Factorisation in integral domains
These notes revise some lectured material and have some questions for assessment—
either fill in the boxes or attach your solutions on extra sheets of paper.
What you already know about factorisation
Factorisation of integers
Let a, b ∈ Z. Then b is a factor of a if and only if there is some c ∈ Z for which a = bc. A
prime number is a positive integer whose only positive factors are 1 and itself. (Warning:
later we define irreducible and prime elements in a ring, and these definitions are not
exactly the same as prime number.) The essential theorem is:
Theorem 1 (Fundamental Theorem of Arithmetic) If a ∈ Z, a > 0, then there
exist prime numbers p1 , . . . pn (not necessarily distinct) so that a = p1 p2 · · · pn .
Moreover, if p1 , . . . , pn and q1 , . . . , qm are prime numbers for which
p1 p2 · · · pn = q1 q 2 · · · qm ,
then n = m and, possibly after reordering the qi , pi = qi for every 1 ≤ i ≤ n.
The proof is not hard: you can use an induction on the size of a together with an
application of division with remainder. The theorem can cover negative numbers too,
by allowing −p, for a prime p, to appear as a factor.
Factorisation of complex polynomials
A polynomial f ∈ C[t] is called irreducible if deg f > 0 and f cannot be written as a
product of polynomials gh where both deg g > 0 and deg h > 0. The essential theorem is:
Theorem 2 (Fundamental Theorem of Algebra) If f ∈ C[t] and deg f > 0, then
there is some α ∈ C such that f (α) = 0.
This theorem is not at all easy to prove: despite its name, its proof needs substantial input
from analysis. However the following corollary follows by a simple induction together with
an application of the division algorithm.
Corollary 3 If f ∈ C[t] and deg f > 0, then there are α1 , . . . , αn ∈ C, where n = deg f ,
and some nonzero c ∈ C such that f = c(t − α1 ) · · · (t − αn ).
Moreover, if c, d, αi , βj ∈ C for which
c(t − α1 ) · · · (t − αn ) = d(t − β1 ) · · · (t − βm ),
then c = d, n = m and, possibly after reordering the βi , αi = βi for every 1 ≤ i ≤ n.
This looks more like integer factorisation once you see that the linear factors are
actually irreducible.
Corollary 4 If f ∈ C[t] is irreducible, then f = at + b for a, b ∈ C with a 6= 0.
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Units in a ring
Let R be a commutative ring (with a multiplicative identity 1R ∈ R, as always.) Recall
that u ∈ R is a unit if and only if there is some v ∈ R such that uv = 1R . Equivalently,
u is a unit if and only if u divides 1R . We denote the set of all units in R by R∗ .
For example, the set of units in the integers is Z∗ = {1, −1}. And the set of units in
C is C∗ = C \ {0}. And the set of units in C[t] is also C \ {0}, regarded as polynomials of
degree 0. In particular, the statement of factorisation in C[t] is very much like factorisation
in Z (including negative numbers) but with irreducibles replacing prime numbers.
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Irreducible elements in a ring
Throughout this booklet, R denotes an integral domain, that is, R is a commutative ring
(with 1R as always) that has no zero divisors.
Definition 5 An element p ∈ R is irreducible if and only if p 6= 0, p is not a unit, and
whenever a, b ∈ R satisfy p = ab then either a is a unit or b is a unit.
You often think of the last condition as saying that ‘p has no non-trivial factors’.
Question 1 Which of 5, −1, 6, 0, 1, −12 are not irreducible elements of Z? In each
case, explain which part of the definition is not satisfied.
Question 2 Show that if f ∈ C[t] is irreducible then deg f > 0.
In particular, the definition we had for an irreducible in C[t] at the beginning is the same
as the definition here when R = C[t].
Definition 6 Let a ∈ R. A factorisation of a into irreducibles in R is an expression
a = p1 p2 · · · pn where p1 , . . . , pn ∈ R are irreducible.
