Download 3 Factorisation into irreducibles

3
Factorisation into irreducibles
Consider the factorisation of a non-zero, non-invertible integer n as a product
of primes: n = p1 × · · · × pt . If you insist that primes should be positive
then, since n could be negative, you have to allow multiplication by a unit,
n = u × p1 × · · · × pt where the pi are primes and u is invertible (i.e. u = ±1).
There are a number of points to note:
• existence of a prime decomposition (i.e. there’s at least one way of splitting n
as a product of primes);
• uniqueness of prime decomposition - of course you can write the primes in a
different order but, since the ring is commutative, that’s an inessential variation.
You probably knew this fact (that each integer has just one prime decomposition) long before you saw a proof of it if, indeed, you have seen a proof;
anyway, we’ll prove it here (and in a more general context).
• The primes appearing can be multiplied by any unit, but that’s also an inessential variation. For instance, we will count −3 as a prime and as being essentially
the same prime as 3. In the case of the ring of integers we could insist that only
positive integers can count as primes but, in more general cases, there may be
no canonical choice like this, so we just have to accept that, e.g., −3 is as good
as +3. In the terminology that we introduced in the previous section, −3 and 3
are associates (in particular they generate the same ideal). Allowing associates
means we can always absorb the unit u (appearing above) in any of the primes
in the decomposition of a non-zero non-unit.
One important point which is rather hidden in the case of the ring of integers
is that there are two concepts bundled together in the notion of prime integer.
The first is irreducibility, meaning that the element can’t be factorised in any
non-trivial way; the second is primeness - if p divides a product then it divides
one of the factors. In general these are different!
There are two main sources of examples of the kinds of ring we consider
in Section 3: rings of integers in number fields and polynomial rings. Number
fields are what we get by taking the field Q of rationals and adding roots of
polynomials;
sitting inside
√
√ such a field is an analogue of the integers: compare
Z[ 2] sitting inside Q[ √2] with Z sitting inside Q (more generally, √
the ring of
integers in the field Q[
d],
where
d
is
a
square-free
integer,
is
Z[
d] if d 6=
√
1 mod 4 and is Z[ 12 + 2d ] otherwise). These, and related rings, are the topic of
algebraic number theory.
Polynomial rings of course you know. If K is a field then any polynomial
in one indeterminate X with coefficients in K can be factorised as a product
of irreducible polynomials, and this decomposition is essentially unique. For
instance if a, b, c ∈ K then aX 2 + bX + c either is irreducible or is a product
of two (necessarily irredicible) linear factors (X − α)(X − β) where α, β are the
roots of the quadratic polynomial (and these can be found using the quadratic
formula). Did you ever wonder whether there could be two essentially different
factorisations of such a polynomial? What about polynomials in more than one
variable? - take a polynomial in, say, X, Y, Z with, say, rational coefficients:
does it necessarily have a factorisation into irreducibles? (The answer is “yes”
and that’s easy to prove.); but is such a factorisation essentially unique? (The
answer again is “yes” but this is not at all obvious and is quite hard to prove but we’ll do it.)
8
3.1
Unique Factorisation Domains
Suppose throughout that R is a commutative domain.
A non-zero, non-unit a ∈ R is irreducible if whenever we have a = bc then
either b or c is invertible (so the other is, by definition of associate, an associate
of a). It follows that a is irreducible iff whenever we have a factorisation a = bc,
one of b, c is an associate of a (it follows that the other is a unit by (the argument
of) 2.6).
For instance, if K is a field then a non-constant (note that implies, non-zero,
non-unit) polynomial f ∈ K[X] = K[X1 , . . . , Xt ] is irreducible if f cannot be
written as a product of two polynomials of smaller degree: f = gh implies g or
h is a constant (i.e. an element of K). The, easy, proof of this uses the fact (8.9)
that if f = gh then deg(f ) = deg(g) + deg(h). Note that non-zero polynomials
f and g have the same degree if they are associates.
Examples 3.1. 1. Let K = R, f = X 2 + 1 ∈ R[X]. Since the square of a real
number is always positive, X 2 + 1 has no real root, hence has no linear factor.
Hence f is an irreducible element of R[X].
2. Let K = C, f = X 2 + 1 ∈ C[X]. Now we have X 2 + 1 = (X − i)(X + i)
where i is a square root of -1. So f is not irreducible as an element of C[X].
