## Unique Factorisation: Basics

Throughout this post, let R be an integral domain; recall that this means R is a commutative ring such that whenever ab=0, either a=0 or b=0. The simplest example of an integral domain is Z, the ring of integers. What’s of interest to us is the following property.

Unique Factorisation of Z. In Z, every non-zero integer n can be expressed as a product:

$n =\pm p_1 p_2 \ldots p_r$,

where each pi is prime. Furthermore, the expression is unique up to a permutation of the list of primes.

In order to state this property more generally, we need to go through a few definitions. Some integral domains to keep in mind as we proceed:

• Z[i] = {a+biab are integers};
• Z[√2] = {a+b√2 : ab are integers};
• Z[√-5] = {a+b√-5 : ab are integers};
• R[x] = set of polynomials in x with real coefficients.

Definitions. Let x be a non-zero element of R.

1. Recall that x is a unit if there exists y such that xy=1. Note that the set of units forms an abelian group U(R).
2. If xuy for some unit u, then we say x and y are associates. Since the set of units forms a group, this association gives an equivalence relation.
3. Suppose x is a non-unit. We say x is reducible if we can write xyz for some non-units y and z. Otherwise, it is irreducible.
4. We say x divides y (or y is divisible by x) if we can write yxz for some z in R. Useful tip: this is equivalent to saying $y\in \left$, or $\left \subseteq \left$.
5. We say x is prime if <x> is a prime ideal. Equivalently, whenever yz is divisible by x, either y or z is divisible by x.

Examples

1. Units:
1. In the ring Z[i], the only units are +1, -1, +i and –i. The group of units is isomorphic to Z/2 × Z/2.
2. In Z[√2], a = 1+√2 is a unit since (1+√2)(-1+√2) = 1. It can be shown that every unit is of the form ±an, but we won’t prove it here.
3. In R[x], the units are the non-zero constants R*.
2. Associates: take Z[i].
1. Elements (1+i) and (1-i) are associates since (1+i) = i(1-i).
2. Elements (1+2i) and (1-2i) are not associates since (1+2i)/(1-2i) = (-3+4i)/5 which is not in Z[i].
3. Divisibility: take Z[i].
1. We have (1+i) | (3+5i) since (3+5i)/(1+i) = 4+i is in Z[i]. And because (1+i) and (1-i) are associates, we also have (1-i) | (3+5i).
2. We have (1+2i) | (2+9i) since (2+9i)/(1+2i) = 4+i. On the other hand (1-2i) does not divide (2+9i).

Before we continue, let’s prove a generic result.

Proposition. A prime element in R is irreducible.

Proof. Suppose x is prime and x = yz, where y and z are in R. Thus $yz\in\left$ so $y\in\left$ or $z\in\left$. In the first case, y = tx, so x = xtz. Since x ≠ 0 by assumption, tz = 1 so z is a unit. By symmetry, in the second case y is a unit. ♦

## Finite Factorisation

Let’s get into the main topic: when does unique factorisation hold in R? First, if $x\in R$, let’s try to factor x. Either x=0, x=unit, x=irreducible or x=reducible. For the first three cases, there’s nothing to factor; for the last case, write x=yz where y and z are not units. Again, for each of y and z, we check if it’s irreducible; if yes, we’re done; otherwise, we factor it. Rinse and repeat.

The question is must the process terminate? One shouldn’t be surprised to hear the answer is no: since we’re looking at general integral domains, pathological examples ought to come up.

Pathological Example. Consider the ring R of all expressions of the form:

$R = \{c_1 x^{r_1} + c_2 x^{r_2} + \ldots + c^n x^{r_n} : c_i\in\mathbf{R}, r_i\in\mathbf{Q}, r_i\ge 0\}.$

Thus, we’re looking at linear combinations of rational powers of a variable x (with finitely many terms). This is an integral domain since if fg = 0, then f and g belong to R[x1/n] for some n>0, which is an integral domain since it’s isomorphic to R[x]. Thus, f=0 or g=0. [ In short, R is the union of all R[x1/n]; a more high-level way of putting it: R is the “direct limit” of the R[x1/n]’s. ]

In this ring, we have $x = x^{1/2}x^{1/2} = (x^{1/4} x^{1/4})(x^{1/4}x^{1/4})\ldots$ and the process clearly doesn’t end. And none of the terms is a unit. So this is one ring where we don’t have finite factoring! ♦

To avoid this, let’s set a technical condition.

