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Polynomial Rings, Irreducibility and Unique Factorisation

Polynomials over a field divide with remainder, and that single fact forces unique factorisation into irreducibles. Deciding whether a given polynomial is irreducible over the rationals needs tests rather than inspection.

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What a learner can do afterwards

  • Divide polynomials with remainder and read off what it says about roots
  • Apply Eisenstein's criterion and reduction modulo a prime
  • Trace the chain from division with remainder to principal ideals to unique factorisation

1 · Read

Over a field F, divide f by a nonzero polynomial g by repeating four actions: divide the leading terms, add that term to the quotient, multiply g by it, then subtract. Stop when the remainder is zero or has smaller degree than g. The result f = qg + r is unique. Field coefficients let you divide by g's nonzero leading coefficient.

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Divide x^2 + 6x + 8 by x + 3. Leading terms give x. Subtract x(x + 3), leaving 3x + 8. The next quotient term is 3. Subtract 3(x + 3), leaving -1. Thus the quotient is x + 3 and the remainder is -1. A nonzero remainder means the divisor is not a factor.

Division by x - a leaves a constant remainder. Substituting a into f = q(x)(x - a) + r gives r = f(a), the Remainder Theorem. Hence x - a is a factor exactly when f(a) = 0. A quadratic over a field is reducible exactly when it has a root there. Over the reals, a quadratic with no real root is irreducible; every real cubic has a real root and therefore a real linear factor.

x^2+6x+8 = x(x+3)+3x+83x+8 = 3(x+3)-1Quotient x+3; remainder -1
Cancel the leading term, subtract and repeat until the remainder degree is below the divisor degree.
Try it together

Grouping reverses distributivity. For x^3 + 2x^2 + 6x + 12, the pairs become x^2(x + 2) + 6(x + 2). Factor the common x + 2 to obtain (x^2 + 6)(x + 2). Expanding a proposed factorization checks every coefficient. State the coefficient field when claiming that the remaining factors are irreducible.

For an integer-coefficient polynomial, a rational root u/v in lowest terms has u dividing the constant term and v dividing the leading coefficient. Test all such candidates. For degree 2 or 3 over Q, no rational root means irreducible; higher degrees can split without a linear factor. Irreducibility depends on the field: a rationally irreducible cubic still has a real root and factors over R.

An integer polynomial is primitive when its coefficients have gcd 1. For a positive-degree integer polynomial, Eisenstein's criterion applies if a prime p does not divide the leading coefficient, divides every other coefficient, and p^2 does not divide the constant coefficient. Then the polynomial is irreducible over Q. Missing coefficients are zero, which is divisible by p. Check all three divisibility conditions.

f = x^3 + 10x + 5p = 5; 5 does not divide 15 divides 0, 10, 525 does not divide 5
These checks establish Eisenstein's criterion for this primitive cubic.

For primitive f in Z[x], reduce coefficients modulo a prime p. If the degree is preserved and the reduction is irreducible over F_p, then f is irreducible over Q. For reduced degree 2 or 3, test all field elements for roots. Higher degrees need other factor tests. For example, x^2 + x + 1 has value 1 at both 0 and 1 modulo 2, so its reduction is irreducible. A degree drop invalidates this test.

g = x^2 + x + 1 mod 2g(0) = 1g(1) = 1No root; degree 2 retained
For a quadratic, the complete root check proves irreducibility of the reduction.

A principal ideal domain is a nonzero commutative domain in which every ideal has one generator. F[x] is one: in a nonzero ideal choose a nonzero g of least degree. Divide any f in the ideal by g. Its remainder also belongs to the ideal, so minimality forces it to be zero. Thus the ideal is (g); the zero ideal is (0). Repeated division also computes gcds, and the gcd generates the ideal (f, g).

In F[x], gcd division gives Bezout identities. If irreducible p does not divide a, gcd(p, a) = 1, so up + va = 1. If p divides ab, multiply this identity by b to see that p divides b. Thus irreducibles are prime. Repeated positive-degree splitting terminates, giving each nonzero polynomial a nonzero constant times irreducible factors. Primeness makes the factors unique up to order and nonzero constant multiples. Zero has no such factorization.

Division with remainderGcd and Bezout identitiesIrreducibles are primeNonzero unique factors
Division proves the principal-ideal and prime-factor properties used in unique factorization.

Division gives remainders for testing factors and leads to unique factorisation of polynomials.

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Readings and extra questions written for this dataset.

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Polynomial Rings, Irreducibility and Unique Factorisation · Mathematics, ages 22 to 24 · LightMySky