3. Let f = X 2 + Y 2 − 1 ∈ K[X, Y ] where K is any field of characteristic 0 (such
as Q, R or C). Is f irreducible? Suppose not, say f = gh with neither g nor h a
constant. Since deg(f ) = 2, it must be that deg(g) = 1 = deg(h). Write each of
g, h as a sum of its homogeneous parts: g = g1 + g0 and h = h1 + h0 . Then we
have X 2 + Y 2 − 1 = (g1 + g0 )(h1 + h0 ) = g1 h1 + (g1 h0 + g0 h1 ) + g0 h0 where g1 h1
is homogeneous of degree 2, g1 h0 + g0 h1 is homogeneous of degree 1 and g0 h0 is
homogeneous of degree 0. Recall, 8.1, that the decomposition of a polynomial
into homogeneous components is unique so we obtain: X 2 + Y 2 = g1 h1 , 0 =
g1 h0 + g0 h1 , −1 = g0 h0 . From the last equation we get h0 6= 0 so we can
divide by h0 in the second equation, re-arrange and obtain g1 = (−g0 /h0 )h1 .
Then substitute this into the first equation to get X 2 + Y 2 = (−g0 /h0 )h21 .
Now, h1 = aX + bY for some a, b ∈ K and, if we square this and multiply by
the scalar −g0 /h0 we cannot get terms X 2 and Y 2 without getting a non-zero
“cross-term” (scalar)XY (write this out carefully to see that, at this point, we’re
using that the characteristic of K is 0, at least, that it’s not 2) - contradiction.
So X 2 + Y 2 − 1 is irreducible (over any field of characteristic 6= 2).
Exercise 3.2. Show that X 2 + Y 2 − 1 ∈ Z2 [X] is not irreducible.
√
√
4. In the ring Z[ 2] the element 1 + 2 2 is irreducible. We show this by
considering
the norm of an element
which, in this ring, is defined by N (a +
√
√
b 2) = a2 − 2b2 , so N (1 + 2 2) = −7. It is easily checked that√the norm is
multiplicative in the sense√that N (rs) = N (r)N (s) for r, s ∈ Z[ 2], so if we
have a factorisation 1 + 2 2 = rs then N (r)N (s) = −7. Since −7 is prime
(rather, irreducible) in Z, that implies that, say, r has norm ±1. If we show
√
that N (r) = ±1 implies r is a unit then we will be done. Suppose r = a + b 2
with a, b ∈ Z.
√
√
√
If N (r) = 1 then a2 − 2b2 = 1, so (a + b 2)(a − b 2) = 1, so ra + b 2 is a
unit.
If N (r) = −1 then a2 − 2b2 = −1. In this case it’s less obvious how to
proceed and there are actually non-trivial solutions to the equation N (r) = −1:
9
√
e.g. N (7 + 5 2) = −1. But note that if N (r) = −1 then N (r2 ) = (N (r))2 = 1
so, as seen just above, r2 = ±1. Thus r2 is a unit, hence (note, this is a general
point) r is a unit, as required.
We will show (3.24) that every polynomial in any number of variables has
an essentially unique decomposition as a product of irreducible factors.
A commutative domain R is said to be a unique factorisation domain if every
non-zero, non-invertible element of R has an essentially unique factorisation as
a product of irreducible elements. More formally: the commutative domain
R is a unique factorisation domain, or UFD for short, if for every nonzero r ∈ R which is not invertible there are irreducible elements r1 , . . . , rk ∈ R
such that r = r1 × · · · × rk (existence of irreducible factorisation) and, if also
r = s1 × · · · × sl where the sj are irreducible elements of R, then k = l and
there is a permutation σ of {1, . . . , k} such that, for each i, sσ(i) is associated
to ri (uniqueness of irreducible factorisation).
Example 3.3. Z is a unique factorisation domain. For instance take r = −24.
Then −24 = (−2) × 2 × (−2) × (−3) is one irreducible factorisation. There are
others, such as −24 = 3 × (−2) × 2 × 2 but you can surely see how to permute
the factors so that they match up as associated elements.
√
Example 3.4. The ring Z[ −5], though
a commutative
domain, is not a UFD.
