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ArticlePublished 7 Aug 20262 min readBy Kevin Joginmaximal orderring of integersintegral basisp-maximality
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Maximal Orders and Decomposition II

The Maximal Order Problem

The maximal order problem, its reduction to a prime-by-prime question, and the factoring obstruction that limits it.

Engineering / MathematicsMaximal Orders and Decomposition II2 min readKV-MATH-0610

Computing the maximal order is the first of the four core tasks and the prerequisite for every other. The problem reduces cleanly to a question about individual primes, but the reduction itself requires a factorisation.

The problem

Given a defining polynomial, find an integral basis for the ring of all algebraic integers in the field.

Input: T(X) irreducible over Q. Output: a basis of O_KExpressed relative to the power basis, with a common denominator.

Reduction to primes

An order is maximal exactly when it is maximal at every prime. Since the index appears squared in the discriminant relation, only primes whose square divides the polynomial discriminant can be involved.

disc(T) = index^2 * disc(K)So p divides the index only if p squared divides disc(T).

Key point

This is what makes the problem tractable. Instead of an unbounded search, there is a finite explicit list of primes to check, and each check is local.

The maximal order computation

  1. Compute the polynomial discriminantBy resultant with the derivative.
  2. Find the square divisorsRequires factoring, or at least finding the square part.
  3. Test p-maximalityFor each candidate prime — see the Dedekind criterion.
  4. Enlarge where neededBy Round 2.
  5. VerifyThe index squared must divide the polynomial discriminant exactly as expected.

The factoring obstruction

Caution

Determining maximality in general requires knowing the square factors of the polynomial discriminant, which in the worst case requires factoring a large integer. The maximal order problem is therefore not known to be easier than integer factorisation.

Partial maximality

If the discriminant cannot be fully factored, an order maximal at all known primes can still be produced, with the unfactored cofactor recorded. Any computation involving only the known primes remains valid.

Working with a partially factored discriminant
SituationWhat remains valid
Fully factored discriminantEverything
Square-free cofactor provedEverything — a squarefree cofactor cannot divide the index
Unfactored cofactorPrime decomposition at known primes; class group results are unsafe

Key point

Proving the cofactor squarefree is enough — the full factorisation is not needed. Since only square divisors matter, a squarefree cofactor settles maximality without being factored at all.

The single best mitigation

Cost

Reducing the defining polynomial first shrinks the discriminant, often dramatically, making it easier to factor. See polynomial reduction. This is the highest-return preprocessing step available.

Source. Henri Cohen, A Course in Computational Algebraic Number Theory, Springer GTM 138 — 6.1. Structural reference unverified: the source file was not available during authoring; chapter and section numbers are taken from the published edition and have not been checked against a physical copy.

Related pages

  • Discriminants and Integral Bases
  • Orders in Number Fields
  • The Pohst-Zassenhaus Theorem

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