properties of discriminant in algebraic number field

Theorem 1.  Let ${\alpha }_1,{\alpha }_2,\mathrm{\dots },{\alpha }_n$ and ${\beta }_1,{\beta }_2,\mathrm{\dots },{\beta }_n$ be elements of the algebraic number field $\mathbb{Q}(\vartheta )$ of degree (http://planetmath.org/NumberField) $n$.  If they satisfy the equations

$${\alpha }_i=\sum _{j=1}^n{c}_{ij}{\beta }_j\mathit{\hspace{1em}}\text{for}\mathit{\hspace{1em}}i=\mathrm{\hspace{0.17em}1},\mathrm{\hspace{0.17em}2},\mathrm{\dots },n,$$

where all coefficients $c_{ij}$ are rational numbers, then the http://planetmath.org/node/12060discriminants are via the equation

$$\mathrm{\Delta }({\alpha }_1,{\alpha }_2,\mathrm{\dots },{\alpha }_n)=det{(c_{ij})}^2\cdot \mathrm{\Delta }({\beta }_1,{\beta }_2,\mathrm{\dots },{\beta }_n).$$

As a special case one obtains the

Theorem 2.  If

${\alpha }_i=c_{i1}+c_{i2}\vartheta +\mathrm{\dots }+c_{in}{\vartheta }^{n-1}\mathit{\hspace{1em}}\text{for}\mathit{\hspace{1em}}i=\mathrm{\hspace{0.17em}1},\mathrm{\hspace{0.17em}2},\mathrm{\dots },n$(1)

are the canonical forms of the elements ${\alpha }_i$ in $\mathbb{Q}(\vartheta )$, then

$$\mathrm{\Delta }({\alpha }_1,{\alpha }_2,\mathrm{\dots },{\alpha }_n)=det{(c_{ij})}^2\cdot \mathrm{\Delta }(1,\vartheta ,\mathrm{\dots },{\vartheta }^{n-1}),$$

where $\mathrm{\Delta }(1,\vartheta ,\mathrm{\dots },{\vartheta }^{n-1})$ is a Vandermonde determinant thushaving the product form

$\mathrm{\Delta }(1,\vartheta ,\mathrm{\dots },{\vartheta }^{n-1})={\left({\displaystyle \prod _{1\le i\le j}}({\vartheta }_j-{\vartheta }_i)\right)}^2={\displaystyle \prod _{1\le i\le j}}{({\vartheta }_i-{\vartheta }_j)}^2$(2)

where ${\vartheta }_1=\vartheta ,{\vartheta }_2,\mathrm{\dots },{\vartheta }_n$ are the algebraic conjugates of $\vartheta$.

Since the (2) is also the polynomial discriminant of the irreducible minimal polynomial of $\vartheta$, the numbers ${\vartheta }_i$ are inequal.  It follows the

Theorem 3.  When (1) are the canonical forms of the numbers ${\alpha }_i$, one has

$$\mathrm{\Delta }({\alpha }_1,{\alpha }_2,\mathrm{\dots },{\alpha }_n)=\mathrm{\hspace{0.33em}0}\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}det(c_{ij})=\mathrm{\hspace{0.33em}0}.$$

The powers $1,\vartheta ,\mathrm{\dots },{\vartheta }^{n-1}$ of the primitive element (http://planetmath.org/SimpleFieldExtension) form a basis (http://planetmath.org/Basis) of the field extension $\mathbb{Q}(\vartheta )/\mathbb{Q}$ (see the canonical form of element of number field).  By the theorem 3 we may write the

Theorem 4.  The numbers ${\alpha }_1,{\alpha }_2,\mathrm{\dots },{\alpha }_n$ of $\mathbb{Q}(\vartheta )$ are linearly independent over $\mathbb{Q}$ if and only if  $\mathrm{\Delta }({\alpha }_1,{\alpha }_2,\mathrm{\dots },{\alpha }_n)\ne \mathrm{\hspace{0.33em}0}$.

Theorem 5. $\mathbb{Q}(\alpha )=\mathbb{Q}(\vartheta )⟺\mathrm{\Delta }(1,\alpha ,{\alpha }^2,\mathrm{\dots },{\alpha }^{n-1})\ne 0$.  Here, the the discriminant is the discriminant of the algebraic number (http://planetmath.org/DiscriminantOfAlgebraicNumber) $\alpha$.