√
√
For instance 21 = 3 × 7 = (1 + 2 −5)(1 − 2 −5) are, one may check, two
factorisations into irreducibles but, again one may check, the irreducible factors
do not match up
pairs. Another example in the same ring is
√ into associated
√
6 = 2.3 = (1 + −5)(1 − −5).
To check that these really are different factorisations into √
irreducibles
√ we
have to show, say in the second example, that all of 2, 3, 1 + −5, 1 − √−5
are irreducible
and also that, for instance, 2 is an associate
of neither 1 + −5
√
√
nor 1 −
−5.
This
is
easy
if
we
use
the
“norm”
on
Z[
−5]:
this is the function
√
√
N : Z[ −5] → Z which is defined by N (a + b −5) = a2 + 5b2 √
.
It is easy to
check
that
N
(rs)
=
N
(r)N
(s)
for
all
r,
s
∈
Z[
−5]. Note√also
√
√
√
that if N (a+b −5) = a2 +5b2 = 1 then (a+b −5)(a−b −5) = 1 so a+b −5
is invertible. Now, N (2) = 4 so if 2 factorises, say as 2 = rs with neither r nor s
invertible, then N (r) = 2 = N (s) which is impossible (the equation a2 + 5b2 = 2
clearly√has no integer solutions). So we’ve shown that 2 √
is irreducible. Turning
to 1+ −5: this has norm 6 so a proper factorisation 1+ −5 = rs with neither
of r, s invertible would be possible only if N (r) = 2 and N (s) = 3 or vice√versa,
but we’ve seen already that N (r)√= 2 is impossible. Therefore 1 + −5 is
irreducible and the proof for 1 − −5√is almost identical.
Finally we have to
√
check that 2 doesn’t divide either 1 − −5 or √
1 − −5: if √
it did then the norm
of 2, which is 4, would divide the norm of 1 + −5 or 1 − −5 - both of which
equal 6 - contradiction, as required.
(By the way,
√ the general formula for the appropriate
√ norm to use in a ring
of the form Z[ d] where d ∈ Z is squarefree is N (a + b d) = a2 − db2 ; you can
check this has the required multiplicative property N (rs) = N (r)N (s). Recall
that such a norm was used in the motivating example I gave at the start of the
course: a semiproof for Fermat’s assertion that the equation y 2 + 2 = x3 has
for√integer solutions only x = 3, y = ±5, and which involved moving to the ring
Z[ −2]. That example can be found, with details sketched, on pp. 80/81 of
Stewart and Tall; Algebraic Number Theory.)
10
Lemma 3.5. The ring Z of integers is a unique factorisation domain. If K is
a field then K[X] is a unique factorisation domain.
We have seen already (2.2, 2.7) that these are PIDs, so this will follow from
3.14. We will show that if K is a field then K[X1 , . . . , Xt ] is also a unique
factorisation domain but the proof is considerably harder. Existence of a factorisation into irreducibles is not a problem - the proof is given next - it’s the
uniqueness of factorisation which is difficult.
Theorem 3.6. Let f ∈ K[X]. Then f = f1 . . . fn where each fi is irreducible.
Proof. If f is irreducible then we’re already finished. Otherwise f = g1 g2 with
deg(g1 ), deg(g2 ) < deg(f ). By induction on the total degree of a polynomial
(see 8.9), we may assume that each of g1 and g2 can be written as a product
of irreducible polynomials. So, putting together these expressions for g1 , g2 we
get an expression for f as a product of irreducible polynomials.
[Note how saying “by induction on the degree” shortcuts the “keep splitting
any reducible factors” idea behind the proof.]
The greatest common divisor of non-zero elements a, b ∈ R (a commutative domain) is an element c = gcd(a, b) ∈ R such that c | a, c | b and, for
any d ∈ R with d | a, d | b we have d | c. It is easy to check (exercise) that if a
greatest common divisor of a and b exists then it is unique up to multiplication
by an invertible element of R: that is, if c, d are both greatest common divisors
of a and b then c ∼ d.
Lemma 3.7. Let R be a unique factorisation domain. Then:
(1) for all non-zero a, b ∈ R, gcd(a, b) exists and is unique up to association;
(2) if a ∈ R is irreducible and if a | bc then a | b or a | c.
