Unique Factorisation and Ascending Chain Conditions

Elements a and b in a commutative ring R are associates if there exists a unit u ∈R with b = ua. For example, in Z, the units are ±1, and so the only associates of an integer m are ±m; in k[x], where k is a field, the units are the nonzero constants, and so the only associates of a polynomial f (x) ∈k[x] are the polynomials u f (x), where u ∈k and u ̸= 0. In any commutative ring R, associates a and b generate the same principal ideal; the converse may be false if R is not a domain.

A domain R is a unique factorization domain (UFD) if (i) every r ∈R, neither 0 nor a unit, is a product2 of irreducibles; (ii) if up1 · · · pm = vq1 · · · qn, where u and v are units and all pi and q j are irreducible, then m = n and there is a permutation σ ∈Sn with pi and qσ(i) associates for all i. When we proved that Z and k[x], for k a field, have unique factorization into irreducibles, we did not mention associates because, in each case, irreducible elements were always replaced by favorite choices of associates: In Z, positive irreducibles (i.e., primes) are chosen; in k[x], monic irreducible polynomials are chosen.

A polynomial f (x) = anxn + · · · + a1x + a0 ∈R[x], where R is a UFD, is called primitive if its coefficients are relatively prime; that is, the only common divisors of an, . . . , a1, a0 are units. Of course, every monic polynomial is primitive. Observe that if f (x) is not primitive, then there exists an irreducible q ∈R that divides each of its coefficients: If the gcd is a nonunit d, then take for q any irreducible factor of d.

Let R be a UFD with Q = Frac(R). If f (x) ∈Q[x], there is a factorization f (x) = c( f ) f ∗(x), where c( f ) ∈Q and f ∗(x) ∈R[x] is primitive. We call c( f ) the content of f (x) and f ∗(x) the associated primitive polynomial. In light of Lemma 6.24(i), both c( f ) and f ∗(x) are essentially unique, differing only by a unit in R.

A commutative ring R satisfies the ACC, the ascending chain condition, if every ascending chain of ideals I1 ⊆I2 ⊆· · · ⊆In ⊆· · · stops; that is, the sequence is constant from some point on: there is an integer N with IN = IN+1 = IN+2 = · · · . ACC rings

If X is a subset of a commutative ring R, then the ideal generated by X is the set of all finite linear combinations I = (X) = finite riai : ri ∈R and xi ∈X . We say that I is finitely generated, often abbreviated to f.g., if X = {a1, . . . , an}; that is, every element in I is an R-linear combination of the ai. We write I = (a1, . . . , an), and we call I the ideal generated by a1, . . . , an. A set of generators a1, . . . , an of an ideal I is sometimes called a basis of I (even though this is a weaker notion than that of a basis of a vector space, for we do not assume that the coefficients ri in the expression c = riai are uniquely determined by c). Of course, every ideal I in a PID is finitely generated, for it can be generated by one element.

If A is a set, let P(A)# denote the family of all its nonempty subsets. The axiom of choice states that if A is a nonempty set, then there exists a function β : P(A)# → A with β(S) ∈S for every nonempty subset S of A. Such a function β is called a choice function. Informally, the axiom of choice is a harmless looking statement; it says that we can simultaneously choose one element from each nonempty subset of a set. The axiom of choice is easy to accept, and it is one of the standard axioms of set theory. However, the axiom is not convenient to use as it stands. There are various equivalent forms of it that are more useful, the most popular of which are maximality principle and the well-ordering principle. Recall that a set X is a partially ordered set if there is a relation x ⪯y defined on X that is reflexive, antisymmetric, and transitive. We introduce some definitions to enable us to state the well-ordering principle.

A partially ordered set X is well-ordered if every nonempty subset S of X contains a smallest element; that is, there is s0 ∈S with s0 ⪯s for all s ∈S. Well-ordering principle. Every set X has some well-ordering of its elements.

Let R be a domain and let a, b ∈R. (i) a | b and b | a if and only if a and b are associates. (ii) The principal ideals (a) and (b) are equal if and only if a and b are associates.

Let R be a domain in which every r ∈R, neither 0 nor a unit, is a product of irreducibles. Then R is a UFD if and only if (p) is a prime ideal in R for every irrreducible element p ∈R.3

If R is a UFD, then a gcd of any finite set of elements a1, . . . , an in R exists.

If R is a UFD and f (x), g(x) ∈R[x] are both primitive, then their product f (x)g(x) is also primitive.

