canonical form of element of number field

Theorem.  Let $\vartheta$ be an algebraic number of degree (http://planetmath.org/DegreeOfAnAlgebraicNumber) $n$.  Any element $\alpha$ of the algebraic number field $\mathbb{Q}(\vartheta )$ may be uniquely expressed in the canonical form

$\alpha =c_0+c_1\vartheta +c_2{\vartheta }^2+\mathrm{\dots }+c_{n-1}{\vartheta }^{n-1}$(1)

where the numbers $c_i$ are rational.

Proof.  We start from the fact that $\mathbb{Q}(\vartheta )$ consists of all expressions formed of $\vartheta$ and rational numbers using arithmetic operations (no divisor (http://planetmath.org/Division) must vanish); such expressions lead always to the form

$\alpha ={\displaystyle \frac{a(\vartheta )}{b(\vartheta )}}$(2)

where the numerator and the denominator are polynomials in $\vartheta$ with rational coefficients (which can, in fact, be chosen integers).

So, let $\alpha$ in (2) an arbitrary element of the field $\mathbb{Q}(\vartheta )$.  Denote by $f(x)$ the minimal polynomial of $\vartheta$ over $\mathbb{Q}$.  Since  $b(\vartheta )\ne 0$,  the polynomial $f(x)$ does not divide (http://planetmath.org/DivisibilityInRings) $b(x)$, and since $f(x)$ is irreducible (http://planetmath.org/IrreduciblePolynomial2), the greatest common divisor (http://planetmath.org/PolynomialRingOverFieldIsEuclideanDomain) of $f(x)$ and $b(x)$ is a constant polynomial, which can be normed to 1.  Thus there exist the polynomials $\phi (x)$ and $\psi (x)$ of the ring $\mathbb{Q}[x]$ such that

$$\phi (x)f(x)+\psi (x)b(x)\equiv \mathrm{\hspace{0.33em}1}.$$

Especially

$$\phi (\vartheta )\underset{=\mathrm{\hspace{0.17em}0}}{\underbrace{f(\vartheta )}}+\psi (\vartheta )b(\vartheta )=\mathrm{\hspace{0.33em}1},$$

whence

$$\frac{1}{b(\vartheta )}=\psi (\vartheta )$$

and consequently

$$\alpha =\frac{a(\vartheta )}{b(\vartheta )}=a(\vartheta )\psi (\vartheta ):={\psi }_1(\vartheta ).$$

Hence, $\alpha$ is a polynomial in $\vartheta$ with rational coefficients.

Let now

Denote

$$r(x):=c_0+c_{1}x+\mathrm{\dots }+c_{n-1}{x}^{n-1}\in \mathbb{Q}[x].$$

It follows that

$$\alpha =r(\vartheta )=c_0+c_1\vartheta +\mathrm{\dots }+c_{n-1}{\vartheta }^{n-1},$$

whence (1) is true.

Suppose that we had also

$$\alpha =s(\vartheta )=d_0+d_1\vartheta +\mathrm{\dots }+d_{n-1}{\vartheta }^{n-1}$$

with every $d_i$ rational.  This implies that

$$(c_{n-1}-d_{n-1}){\vartheta }^{n-1}+\mathrm{\dots }+(c_1-d_1)\vartheta +(c_0-d_0)=\mathrm{\hspace{0.33em}0},$$

i.e. that $\vartheta$ satisfies the equation

$$(c_{n-1}-d_{n-1})x^{n-1}+\mathrm{\dots }+(c_1-d_1)x+(c_0-d_0)=\mathrm{\hspace{0.33em}0}$$

with rational coefficients and degree less than $n$.  Because the degree of $\vartheta$ is $n$, it is possible only if all differences $c_i-d_i$ vanish.  Thus

$$d_0=c_0,d_1=c_1,\mathrm{\dots },d_{n-1}=c_{n-1},$$

i.e. the (1) is unique.

Note 1.  The polynomial $c_0+c_{1}x+\mathrm{\dots }+c_{n-1}{x}^{n-1}$ is called the canonical polynomial of the algebraic number $\alpha$ with respect to the primitive element (http://planetmath.org/SimpleFieldExtension) $\vartheta$.

Note 2.  The theorem allows to denote the field $\mathbb{Q}(\vartheta )$ similarly as polynomial rings:$\mathbb{Q}[\vartheta ]$.

Note 3.  When allowed, unlike in (1), higher powers of the primitive element $\vartheta$ (whose minimal polynomial is $x^n+a_1{x}^{n-1}+\mathrm{\dots }+a_n$), one may unlimitedly write different sum of $\alpha$, e.g.

$\alpha$$\mathrm{\hspace{0.33em}}=(c_0+c_1\vartheta +\mathrm{\dots }+c_{n-1}{\vartheta }^{n-1})+({\vartheta }^n+a_1{\vartheta }^{n-1}+\mathrm{\dots }+a_n)$

$\mathrm{\hspace{0.33em}}=(c_0+a_n)+\mathrm{\dots }+(c_{n-1}+a_1){\vartheta }^{n-1}+{\vartheta }^n.$