Proof. The general, and key, point is that if x|y, say y = xz, then an irreducible
factorisation of y is obtained by placing an irreducible factorisation of x next to
an irreducible factorisation of z. In particular, if x|y then each irreducible factor
of x occurs, up to association, as an irreducible factor of y and the multiplicity
of that factor (to association) in the factorisation of x is less than or equal to its
multiplicity (up to association, but let’s stop saying that) in any factorisation
of y.
(1) Factorise each of a and b and then let c be the product of all the factors,
including multiplicity, that they have in common. Then proceed as you would
if R = Z, using the key fact above to deduce that if d|a and d|a then d|c.
(2) Again, with the key fact, this is just as if you were dealing with ordinary
integers. If, say, ad = bc, then write each of d, b, c as a product of irreducible
elements; substitute back into the equation ad = bc; note that a occurs on the
left-hand side hence, by unique factorisation, a occurs on the right-hand side;
hence a is a factor of b or c, as required.
3.2
Prime ideals and prime elements
Let R be a commutative ring.
An element a ∈ R is prime if a is not invertible and whenever a|bc (b, c ∈ R)
then a|b or a|c. This is almost the usual definition in Z, except that in any
domain 0 will count as a prime element.
11
An ideal P of R is prime if whenever a, b ∈ R with ab ∈ P then either
a ∈ P or b ∈ P . It follows (by induction) that if P is a prime ideal and if some
product a1 a2 . . . an lies in P then at least one of the ai lies in P .
Examples 3.8. (1) Recall that in the ring Z of integers, every ideal is principal.
By 3.9 below, hni is a prime ideal iff n is a prime integer or n = 0. For instance
h12i is not a prime ideal because we have 6.4 ∈ h12i yet neither 6 nor 4 is in
h12i. On the other hand, h17i is a prime ideal because if ab ∈ h17i then 17 | ab
and hence, since 17 is prime, 17 | a or 17 | b, that is, a ∈ h17i or b ∈ h17i.
(2) Let R = Q[X, Y, Z]. The ideal hXi is prime but hXY i is not. The ideal
hXY − Z 2 i looks as if it should be prime but you need to check that XY − Z 2
is irreducible. Another prime ideal is hX 2 + Y 2 i (however this is no longer
prime if regarded as an ideal in R = C[X, Y, Z] since in that ring we have the
factorisation X 2 +Y 2 = (X +iY )(X −iY )). The ideal hXY +1, X 2 i is not prime
but the ideal hX + 1, Y + 0.5, Z − 2i is prime. What about hXY − Z 5 , Y 3 − X 5 i?
It’s not always obvious whether or not an ideal in a polynomial ring is prime; the
Gröbner basis techniques that we will see later in the course provide algorithms
to answer this kind of question.
The next result illustrates that prime ideals generalise prime elements. The
notion of an ideal arose exactly because there are rings of integers in number
fields which don’t have a good prime decomposition theory for elements but do
have such a theory for ideals, with prime ideals (and note that a prime ideal is
not required to be a principal ideal) replacing prime elements.
Lemma 3.9. Let R be a commutative ring. Let p ∈ R, p 6= 0; then p is a prime
element iff the ideal, hpi, generated by p is a prime ideal.
Proof. (⇒) Suppose that p is a prime element and that ab ∈ hpi, so ab = pc for
some c ∈ R. Since p is prime either p|a, so a ∈ hpi, or p|b, so b ∈ hpi, showing
that hpi is a prime ideal.
(⇐) Suppose that hpi is a prime ideal and that p|ab, so ab ∈ hpi. Then either
a ∈ hpi, so p|a, or b ∈ hpi, so p|b, showing that p is a prime element.
Lemma 3.10. Let R be a commutative domain. Every non-zero prime element
of R is irreducible.
Proof. If p is prime and p = bc then either p|b or p|c, let’s suppose the former.
Thus we have both b|p and p|b so, by 2.1, hbi = hpi. Since R is a domain 2.6
applies and we deduce that p and b are associates. It follows (see the comment
after the definition of irreducible) that p is irreducible.
In Example
√3.4 we can see that the converse of 3.10 is false: 2 is an irreducible
element
of
Z[
−5]
√
√ but 2 is not a prime(!) in that ring: 2 divides the product
(1 + −5)(1 − −5) but, as shown in 3.4, divides neither of the factors. In
UFDs, however, we do have the converse.
Corollary 3.11. Let R be a unique factorisation domain. Then every irreducible element is prime. That is, in a UFD, irreducible = prime (for non-zero
elements).
