When Dedekind's theorem fails, split the algebra instead
For a prime dividing the index, the factorisation of the defining polynomial modulo p no longer describes the prime ideals. The Buchmann–Lenstra approach works instead inside ℤK/pℤK: quotient out the radical to obtain a separable algebra over Fp, split it into simple factors by finding idempotents, and pull each factor back to a prime ideal. Newton polygons often supply the answer more cheaply when the ramification is simple.
Learning objectives
- Explain why Dedekind's theorem fails at index divisors.
- Read ramification data from a Newton polygon.
- Split a commutative separable algebra over a finite field.
- Recover prime ideals and their ramification indices from the splitting.
- Choose between the polygon route and the algebra route.
Section 01Why the simple method fails
Dedekind's theorem identifies the primes above p with the irreducible factors of T modulo p, but only under the hypothesis that p does not divide [ℤK : ℤ[θ]]. When it does, the two structures genuinely differ: the ideals (p, gi(θ)) need not be prime, and distinct primes may be conflated.
Some fields have primes dividing the index for every choice of generator — the classical example being the prime 2 in certain cubic fields. No change of defining polynomial removes the difficulty, so a general algorithm must handle this case rather than avoid it.
Section 02Newton polygons
The Newton polygon of T at p is the lower convex hull of the points (i, vp(ai)) for the coefficients ai. Its segments correspond to factorisations over the p-adic field, and their slopes give the valuations of the roots.
| Feature | Meaning |
|---|---|
| Number of segments | Number of p-adic factors with distinct root valuations |
| Slope −h/e in lowest terms | Roots of valuation h/e; contributes ramification index divisible by e |
| Horizontal length of a segment | Total degree of the corresponding factor |
| Single segment of slope 0 | T is p-integral with unit roots — unramified case |
| Segment with denominator e = n | Totally ramified at p |
If every segment has a distinct slope with denominator equal to its length, the decomposition is read off immediately — each segment gives one totally ramified prime. Only when segments have residual polynomials that themselves factor is further work needed, and that is where the algebra method takes over.
Section 03Splitting the algebra
The quotient A = ℤK/pℤK is a finite commutative algebra over Fp. Quotienting by its radical leaves a separable algebra, which by the structure theorem is a product of finite fields — one for each prime above p, with the field size giving the residue degree.
- Compute the radical of A and form Ā = A / rad(A). Ā is a product of fields, one per prime above p.
- Compute the kernel of the map x ↦ xp − x on Ā; its dimension is the number of simple factors g.
- Take a non-trivial element of that kernel and use its minimal polynomial to construct orthogonal idempotents.
- Each idempotent projects onto one simple factor; the residue degree fi is its dimension over Fp.
- Pull each factor back to an ideal 𝔽i of ℤK containing pℤK.
- Determine ei by computing valuations, and verify ∑ eifi = n.
Working throughout modulo p keeps every object a finite-dimensional Fp-vector space, so all the operations are linear algebra over a small field. This is what makes the method efficient despite operating in the full ring of integers.
Section 04Method selection
- Does p divide the index?
- No Dedekind's theorem — factor T modulo p; this covers all but finitely many primes.
- Yes Is the maximal order already known?
- Yes Split the algebra — Buchmann–Lenstra inside ℤK/pℤK.
- No Newton polygon first — often resolves the decomposition without computing a full integral basis.
Splitting the algebra requires ℤK, which requires factoring the discriminant. Newton polygons and the Montes algorithm work directly from T and p, so they can decompose primes in fields whose maximal order has not been — and perhaps cannot be — computed.
ReferenceFrequently asked questions
Is the number of primes above p always at most the degree?
Yes, and the fundamental identity makes this precise: the sum of e times f over all primes above p equals n, so there are at most n primes, with equality exactly in the totally split case.
Why does the radical need to be quotiented out?
Because A itself may have nilpotents when p ramifies, and the structure theorem for products of fields applies only to the semisimple quotient. The nilpotents carry the ramification information, which is recovered afterwards through the valuations.
How does the Montes algorithm relate to this?
It develops the Newton polygon approach systematically, using higher-order polygons and residual polynomials to obtain the full decomposition, including ramification indices, without computing the maximal order. It is the most efficient known method for high-degree fields and large ramification.
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