Past exam of the mathematics course of the University of Cambridge / 2005 / ib / Paper 3 / 11C (Groups, Rings and Modules)

A nonzero primitive polynomial in $\mathbb{Z}[x]$ has greatest common divisor of its coefficients equal to one. Suppose primitive polynomials $f,g$ had a product whose coefficients were all divisible by some prime $p$. Their reductions in ${\mathbb{F}}_p[x]$ are both nonzero, because neither polynomial has every coefficient divisible by $p$. A polynomial ring over a field is an integral domain, so their product is nonzero modulo $p$, a contradiction. Hence the product of primitive polynomials is primitive.

To deduce unique factorization, write any nonzero integer polynomial as its positive polynomial content times a primitive polynomial, with its sign treated as a unit. The preceding result gives multiplicativity of content. Factor the primitive part over the Euclidean polynomial ring $\mathbb{Q}[x]$, and rescale every irreducible rational factor to a primitive integer polynomial. If $f=q{\prod }_i{f}_i$ with all polynomials primitive and $q=a/b$ in lowest terms, the equality $bf=a{\prod }_i{f}_i$ has contents $b$ and $|a|$. Thus $b=|a|$, forcing $q=\pm 1$. The rational factorization therefore lifts to an integer factorization.

A primitive integer polynomial is irreducible over $\mathbb{Z}$ exactly when it is irreducible over $\mathbb{Q}$: an integer factorization gives a rational one, and a rational factorization lifts by the same primitive-content argument. Factor the content into integer primes. Existence and uniqueness over the integers and over $\mathbb{Q}[x]$ now give existence and uniqueness of the resulting factorization into constant prime factors and primitive irreducible polynomial factors, up to order and signs. Therefore $\boxed{\mathbb{Z}[x]\text{ is a unique factorization domain}}$. This is the full lifting step in Gauss lemma for polynomials.

The polynomial $p=x^5-4x+2$ satisfies the Eisenstein criterion at the prime two: every nonleading coefficient is even, the leading coefficient is not, and the constant term is not divisible by four. Hence it is irreducible over $\mathbb{Q}$. Its ideal is maximal in $\mathbb{Q}[x]$; explicitly, for any polynomial not divisible by $p$, the Euclidean algorithm gives $ag+bp=1$, furnishing the inverse of its residue class. ThusFor the integer quotient, division by the monic polynomial $p$ gives a unique remainder of degree below five with integer coefficients. If an integer polynomial becomes zero in the rational quotient, its integer remainder is still divisible by $p$ over $\mathbb{Q}$, so must be zero. This gives an injective ring map $\mathbb{Z}[x]/(p)\to \mathbb{Q}[x]/(p)$, proving the monic irreducible integer-polynomial quotient is a domain.

The same unique-remainder argument makes the integer quotient a free abelian group with basis $1,x,x^2,x^3,x^4$. The class of two is nonzero and cannot have an inverse: twice an integer remainder cannot equal the remainder one. Consequently