advanced39 min studyLesson 575 in Mathematics

Splitting and ramification of primes

Track ramification indices and residue degrees, and apply polynomial factorization only away from power-order index primes.

On this page

What you’ll understand

  • Check the sum of ramification index times residue degree.
  • Use Dedekind–Kummer with its order-index hypothesis.
  • Distinguish split, inert, and ramified quadratic examples.

A rational prime can split into several prime ideals, remain prime with a larger residue field, or occur with repeated ideal factors. These are different phenomena measured by separate integers. The ramification index records repetition; the residue degree records the finite field above the rational residue field. Polynomial reduction can reveal both when its generator is valid at that prime, but index primes require additional work. We will calculate all three quadratic patterns and check both the extension-degree identity and the resulting quotient sizes.

Separate e, f, and the number of primes

For a degree-n number field K and rational prime p, write pOK=∏i=1gPieip\mathcal O_K=\prod_{i=1}^g\mathfrak P_i^{e_i} with distinct prime ideals. The residue field OK/Pi\mathcal O_K/\mathfrak P_i has degree fif_i over Fp\mathbb F_p and cardinality pfip^{f_i}. The fundamental degree identity is ∑ieifi=n\sum_i e_if_i=n. In a non-Galois number field, the e-values and f-values need not be identical across the primes; replacing the sum by one product without that hypothesis can give a false count.

Ramification means at least one exponent exceeds one. Complete splitting means every exponent and residue degree is one, while inertness means the rational prime ideal itself stays prime with residue degree n. These definitions concern the full integer ring. A quotient of order p squared need not be the field with p squared elements: ramification can instead produce nilpotents in the quotient by p. The separate prime-ideal quotients are fields, whereas the entire quotient by the rational prime can be a product or a nonreduced ring.

pOK=∏i=1gPiei,N(Pi)=pfi,∑i=1geifi=[K:Q]p\mathcal O_K=\prod_{i=1}^g\mathfrak P_i^{e_i},\qquad N(\mathfrak P_i)=p^{f_i},\qquad\sum_{i=1}^g e_if_i=[K:\mathbb Q]
The primes are distinct, and f_i is the residue-field extension degree. Equal e and f across primes require an additional Galois hypothesis.↔ Scroll wide equations horizontally.

Reduction is reliable away from the index

Let OK\mathcal O_K contain the power order Z[θ]\mathbb Z[\theta] for an integral primitive element θ, with monic minimal polynomial f. If p does not divide the order index, factor its reduction into distinct monic irreducibles f‾=∏igiei\overline f=\prod_i g_i^{e_i}. Dedekind–Kummer identifies the primes as (p,gi(θ))(p,g_i(\theta)), where an integer polynomial lift of each factor is used, and gives residue degree equal to the factor degree. Repeated factors then describe ramification under this index hypothesis.

If p divides the index, neither factor multiplicities nor the resulting ideal list can be trusted without an appropriate refinement. Replacing the generator, computing a maximal-order basis, or using a local method may resolve the issue. Squarefree reduction away from the index gives an unramified prime, but a repeated factor at an index prime can be spurious. The field discriminant supplies an independent ramification check, and the sum of e times f supplies a consistency check after a legitimate decomposition has been obtained.

p∤[OK:Z[θ]],f‾=∏igiei⟹Pi=(p,gi(θ)), fi=deg⁡gip\nmid[\mathcal O_K:\mathbb Z[\theta]],\quad\overline f=\prod_i g_i^{e_i}\quad\Longrightarrow\quad\mathfrak P_i=(p,g_i(\theta)),\ f_i=\deg g_i
The g_i are distinct monic irreducible factors modulo p. The order-index condition is necessary for this direct factorization rule.↔ Scroll wide equations horizontally.

Quadratic residue patterns have different quotient rings

For a quadratic field at an unramified odd prime, a split defining polynomial gives two degree-one residue fields, while an irreducible polynomial gives one degree-two residue field. When the maximal-order polynomial has a repeated root under the valid factorization criterion, one gets one prime with ramification index two. The norm identity remains the same in each case, but the arithmetic is different. Local completions will later reflect these alternatives as separate factors or as an extension with different ramification and residue behavior.

A quadratic congruence test must use the right generator and treat exceptional primes separately. It is also useful to record which root defines which residue map: a generator may reduce to different elements in the two split residue fields. Those maps are not merely names for two copies of the same quotient. Galois automorphisms can exchange the primes and their residue values, and the later Frobenius construction depends on this prime-by-prime viewpoint rather than on an undifferentiated list of polynomial roots.

All three patterns in the Gaussian integers

In Q(i), use O=Z[i] and x²+1 to describe the prime factorizations above five, three, and two, including e, f, g and prime-ideal norms.

  1. Modulo five, x²+1=(x−2)(x+2), with two distinct linear factors. Thus g=2, each e=f=1, and the primes (5,i−2) and (5,i+2) each have norm five.

  2. Modulo three, x²+1 has no root, so it is irreducible of degree two. There is one prime (3), with e=1, f=2, and norm nine. The residue field is F9, not the cyclic ring Z/9Z.

  3. Modulo two, x²+1=(x+1)². The power order is already maximal, so this repeated factor is genuine: (2)=(1+i)², with g=1, e=2, f=1 and prime norm two. In all three cases the sum of e times f is two.

The result

Five splits, three is inert, and two ramifies. A shared field degree does not make the residue fields or quotient rings interchangeable.

Explore the model

Interactive lab

Interactive lab · modular

Compare the two split residue maps

Pause and Reset, then set both controls to modulus m=5 and additive step a=2. Predict residues 0,2,4,1,3: these are integer multiples of the image i=2 in one split quotient. Reset and set m=5 and a=3; predict 0,3,1,4,2 for the image i=−2 in the other quotient. The model only displays additive reduction in F5; it does not factor x²+1 or model the inert field F9 as Z/9Z.

Compare the two split residue mapsAddition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Current readings: Residue r0: 0; Orbit period · m/gcd(m,a): 12; Distinct residues visited: 1 of 12. All parameters are adjustable with the labeled controls below.Repeated addition modulo 12: the current residue is 0, and the orbit period is 12.01234567891011+ 5mod 12An additive orbitn = 0rₙ = 0Period = 12Cycle starts at 0:0 → 5 → 10 → 3 →8 → 1 → 6 → 11 →4 → 9 → 2 → 7 → 0Exact residues · two additions per second
Additive orbitCurrent residue / next edge
Residue r0
0
Orbit period · m/gcd(m,a)
12
Distinct residues visited
1 of 12
t = 0 s

Addition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe two primes above a split rational prime can send the same algebraic generator to different residue values.

Check your understanding

What must be checked before directly reading maximal-order prime factors from a power polynomial modulo p?

Try it yourself

In Q(sqrt(5)) use omega=(1+sqrt(5))/2 and its maximal-order polynomial x²−x−1. Factor modulo eleven and state the prime-ideal residue degrees and ramification indices.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Test roots four and eight modulo eleven. The power order generated by omega is the full integer ring.

Compare with the solution

The product (x−4)(x−8)=x²−12x+32 reduces to x²−x−1 modulo eleven, giving two distinct linear factors. Therefore eleven splits into P1=(11,omega−4) and P2=(11,omega−8), each with e=1 and f=1 and norm eleven. The degree identity is 1+1=2. The field discriminant five is not divisible by eleven, agreeing with the unramified result.

Take it with you

  • Ramification, residue degree, and the number of primes measure different features.
  • Polynomial factorization describes maximal-order splitting only with its index hypothesis.
  • Residue-field cardinality alone does not describe the quotient ring by p.

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