# properties of discriminant in algebraic number field

Theorem 1.  Let $\alpha_{1},\,\alpha_{2},\,\ldots,\,\alpha_{n}$ and $\beta_{1},\,\beta_{2},\,\ldots,\,\beta_{n}$ be elements of the algebraic number field  $\mathbb{Q}(\vartheta)$ of degree (http://planetmath.org/NumberField) $n$.  If they satisfy the equations

 $\alpha_{i}\;=\;\sum_{j=1}^{n}c_{ij}\beta_{j}\quad\mbox{for}\quad i\,=\,1,\,2,% \,\ldots,\,n,$
 $\Delta(\alpha_{1},\,\alpha_{2},\,\ldots,\,\alpha_{n})\;=\;\det\!(c_{ij})^{2}% \cdot\Delta(\beta_{1},\,\beta_{2},\,\ldots,\,\beta_{n}).$

As a special case one obtains the

Theorem 2.  If

 $\displaystyle\alpha_{i}\;=\;c_{i1}+c_{i2}\vartheta+\ldots+c_{in}\vartheta^{n-1% }\quad\mbox{for}\quad i\,=\,1,\,2,\,\ldots,\,n$ (1)

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

 $\Delta(\alpha_{1},\,\alpha_{2},\,\ldots,\,\alpha_{n})\;=\;\det\!(c_{ij})^{2}% \cdot\Delta(1,\,\vartheta,\,\ldots,\,\vartheta^{n-1}),$

where $\Delta(1,\,\vartheta,\,\ldots,\,\vartheta^{n-1})$ is a Vandermonde determinant  thus having the product form

 $\displaystyle\Delta(1,\,\vartheta,\,\ldots,\,\vartheta^{n-1})\;=\;\left(\prod_% {1\leq i\leq j}(\vartheta_{j}-\vartheta_{i})\right)^{\!2}\;=\;\prod_{1\leq i% \leq j}(\vartheta_{i}-\vartheta_{j})^{2}$ (2)

where $\vartheta_{1}=\vartheta,\,\vartheta_{2},\,\ldots,\,\vartheta_{n}$ 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

 $\Delta(\alpha_{1},\,\alpha_{2},\,\ldots,\,\alpha_{n})\;=\;0\quad% \Longleftrightarrow\quad\det(c_{ij})\;=\;0.$

The powers $1,\,\vartheta,\,\ldots,\,\vartheta^{n-1}$ 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 $\alpha_{1},\,\alpha_{2},\,\ldots,\,\alpha_{n}$ of $\mathbb{Q}(\vartheta)$ are linearly independent  over $\mathbb{Q}$ if and only if  $\Delta(\alpha_{1},\,\alpha_{2},\,\ldots,\,\alpha_{n})\;\neq\;0$.

Theorem 5.$\displaystyle\mathbb{Q}(\alpha)=\mathbb{Q}(\vartheta)\;\;\Longleftrightarrow\;% \;\Delta(1,\,\alpha,\,\alpha^{2},\,\ldots,\,\alpha^{n-1})\neq 0$.  Here, the the discriminant is the discriminant of the algebraic number  (http://planetmath.org/DiscriminantOfAlgebraicNumber) $\alpha$.

Title properties of discriminant in algebraic number field PropertiesOfDiscriminantInAlgebraicNumberField 2013-03-22 19:09:28 2013-03-22 19:09:28 pahio (2872) pahio (2872) 12 pahio (2872) Result msc 11R29 MinimalityOfIntegralBasis ConditionForPowerBasis