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This definition is reminiscent of the factorisation theorems of the introduction, although
it makes no mention of uniqueness.
Question 3 Let f = t3 + t + 2. Compute a factorisation of f into irreducibles twice, once
in R[t] and once in C[t].
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Unique Factorisation Domains
We want to think of two factorisations 6 = 2 × 3 = (−2) × (−3) into irreducibles in
Z as being essentially the same. The notion of associates handles this, emphasising in
particular that you should not think of multiplying an irreducible by a unit as a big deal.
Definition 7 Let R be an integral domain. Then p, q ∈ R are associates if and only if
there is a unit u ∈ R such that p = uq.
Question 4 Show that if p ∈ R is irreducible and p, q ∈ R are associates then q is
irreducible too. [Hint: suppose not.]
The next definition characterises rings in which factorisation works just as well as it does
in Z and C[t].
Definition 8 Let R be an integral domain. Then R is a unique factorisation domain
(abbreviated to UFD) if and only if conditions (1) and (2) below are both satisfied:
(1) for all a ∈ R such that a 6= 0R and a is not a unit, there is a factorisation of a into
irreducibles in R.
(2) whenever p1 · · · pn = q1 · · · qm for irreducibles pi , qj ∈ R, then n = m and, possibly
after reordering the qj , pi and qi are associates for every 1 ≤ i ≤ n.
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Think of these two conditions as guaranteeing the existence (1) and uniqueness (2) of
factorisations into irreducibles. The theorems on factorisation in Z and in C[t] show that
these two rings are UFDs. In fact, if F is a field, then F [t] is always a UFD—we sketch
this in the optional last section, although you could already use what you know about
factorisation to check it when F = R.
√
√
−1],
Z[
−2],
However,
most
rings
are
not
UFDs.
We
consider
the
rings
Z[i]
=
Z[
√
√
are all subrings of C, so they are certainly integral domains. We
Z[ −3], Z[ −5]. These √
start by showing that Z[ −5] is not a UFD.
2.1
Norm functions and irreducibles
√
√
Recall that Z[ −5] = a + b −5 | a, b ∈ Z . Define
√
√
N : Z[ −5] −→ Z by N (a + b −5) = a2 + 5b2 .
√
This is√called a norm function on Z[ −5], hence its name N . Note that N (α) ≥ 0 for all
α ∈ Z[ −5], and moreover N (α) = 0 if and only if α = 0.
√
√
Question 5 Compute all α = a + b −5 ∈ Z[ −5] for which N (α) ≤ 11.
N (α)
0
1
2
3
4
5
6
7
8
9
10
11
possible α
N (α)
possible α
√
Question 6 Prove that N (αβ) = N (α)N (β) for all α, β ∈ Z[ −5].
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Question√7 Using the two previous questions, prove that N (α) = 1 if and only if α is a
unit in Z[ −5].
Step 1: Confirm that those α having N (α) = 1 are indeed units.
Step 2: Show that if αβ = 1 then N (α) = 1. Deduce that if α a unit then N (α) = 1.
√The point of the norm function N is to translate some questions of factorisation in
Z[ −5] to questions of factorisation in Z, where we know all the answers. So having seen
from the previous two questions that N is multiplicative and that it can detect units is
obviously useful.
√
√
Question 8 Following √
the examples in the lectures, prove that 3, 2 + −5 and 2 − −5
are all irreducible in Z[ −5]. That is, compute N (α) in each case, then . . .
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Question 9 Using
√ the previous question, exhibit two distinct factorisations of 9 into
irreducible in Z[ −5].
√
√
This proves that Z[ −5] is not a UFD. Similar considerations prove that Z[ −3] is not
a UFD either.
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Prime elements in a ring
So far, we have talked exclusively about irreducibles rather than primes.
Definition 9 An element p ∈ R is prime if and only if p 6= 0, p is not a unit, and
whenever a, b ∈ R satisfy p = ab then either p|a or p|b.
