Move factorization between a UFD and its fraction field

Try

The coefficients of 6x²+9x+3 share a factor three. Removing it leaves 2x²+3x+1, whose coefficients have gcd one. This separation into scalar content and a primitive polynomial lets us distinguish arithmetic factorization of coefficients from positive-degree polynomial factorization.

Over a UFD R, content is defined up to a unit as a gcd of the nonzero polynomial’s coefficients. A polynomial is primitive when no nonunit divides every coefficient. In Z, choose positive content for a definite representative.

Predict

Could multiplying two primitive integer polynomials make every coefficient divisible by a prime? Coefficients involve sums, so cancellation makes the answer less immediate than multiplying two integers. Predict how reducing both polynomials modulo that prime turns the question into a zero-product problem in a polynomial ring over a domain.

Experiment

Multiply two primitive polynomials and separate a scalar example.

Polynomial	Positive content	Primitive part
6x+9	3	2x+3
10x+15	5	2x+3
Their product	15	4x²+12x+9

The product coefficients are 60,180,135, whose gcd is fifteen. The primitive-part coefficients have gcd one even though the middle coefficient has several prime factors. Content is a common-divisor property across all coefficients, not the size or factorization of one coefficient.

Understand

Gauss’s lemma states that a product of primitive polynomials over a UFD is primitive. Suppose a prime p divided every coefficient of fg. Reduce modulo p. In a UFD every irreducible is prime, so R/(p) is a domain. Primitivity means neither reduced factor is zero, but their product would be the zero polynomial. This contradicts the degree rule for nonzero polynomials over a domain. A nonunit common coefficient divisor would have a prime factor, so excluding primes excludes every nonunit common divisor.

Writing f=c(f)f₀ and g=c(g)g₀ now gives c(fg) associated to c(f)c(g), since f₀g₀ is primitive. This content law controls denominators when passing to K=Frac(R).

Suppose a primitive f factors over K as gh with both factors of positive degree. Clear denominators and extract contents to write g=a g₀ and h=b h₀, with a,b∈K nonzero and g₀,h₀ primitive in R[x]. Then f=ab g₀h₀. The product g₀h₀ is primitive. Comparing prime exponents in coefficients shows a rational scalar multiplying one primitive R-polynomial into another must be a unit of R: a negative prime exponent would leave some coefficient outside R, while a positive exponent would divide every coefficient. Thus ab is a unit, and the factorization can be placed in R[x]. The reverse implication follows by inclusion.

This yields the polynomial UFD theorem. Factor the content in R. Factor the primitive part in K[x], which is Euclidean and hence a UFD. Gauss’s lemma transfers the positive-degree factors back to primitive factors in R[x]. Uniqueness in K[x], combined with the content law and uniqueness in R, supplies uniqueness up to units and order.

Apply

Explain why primitivity is required when calling an integer polynomial irreducible over Q. A polynomial such as 2x+2 factors by the nonunit constant two in Z[x], while that constant becomes a unit in Q[x]. Positive-degree factorization and coefficient content must be treated separately.

Audit a proof that “clearing denominators obviously gives integer factors.” Write the scalar contents explicitly and identify the content law that cancels them. The conclusion is correct under the stated UFD and primitive hypotheses, but denominator manipulation alone does not prove it.

Continue

Irreducibility tests next use this bridge. Eisenstein and modular reduction are sufficient tests with specific hypotheses; failure of a test does not by itself prove a polynomial reducible.