Important: The numtheory package has been deprecated.  Use the superseding command NumberTheory[IntegralBasis] instead.

This procedure computes an integral basis for an algebraic number field K. There are several ways to specify the algebraic number field under consideration:

by a RootOf or a set of RootOfs representing algebraic numbers. The RootOfs must be independent.

by a univariate polynomial over the rational numbers or over an algebraic number field in RootOf or radical notation. The polynomial must be irreducible over its coefficients field and the RootOf appearing in the coefficients must be independent. The field represented by the polynomial is the field generated by one of its roots, up to isomorphism.

by a radical number or a set of radical numbers.

The output is a list of elements of K which generate the Z-module of integral elements of K, where Z denotes the ring of integers. An element of K is called integral over Z if it annihilates a monic univariate polynomial with coefficients in Z.

If the input is in RootOf notation, then the output will also be in RootOf notation.

If the input is in radical notation, then the output will be also in radical notation, but the output radicals may differ from the input radicals.

If the input is a polynomial, then the output will be expressed in terms of the variable.

This function supports options of the form option=value:

primes=S : In this case, S must be a prime number or a set of prime numbers. A local integral basis at these primes is computed instead of an integral basis over Z.

method=roundfour : The Ford & Zassenhaus Round Four algorithm is used. This is the default. See "Implementing the Round Four Maximal Order Algorithm", by D. Ford & P. Letard, Journal de Theorie des Nombres de Bordeaux, 1994, 6, p. 39-80.

method=roundtwo : The Zassenhaus Round Two algorithm is used. See "A Course in Computational Algebraic Number Theory" by Henri Cohen, Springer.

method=p_adic : With this option the algorithm will try to compute roots of the polynomial f in the form of p-adic numbers. If at a prime p no wild ramification occurs then the integral basis at the prime p is computed using these p-adic roots, see  "An algorithm for computing an integral basis in an algebraic function field", J. of Symbolic Computation, 1994, 18, p. 353-363. If a wild ramification (i.e. p divides the ramification index) does occur then this method does not work. A warning message will be given and the integral basis at the prime p will be computed using method = roundfour.

By setting infolevel[integral_basis] to a positive value some information will be printed during the computation.

$\mathrm{with}\left(\mathrm{numtheory}\right)\:$

$q≔41296-6864t+1128t^2-336t^3-6t^5+t^6$

$q≔t^6-6t^5-336t^3+1128t^2-6864t+41296$

$\mathrm{integral\_basis}\left(q\right)$

$\left[1\,t\,\frac{t^2}{2}\,\frac{t^3}{4}\,\frac{1}{24}{t}^4+\frac{1}{3}{t}^2+\frac{1}{2}t+\frac{2}{3}\,\frac{1}{1443432}{t}^5+\frac{4513}{360858}{t}^4+\frac{31066}{180429}{t}^3+\frac{36025}{180429}{t}^2+\frac{32903}{180429}t+\frac{8819}{180429}\right]$

$\mathrm{integral\_basis}\left(q\,\mathrm{primes}=2\,\mathrm{method}=\mathrm{p\_adic}\right)$

$\left[1\,t\,\frac{t^2}{2}\,\frac{t^3}{4}\,\frac{1}{8}{t}^4+\frac{1}{2}t\,\frac{t^5}{8}\right]$

$\mathrm{integral\_basis}\left(q\,\mathrm{primes}=\left\{2\,3\right\}\right)$

$\left[1\,t\,\frac{t^2}{2}\,\frac{t^3}{4}\,\frac{1}{24}{t}^4+\frac{1}{3}{t}^2+\frac{1}{2}t+\frac{2}{3}\,\frac{1}{24}{t}^5+\frac{1}{12}{t}^3+\frac{2}{3}t\right]$

$\mathrm{integral\_basis}\left(\left\{{10}^{\frac{1}{3}}\right\}\right)$

$\left[1\,{10}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\,\frac{{10}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{3}+\frac{{10}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{3}+\frac{1}{3}\right]$

$\mathrm{integral\_basis}\left(\left\{6^{\frac{1}{2}}\,{10}^{\frac{1}{2}}\right\}\right)$

$\left[1\,\sqrt{10}\,-\frac{\sqrt{10}}{2}+\frac{\sqrt{6}}{2}\,\frac{\sqrt{10}\sqrt{6}}{2}\right]$

$r≔\mathrm{RootOf}\left(x^3-15x^2+75x-4825\,x\right)$

$r≔\mathrm{RootOf}\left({\mathrm{\_Z}}^3-15{\mathrm{\_Z}}^2+75\mathrm{\_Z}-4825\right)$

$\mathrm{integral\_basis}\left(r\right)$

$\left[1\,\mathrm{RootOf}\left({\mathrm{\_Z}}^3-15{\mathrm{\_Z}}^2+75\mathrm{\_Z}-4825\right)\,\frac{{\mathrm{RootOf}\left({\mathrm{\_Z}}^3-15{\mathrm{\_Z}}^2+75\mathrm{\_Z}-4825\right)}^2}{10}+\frac{1}{2}\right]$