Rings, Modules, Fields

2 Unique Factorization

The reason it is called a prime ideal, is that there is a related notion of a prime element, which is an element that generates a prime ideal. Note we can use the notation \(a|b\) (“\(a\) divides \(b\)") to mean \(b \in (a)\). A unit is an element that divides \(1\). The collection of units form a group under multiplication, denoted \(R^\times \) for the ring \(R\). Then a prime element is one where \(a|bc \implies a|c\) or \(a|b\). There is a related notion, called an irreducible element. A non-unit \(a\) is irreducible if \(bc|a\) means either \(b\) or \(c\) is a unit. We say that two elements are associates if they generate the same ideal, and that \(a\) is a proper divisor of \(b\) if \((a) \supsetneq (b)\). Note that many of these notions such as prime and irreducible make sense in arbitrary commutative monoids.

We would like to study the multiplicative structure of a ring, and in particular we would like to be able to tell when nonzero elements can be factorized into irreducibles.

Lemma 2.1. The set of zero divisors is closed under multiplication by any element of the ring, and the set of non zero divisors is closed under multiplication.

Proof. If \(a\) is a zero divisor let \(b\neq 0\) such that \(ab = 0\). Then for any \(r\), \((ar)b = 0\), so \(ar\) is a zero divisor. Now suppose that \(ab\) is a zero divisor. Then choose \(c \neq 0\) such that \(abc = 0\). If \(bc = 0\), then \(b\) is a zero divisor, but if not, then \(a\) is a zero divisor. □

The simplest way to understand the multiplicative structure of a ring \(R\) is to restrict to nonzero divisors (to avoid cases like \(2 \in \ZZ /6\ZZ \), for which \(2*4=2\)). We will call the monoid of nonzero divisors (in a nonzero ring) \(R_z\). Note that by the previous lemma, any factorization of elements in \(R_z\) remains entirely in \(R_z\).

Lemma 2.2. If \(a \in R_z\) and \((a) = (b)\) then \(a = ub\) for some unit. Moreover, if \(a = bc\), where \(b,c\) aren’t units, then \(b,c\) are proper divisors of \(a\).

Proof. If \((a) = (b)\), then \(a = cb, b = da\), so \((cd-1)a = 0 \implies cd = 1\), so \(c\) is a unit. Now if \(a = bc\) where \(b,c\) are not units, then \(b,c\) are not zero divisors as \(a\) isn’t. \((a) \supset (b)\), but were they equal, □

One way to get factorization into irreducibles is to require our ring be Noetherian. First, here are some conditions equivalent to being Noetherian.

Proposition 2.3. The following notions are equivalent for an \(R\)-module:
- • Ascending chain condition (ACC): any ascending chain of submodules stabilizes.
- • Every nonempty set of submodules of \(M\) has a maximal element.
- • Every submodule is finitely generated.

Proof. The first two are equivalent for any poset. The third implies the first since the union of an ascending chain of submodules is finitely generated, hence is contained in one of the elements of the chain. The first implies the third since given a submodule, we can construct a finite generating set by adding elements not already generated, and since this must stabilize, the submodule is finitely generated. □

Definition 2.4. A Bezout ring is one where finitely generated ideals are principle.

Proposition 2.5. If \(R\) is Noetherian, every non-unit of \(R_z\) can be factored into irreducibles.

Proof. If not, choose a maximal proper \((a)\) such that \(a \in R_z\) doesn’t factor into irreducibles. Then \(a\) cannot be irreducible so \(a = bc\), where \(b,c \in R_z\) aren’t units. But then \((b),(c)\) are strictly larger ideals, so \(b,c\) can be factored into irreducibles, a contradiction. □

Sometimes what lets us have unique factorization is a division algorithm:

Definition 2.6. A Euclidean ring is a ring \(R\) with a function \(\rho :R\to W\) where \(W\) is well ordered, such that only \(0\) is sent to the minimal element, and if \(a,b \in R, b \neq 0\), then there is a \(q,r\) such that \(a = bq+r\) with \(\rho (r)<\rho (b)\).

Definition 2.7. A principle ideal ring (PIR) is one where all ideals are principle (i.e. generated by \(1\) element). In other words, it is a Noetherian Bezout ring.

Definition 2.8. We say that a commutative monoid \(M\) has unique factorization if \(M/M^\times \) is a free monoid. We say \(M\) has factorization if every element can be factored into irreducibles.

When there is unique factorization, greatest common divisors (gcd) and least common multiples (lcm) always exist between finite collections of elements of \(M/M^\times \).

Definition 2.9. A unique factorization ring (UFR) is a nonzero ring where \(R_z\) has unique factorization.

Proposition 2.10. \(\ZZ \) is a Euclidean domain.

Proof. \(\rho \) is the absolute value function. Then given \(a,b\), choose \(q\) to be the integer such that \(a-qb\) is the smallest non negative number since \(\NN \) is well ordered. □

Let \(K[x]\) denote the ring of polynomials with coefficients in \(K\) in the variable \(x\).

Proposition 2.11. \(K[x]\) where \(K\) is a field is a Euclidean domain.

Proof. \(\rho \) is the degree of a polynomial. Given two polynomials with \(ax^k,bx^l\) as the leading term, we can choose \(q\) to be \(\frac a b x^{k-l}\) if \(k\geq l\) and \(0\) otherwise. □

Proposition 2.12. Every Euclidean ring is a PIR.

Proof. First note that it is Noetherian, as given a set of generating elements \(A\), we can construct a finite set of generators \(a_i\) as follows: pick \(a'_i\) to be an element not generated by \(a_j, j<i\) if it exists, and then apply the property of \(\rho \) to \(a'_i\) and each of the \(a_j\) to create a \(a_i\) such that \(\rho (a_i)<\rho (a_j)\) for \(j<i\). Now since \(W\) is well ordered, some \(a_i\) must be \(0\), so the \(a_j,j<i\) generate the ideal.

Thus it suffices to show that \((a,b) = (c)\) for some \(c\). Indeed choosing \(c\) be a non zero element of \((a,b)\) with \(\rho (c)\) minimal works. □

Definition 2.13. A commutative monoid has cancellation if \(ax = ay \implies x = y\).

Lemma 2.14. If \(M\) is a commutative monoid, \(M\) has unique factorization iff it has cancellation, factorization, and irreducibles are prime.

Proof. If \(M\) has unique factorization, it has clearly has cancellation and factorization, and if \(m\) is irreducible, and \(m|ab\), \(m\) is in the factorization of \(a\) or \(b\), so \(m|a\) or \(m|b\). Conversely, given two factorizations, any irreducible on one side must divide one on the other side, so they agree up to a unit. By cancellation and induction, the factorization is unique. □

Proposition 2.15. Every PIR is a UFR.

Proof. We can show that irreducibles in \(R_z\) are prime. Indeed, if \(a \in R_z\) is irreducible, and \(a|bc, b,c \in R_z\), \(a\nmid b\), then \((a,b) = (d)\), so \(dx = a\). Now \(x\) cannot be a unit so \(d\) must be, and \((a,b) = R\). Thus by the previous lemma we are done. □

In particular, since \(\ZZ \) is a PID the kernel of the unique map to any ring \(R\) is an ideal \((n)\), for some \(n\geq 0\), and \(n\) is called the characteristic of \(R\).