Definition. R is said to satisfy the ascending chain condition (a.c.c.) on principal ideals if, whenever we have:

$\left \subseteq \left \subseteq \left \subseteq \ldots$,

there exists an N such that $\left = \left = \left = \ldots$.

Proposition.

When R satisfies a.c.c. on principal ideals, we can always factor a non-zero non-unit x in R as a finite product of irreducibles.

Proof.

Suppose x is reducible. We first claim x has an irreducible factor. Write xyz where y and z are not units; if y is irreducible we’ve proven our claim. Otherwise, write again y = y2z2; if  yis irreducible, we’ve proven our claim. Etc. If the process terminates after finitely many steps, good. Otherwise, we get an infinite sequence

$\left \subsetneq \left \subsetneq \left \subsetneq \ldots$

which contradicts our assumption. This proves our claim.

So x has an irreducible factor y. Write xyx2, where x2 is not a unit (otherwise y is reducible). Now repeat the same thing with x2: it must have an irreducible factor. Etc. This gives a sequence of factorisations:

$x = yx_2 = (yy_2)x_3 = \ldots = (yy_2 y_3\ldots y_n)x_n$, each $y_i$ is irreducible.

Once again, we get an increasing sequence of principal ideals $\left \subsetneq \left \subsetneq \left \subsetneq \ldots$ which must eventually terminate. Hence, the factorisation must end eventually. ♦

Note

1. The reader may wonder why the proof is so long for a seemingly intuitive result. However, we can’t find a shorter rigourous proof. The intuitive approach to the proof is fraught with subtle issues.
2. One naturally asks how we can check if a ring R satisfies a.c.c. on principal ideals. It turns out, from a later theory, that for most rings this can be considered “obvious”. In fact, almost all rings we’re dealing with satisfy a.c.c. on all ideals. Such rings, called Noetherian rings, play a major role in commutative algebra. Anyway, for now, we urge the reader to bear with us for using rather ad-hoc methods to prove the a.c.c. condition.

## Unique Factorisation: Conditions

Now we’re ready to state the condition for unique factorisation. But first, we should be clearer about what unique factorisation means. After all, each irreducible or prime has many associates. Even in Z, we can factor -12 = 2 × (-2) × 3 and -12 = (-2) × (-2) × (-3) since under our definition, negative prime numbers are also primes.

Definition. Let R be a ring satisfying a.c.c. and $x\in R$ be non-zero, non-unit. For every factorisation

$x = y_1 y_2 \ldots y_n$,

of x into irreducibles, we can obtain another by permuting the terms and/or replacing each term by an associate. We say R is a unique factorisation domain (UFD) if any two such expressions must be related in the above manner.

Main Theorem. R is a UFD if and only if every irreducible is also prime.

Proof.

First, suppose every irreducible is prime. To prove that R is a UFD, first we show that in two factorings of x, if y occurs in one, then one of its associates must occur another:

$x = y y_2 y_3 \ldots y_n =z_1 z_2 \ldots z_m.$

So y divides the RHS. Since y is irreducible, it is prime. And since y divides the RHS, it must divide one of the terms, say z1. Writing yuz1, the fact that y is irreducible tells us either u or z1 is a unit. Since z1 is irreducible, it’s not a unit so u is. Ergo, y and z1 are associates.

Dividing the two expressions by y and letting the remaining terms absorb the unit, we repeat the reasoning to prove that both expressions are identical, up to permutation and associates.

Now for the converse, which is easier. If R is a UFD and x is an irreducible, suppose x divides yz. Write wxyz. Write wy and z as products of irreducible elements; this gives a factorisation of the element wxyz. By the property of unique factorisation, x must occur in one of the terms on the RHS, either in the factoring of y or z. Thus, x|y or x|z. This shows x is prime. ♦

Conclusion. In a nutshell, assuming some finiteness condition on the ring, unique factorisation property is equivalent to the fact that {primes} = {irreducibles}.

We’ve done the harder part of the theory. However, examples are sorely lacking as of now, since the above theorems don’t lend themselves to computations. We’ll try to redeem ourselves with some computational examples in the next two installations.

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