Engineering / Mathematics — Fields, Series and Factorisation
Polynomial Congruences
Congruences of polynomials modulo a fixed polynomial, and the parallel with integer congruences.
Executive summary
Congruence modulo a polynomial is defined exactly as for integers, with divisibility of polynomials replacing divisibility of integers. Every structural result transfers.
The resulting quotient ring is a field precisely when the modulus is irreducible, mirroring the prime modulus case.
Learning objectives
- Define polynomial congruence and verify arithmetic compatibility.
- Solve linear polynomial congruences.
- State the condition for the quotient to be a field.
01Definition and the parallel
Polynomial congruence
For f, g, h ∈ F[X] with h ≠ 0, write f ≡ g (mod h) if h | (f − g).
| Integers | Polynomials over a field |
|---|---|
| Z | F[X] |
| Absolute value | Degree |
| Division with remainder | Division with remainder |
| Prime | Irreducible |
| Z_n | F[X]/(h) |
| Z_p is a field | F[X]/(h) is a field iff h is irreducible |
| Chinese remainder theorem | Chinese remainder theorem for coprime moduli |
The dictionary is exact because both rings are Euclidean domains. Every algorithm resting only on division with remainder transfers verbatim, which is why polynomial Euclid, polynomial CRT and rational function reconstruction all exist and look identical to their integer counterparts.
02Solving linear congruences
Solvability
af ≡ b (mod h) has a solution if and only if gcd(a, h) divides b. When gcd(a, h) = 1 the solution is unique modulo h.
Solve a linear polynomial congruence
polynomials a, b, h over a fieldall solutions f, or a report of insolubility- Run extended Euclid on (a, h) to obtain d = gcd(a, h) and coefficients s, t with as + ht = d.
- If d does not divide b, report no solution.
- Set a' = a/d, b' = b/d, h' = h/d.
- The inverse of a' modulo h' is s reduced mod h'.
- Return f = b' · s mod h', with the full solution set obtained by adding multiples of h'.
O(deg(h)²) field operationsThis is the polynomial congruence page's integer counterpart line by line. The only change is that the size measure is degree rather than absolute value.
03When the quotient is a field
Field criterion
F[X]/(h) is a field if and only if h is irreducible over F.
If h is irreducible, every non-zero residue is coprime to it and hence invertible by Bezout. If h factors as h = h₁h₂ with both of smaller degree, then the classes of h₁ and h₂ are non-zero with zero product.
04Frequently asked questions
Does the Chinese remainder theorem hold for polynomials?
Yes, for pairwise coprime moduli, with the same statement and proof. Its most familiar instance is Lagrange interpolation, where the moduli are the linear polynomials X − aᵢ.
Is the parallel with integers exact?
For the Euclidean structure, yes. It breaks where the specific arithmetic matters: polynomial degrees add under multiplication whereas integer sizes only roughly add, which makes degree bounds cleaner than size bounds.
What is the residue representation?
Each class has a unique representative of degree less than the modulus, obtained by division with remainder. So F[X]/(h) is a vector space of dimension deg(h) over F, with the powers of X as a basis.
Sources and method
Structural reference: Victor Shoup, A Computational Introduction to Number Theory and Algebra, Version 1, Cambridge University Press, 2005 — book pages 371-374.
This page carries the durable method layer only: definitions, constructions, algorithms, complexity results and selection criteria, authored originally for KEVOS. No text is transcribed or paraphrased from the source, and no numeric tables or benchmark data are reproduced — these are routed to live authoritative sources instead.
Author: Kevin Jogin. Last reviewed 2026-08-07.