As usual, your first thought is to understand this definition in Z or in C[t]. In either
case, you will see that primes and irreducibles are the same thing. (Note that according
to this definition −5 is prime in Z. That is why we referred to ‘prime numbers’ when
referring to the usual primes p > 1 in Z.)
The definition of prime is very similar to that of irreducible. In fact, it is a bit stronger.
Question 10 Prove that if p ∈ R is prime in R, then p is irreducible in R.
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√
√
Question 11 Show that√3 does not
divide
either
of
2
±
−5
in
Z[
−5]. Show that 3
√
divides the product (2 + −5)(2 − −5), and conclude that 3 is not a prime.
√
We proved earlier that 3 is irreducible in Z[ −5], so this is an element that is irreducible
but not prime.
The difference between irreducibles and primes is the whole point of uniqueness of
factorisations.
Question 12 Let R be a UFD. Show that if p ∈ R is irreducible in R, then p is prime
in R.
√
We proved earlier that 3 is irreducible in Z[ −5], so this is an element that is irreducible
but not prime.
√
√
For extra
√revision, you can attempt similar questions in Z[ −3], Z[ −7] and so on.
However Z[ −2] and Z[i] are in fact UFDs, so in those rings we will not find different
factorisations, and we cannot find irreducibles that are not prime. We do not prove this
in the course, but the (easy) ideas are in the next optional section.
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4.1
The punchlines (optional and non-examinable)
Division with remainder and UFDs
In fact, it’s easy to prove that if you have sensible division-with-remainder in a ring, then
it has to be a UFD. Recalling the division algorithm, the point is to divide f by g 6= 0 to
get an expression
f = qg + r
where either r = 0, or r 6= 0 and “r is less than g”.
In Z, “less than” means 0 < r < |g|. In R[t], “less than” means deg r < deg g. The main
issue in other rings is to have a notion of “less than”. This is given by a degree function,
that is, some ∂ : R \ {0} → Z≥0 so that if a divides b then ∂(a) ≤ ∂(b), and for which a
division-with-remainder statement holds with “r = 0, or r 6= 0 and ∂(r) < ∂(g)”.
A ring with a suitable degree function is called a Euclidean domain. You can check
that Z[i] is a Euclidean domain with degree function ∂(a + ib) = a2 + b2 —this is the same
as the natural norm function, in fact.
Theorem 10 If R is a Euclidean domain, then R is a UFD.
We don’t prove this, but it’s easy to give the idea. One proves the existence of factorisations by contradiction. Using ∂, one can focus attention on a smallest element, a ∈ R
say, that does not factorise into irreducibles. Since a cannot be irreducible (for then it
would be factorised already), it must split into smaller factors—and one can combine
factorisations of these to get a factorisation of a. To prove uniqueness, one shows easily
that in a Euclidean domain irreducibles are prime, and then exactly the same first-year
proof that shows uniqueness of factorisations in Z works.
Corollary 11 If F is a field, then F [t] is a UFD. Also, Z[i] is a UFD.
√
−2], so that you can
To complete the picture, you can find
a
degree
function
on
Z[
√
do division-with-remainder there. So Z[ −2] is also a Euclidean domain, and so it is also
a UFD by the theorem.
4.2
Primes and UFDs
As we said earlier, the difference between irreducibles and primes is the whole point of
uniqueness of factorisations. The main theorem is this; compare it with the Definition 8
of UFDs. Its proof is, again, similar to the proof of uniqueness of factorisations in Z.
Theorem 12 Let R be an integral domain. Suppose R satisfies conditions (1–2) below:
(1) for all a ∈ R such that a 6= 0R and a is not a unit, there is a factorisation of a into
irreducibles in R.
(2) whenever p ∈ R is irreducible in R, then p is also prime in R.
Then R is a UFD.
In other words, if you have existence of factorisations into irreducibles, and you know also
that irreducibles are prime, then you get unique factorisation (in the sense of Definition 8).
Gavin Brown, Kent, February 2006
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