now applies: R[x] is a UFD if (p(x)) is a prime ideal for every irreducible p(x) ∈R[x]; that is, if p | f g, then p | f or p | g. Let us assume that p(x) ∤f (x). Case (i). Suppose that deg(p) = 0. Write f (x) = c( f ) f ∗(x) and g(x) = c(g)g∗(x), where c( f ), c(g) ∈R, and f ∗(x), g∗(x) are primitive. Now p | f g, so that p | c( f )c(g) f ∗(x)g∗(x). Since f ∗(x)g∗(x) is primitive, Lemma 6.24(ii) says that c( f )c(g) an associate of c( f g). However, if p | f (x)g(x), then p divides each coefficient of f g; that is, p is a common divisor of all the coefficients of f g, and hence in R, which is a UFD, p divides the associates c( f g) and c( f )c(g). But Proposition 6.17 says that (p) is a prime ideal in R, and so p | c( f ) or p | c(g). If p | c( f ), then p divides c( f ) f ∗(x) = f (x), a contradiction. Therefore, p | c(g) and, hence, p | g(x), as desired. Case (ii). Suppose that deg(p) > 0. Let (p, f ) = { s(x)p(x) + t(x) f (x): s(x), t(x) ∈R[x] } ; of course, (p, f ) is an ideal containing p(x) and f (x). Choose m(x) ∈(p, f ) of minimal degree. If Q = Frac(R) is the fraction field of R, then the division algorithm in Q[x] gives polynomials q′(x),r′(x) ∈Q[x] with f (x) = m(x)q′(x) + r′(x), where either r′(x) = 0 or deg(r′) < deg(m). Clearing denominators, there are polynomials q(x),r(x) ∈R[x] and a constant b ∈R with bf (x) = q(x)m(x) + r(x), where r(x) = 0 or deg(r) < deg(m). Since m ∈(p, f ), there are polynomials s(x), t(x) ∈ R[x] with m(x) = s(x)p(x) + t(x) f (x); hence r = bf −qm ∈(p, f ). Since m has minimal degree in (p, f ), we must have r = 0; that is, bf (x) = m(x)q(x), and so bf (x) = c(m)m∗(x)q(x). But m∗(x) is primitive, and m∗(x) | bf (x), so that m∗(x) | f (x), by

A similar argument, replacing f (x) by p(x) (that is, beginning with an equation b′′ p(x) = q′′(x)m(x) + r′′(x) for some constant b′′), gives m∗(x) | p(x). Since p(x) is irreducible, its only factors are units and associates. If m∗(x) were an associate of p(x), then p(x) | f (x) (because p(x) | m∗(x) | f (x)), contrary to the hypothesis. Hence, m∗(x) must be a unit; that is, m(x) = c(m) ∈R, and so (p, f ) contains the nonzero constant c(m). Now c(m) = sp + t f , and so c(m)g(x) = s(x)p(x)g(x) + t(x) f (x)g(x). Since p(x) | f (x)g(x), we have p(x) | c(m)g(x). But p(x) is primitive, because it is irreducible, by Example 6.22, and so Lemma 6.24(iv) gives p(x) | g(x). •

shows that every algebraic integer α has a unique minimal polynomial m(x) ∈Z[x], namely, m(x) = irr(α, Q), and m(x) is irreducible in Q[x]. Remark. We define the (algebraic) conjugates of α to be the roots of irr(α, Q), and we define the norm of α to be the absolute value of the product of the conjugates of α. Of course, the norm of α is just the absolute value of the constant term of irr(α, Q), and so it is an (ordinary) integer. We have also considered them in the proof of polynomial finite-generation’s Theorem 90, which was used to prove that if the field-automorphism groups of a polynomial f (x) ∈k[x] is solvable, where k has characteristic 0, then f (x) is solvable by radicals. ◀ The next criterion uses the integers mod p.

Let f (x) = a0 + a1x + a2x2 + · · · + xn ∈Z[x] be monic, and let p be a prime. If f (x) is irreducible mod p, that is, if Δf (x) = [a0] + [a1]x + [a2]x2 + · · · + xn ∈Fp[x], is irreducible, then f (x) is irreducible in Q[x].

Let g(x) ∈Z[x]. If there is c ∈Z with g(x + c) irreducible in Z[x], then g(x) is irreducible in Q[x].

Let R be a UFD with Q = Frac(R), and let f (x) = a0 +a1x +· · ·+anxn ∈R[x]. If there is an irreducible element p ∈R with p | ai for all i < n but with p ∤an and p2 ∤a0, then f (x) is irreducible in Q[x].

Let k be a field and let f (x1, . . . , xn) be a primitive polynomial in R[xn], where R = k[x1, . . . , xn−1]. If f cannot be factored into two polynomials of lower degree in R[xn], then f is irreducible in k[x1, . . . , xn].

If k is a field and g(x1, . . . , xn), h(x1, . . . , xn) ∈k[x1, . . . , xn] are relatively prime, then f (x1, . . . , xn, y) = yg(x1, . . . , xn) + h(x1, . . . , xn) is irreducible in k[x1, . . . , xn, y].

If I is a proper ideal in a ACC rings R, then there exists a maximal ideal M in R containing I. In particular, every ACC rings has maximal ideals.5

(i) If k is a field, then k[x1, . . . , xn] is ascending-chain-finite. (ii) The ring Z[x1, . . . , xn] is ascending-chain-finite. (iii) For any ideal I in k[x1, . . . , xn], where k = Z or k is a field, the quotient ring k[x1, . . . , xn]/I is ascending-chain-finite.

Let k be a field and let R be the subring of k[x] consisting of all polynomials f (x) ∈k[x] having no linear term; that is, f (x) = a0+a2x2+· · ·+anxn. It now follows from Proposition 6.20 that R is not a UFD.

It is easy to see, for every positive integer n, that In = { f : R →R : f (x) = 0 for all x ≥n} is an ideal and that In ⊊In+1 for all n. Therefore, R does not satisfy the ACC, and so R is not ascending-chain-finite. ◀ Here is an application of the maximum condition.