Algebraic Integers. Fields

If $\alpha$ is an integer, then $\sqrt\alpha$ is also an integer.

If $r$ and $s$ are rational numbers and $r+s\sqrt{-1}$ is an integer, then $r$ and $s$ must be rational integers.

If $r$ and $s$ are rational numbers and $r+s\sqrt{-5}$ is an integer, then $r$ and $s$ must be rational integers.

If $r$ and $s$ are rational numbers and $r+s\sqrt{-3}$ is an integer, then $2r$ and $2s$ must be rational integers and $2r\equiv 2s \pmod2$; $r$ and $s$ need not necessarily be rational integers.

If both $\vartheta /\pi$ and $\cos \vartheta$ are rational, then $\cos \vartheta$ must have one of the following five values:

Consider the interior angle included between two contiguous faces of a regular solid and call it

according as the solid is a tetra-, hexa-, octa-, dodeca-, or icosahedron (a hexahedron is a cube). Show that the ratios

are irrational.

The numbers $\alpha_1,\alpha_2,\ldots ,\alpha_m$ are called linearly dependent if there exist rational integers $a_1,a_2,\ldots ,a_m$ not all zero such that $a_1\alpha_1+a_2\alpha_2+\cdots +a_m\alpha_m=0$. If no relation of this kind exists then $\alpha_1,\alpha_2,\ldots ,\alpha_m$ are called linearly independent. Thus, if $\beta \neq 0$, the numbers $\alpha$ and $\beta$ are linearly dependent or independent according as $\alpha /\beta$ is rational or irrational. The number $\alpha$ is algebraic if there exists an $n$ (a positive rational integer) such that $1,\alpha ,\alpha^2,\ldots ,\alpha^n$ are linearly dependent.

The numbers (angles) $T,H$ and $O$ are linearly dependent; in fact

(see 197.2; it can be also proved by elementary geometry). Yet $T$ and $O$ are linearly independent.

(continued). Show that $T,H,D$ and $I$ are linearly independent. Also $H,O,D$ and $I$ are linearly independent, and any linear homogeneous relation with rational coefficients between $T,H,O,D$ and $I$ can differ only trivially (that is, only by a numerical factor) from the one given in 197.3.

If both $\vartheta /\pi$ and $\tan^2\vartheta$ are rational then $\tan^2\vartheta$ must have one of the following five values (regarding $\infty =1/0$ as ``rational''):

With the exception of the four cases $n=1,2,4$, and 8, $\tan 2\pi /n$ is irrational for all positive rational integers $n$.

There are only finitely many integers of given degree $n$ that together with their conjugates lie in a fixed circle $\left|z\right|<k$ of the complex plane.

The only integer that together with all its conjugates lies in the open unit circle is the number 0.

Let all the zeros of a polynomial with rational integral coefficients and leading coefficient 1 lie in the closed unit circle. Such a zero must lie either at the center or on the boundary of the unit circle and in the latter case it must be a root of unity. [I.e. it must satisfy an equation of the form $x^h=x^k$, where $h,k$ are rational integers, $0<h<k$; 198.]

If the integer $\alpha$ together with all its conjugates is real and in absolute value $<2$, then $\alpha =2\cos 2\pi p/q$, where $p$ and $q$ are rational integers.

The irrational integers $\sqrt{-1}, \sqrt{-3}, \sqrt{-5}$ generate three distinct complex quadratic number fields, in which the integers have the forms

respectively, where $a$ and $b$ are rational integers.

A set $\mathfrak S$ of complex (or real) numbers is called a ``discontinuous integral domain'' if it has the following two properties:

1. If $\zeta '$ and $\zeta ''$ belong to the set $\mathfrak S$, then $\zeta '+\zeta ''$, $\zeta '-\zeta ''$, and $\zeta '\zeta ''$ also belong to the set $\mathfrak S$.

2. The point 0 is not a point of accumulation of the set of numbers $\mathfrak S$. (By (1) the number $0=\zeta '-\zeta '$ belongs to the set $\mathfrak S$.) A discontinuous integral domain consists either of all or of some of the integers of a complex quadratic number field. (E.g. only of the numbers $0,\pm 6,\pm 12,\pm 18,\ldots$ or only of the single number 0.)