Choose a valid test and respect its limits

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The polynomial x²+1 is irreducible over R but factors over C. Irreducibility always names a coefficient ring or field. In F₂ it equals (x+1)² because two is zero. The same written coefficients can therefore describe different factorization behavior in different settings.

Over a field, a positive-degree polynomial is reducible when it is a product of two positive-degree polynomials. Nonzero constants are units and do not count as nontrivial factors.

Predict

Does having no root prove irreducibility? Predict why it works in degrees two and three but fails in degree four. A nontrivial factorization of a cubic must include a linear factor; a quartic can instead split as two quadratics, neither with a root in the base field.

Experiment

Reduce f=x³+x+1 modulo two and test every field element.

Input in F₂	Cubic value
0	1
1	1

There is no root, so the cubic is irreducible over F₂. Because the integer polynomial is monic and primitive, a rational factorization would transfer to an integer factorization with degree preserved modulo two, contradicting that result. The finite root check is complete because the field has exactly two elements.

For contrast, x⁴+3x²+2=(x²+1)(x²+2) has no real roots and therefore no rational roots, but is reducible over Q. Identify the missing linear-factor guarantee in the quartic case.

Understand

The rational-root theorem restricts a rational root u/v in lowest terms of an integer polynomial: u divides the constant term and v divides the leading coefficient. Substitution and clearing denominators prove these divisibilities using coprimality. For a monic polynomial, every rational root is an integer. This gives a finite test, but beyond degree three absence of rational roots does not settle irreducibility.

Eisenstein’s criterion gives a different sufficient condition. Let f be primitive in Z[x]. If a prime p divides every coefficient below the leading term, does not divide the leading coefficient, and p² does not divide the constant term, then f is irreducible over Q.

To prove it, suppose Gauss’s lemma supplies a positive-degree factorization f=gh in Z[x]. The leading coefficients of both factors are nonzero modulo p because their product is. Reducing modulo p gives a nonzero scalar multiple of xⁿ. In the UFD F_p[x], each reduced factor must therefore be a monomial. Their positive degrees are preserved, so both constant terms vanish modulo p. Their product, the constant term of f, is divisible by p², contradicting the hypothesis.

For x⁴+3x+6, the prime three meets all conditions: the omitted coefficients are zero and hence divisible by three, the leading coefficient is one, and nine does not divide six. Thus it is irreducible over Q. This example does not require testing possible quartic factorizations individually.

The same proof works over a UFD with a prime element in place of the integer prime and its fraction field in place of Q. This version will justify an inseparability example over a rational-function field later.

Modular reduction is also one-way: an irreducible degree-preserving reduction proves irreducibility over Q for a primitive polynomial. A reducible reduction does not prove the original reducible. For example x²+1 is irreducible over Q but becomes (x+1)² modulo two.

Apply

Choose a test before calculating. Record the coefficient field, primitive content, degree, and any prime used. If the test is inconclusive, say so and change methods; do not turn failure of a sufficient condition into its converse.

For each written task, state why the chosen test applies and which alternative factorizations it excludes. The goal is a proof of irreducibility, not a collection of unsuccessful guesses at factors.

Continue

The Chinese remainder theorem next splits quotients by relatively independent factors. It is useful for calculations, and it also shows why a quotient by a reducible polynomial may have zero divisors instead of being a field.