprocals of [Pisot-Vijayaraghavan numbers](http://en.wikipedia.org/wiki/Pisot%E2%80%93Vijayaraghavan_number), but that's not necessary - your explanation is the most general. – 2011-05-02
For the special case of algebraic integers, the answer is no in an extremely strong sense. Note that if what you say is true, it would imply that if $N(\alpha) = \pm 1$, then $|\alpha_i| = 1$ for all conjugates of $\alpha$.
Proposition: Let $\alpha$ be an algebraic integer all of whose conjugates have absolute value $1$. Then $\alpha$ is a root of unity.
Proof. Let $\alpha = \alpha_1$ have conjugates $\alpha_1, ... \alpha_d$ and let $f_n(x) = \prod_{i=1}^d (x - \alpha_i^n).$
The coefficient of $x^k$ in $f_n(x)$ is an integer of absolute value at most ${d \choose k}$ for all values of $n$, so it follows that $f_n(x)$ can only be one of a finite list of polynomials. Hence there exist $m, n$ such that $f_n(x) = f_m(x)$, and this relation implies that some pair of powers of $\alpha_1$ are identical.
Thus if $\alpha$ is an algebraic integer that is not a root of unity, some conjugate of $\alpha$ has absolute value greater than $1$, and if $N(\alpha) = \pm 1$ then some conjugate of $\alpha$ has absolute value less than $1$.
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2Historical note: the Proposition is due to Kronecker. – 2011-05-03