Proof. The first statement is exactly what 3.7(2) says. Then combine this with
3.10 above.
12
Next we show that every principal ideal domain is a unique factorisation
domain. An ideal is said to be maximal if there is no ideal strictly between it
and the whole ring; we consider these further in §4.3.
Proposition 3.12. Suppose that the commutative ring R is a PID and not a
field. Then a principal ideal hpi of R is maximal iff the element p is prime and
non-zero iff p is irreducible.
Proof. Every maximal ideal is prime: a fact which we will prove later (4.17); so
if hpi is maximal then, by 3.9, p is prime.
For the converse, suppose that p is a prime element. Let r ∈ R \ hpi and
let J = hp, ri denote the ideal generated by p and r together. It will be enough
to show that J = R. Since R is a PID there is some s ∈ J such that J = hsi.
Since p ∈ J we have p = st for some t ∈ R. Thus st ∈ hpi and, by 3.9 the
ideal hpi is prime, so either s ∈ hpi or t ∈ hpi. Since we chose s outside hpi the
first case is impossible, so we have t = up for some u ∈ R. But that gives us
p = st = sup so (1 − su)p = 0 and hence, since R is a domain, s is invertible.
But that implies that J = hp, si is the whole of R, as required.
Every prime element is, by 3.10, irreducible so it remains to show that if p is
irreducible then it is prime. By what has been shown already it will be enough
to show that hpi is a maximal ideal. If not, then there is an ideal, by assumption
principal, say hsi, with hpi < hsi < R. Since hsi =
6 R, s is not invertible. Since
p ∈ hsi we have p = st for some t ∈ R. If t were invertible then p and s
would be associates hence, by 2.6, would generate the same ideal, contradiction.
So we have a proper factorisation of p , contrary to our assumption that p is
irreducible.
Corollary 3.13. If R is a PID then a ∈ R is irreducible iff it is prime.
Theorem 3.14. Every PID is a UFD.
Proof. Suppose that R is a PID.
Existence of irreducible factorisations: suppose that a ∈ R is neither 0 nor
invertible. If a is irreducible then we’re done; if not, write a = bc where neither
b nor c is an associate of a (nor invertible). By 2.1 we have hai ⊂ hbi and
hai ⊂ hci, with both inclusions being proper (by 2.6). If b is not irreducible then
write it as b = b′ b′′ with neither an associate of b, and similarly for c. Continue.
If this ‘splitting’ process stops at some point then we collect up all the irreducible factors and have an irreducible factorisation of a. If it doesn’t stop then
we obtain a strictly increasing sequence of principal ideals hai ⊂ hbi ⊂ hb′ i ⊂ . . .
say. That contradicts the fact that every ideal of R is finitely generated - that is,
R is a noetherian ring, since a noetherian ring can have no properly ascending
sequence of ideals - a result, 4.1 which we will prove later. (If R is a Euclidean
domain then we don’t have to wait to finish this proof because, with notation
as in the first paragraph and if ν is as in the definition of Euclidean domain,
we have ν(a) > ν(b), ν(c) so, if the process did not stop we’d get a strictly
decreasing sequence of natural numbers, contradiction.)
Uniqueness of irreducible factorisation: The proof is, now that we have
that irreducible=prime (3.13), exactly as with integers: suppose that r = r1 ×
· · · × rk = s1 × · · · × sl where the ri and sj are irreducible elements of R. Since
r1 is irreducible it is prime so it divides one of the sj and hence, since sj is
irreducible, ri and sj are associates. Cancel r1 from each side of the equation
13
(if sj = ur1 that will leave an extra factor “u” on the RHS but u is a unit, so
can be absorbed into one of the other factors). Continue... In this way we pair
up the factors ri with associate factors sk , as required for uniqueness.
In particular, by 2.7, we deduce the following corollary.
Corollary 3.15. Every Euclidean domain is a UFD.
If R is a UFD then the notions of greatest common divisor and least common multiple may be defined as usual: gcd(a, b) is computed by picking out the
common factors in irreducible factorisations of a and b and, more generally, any
finite set of elements of R has a gcd which may be calculated by using the (easily proved) fact that gcd(a1 , . . . , an ) = gcd(. . . gcd(gcd(a1 , a2 ), a3 ), . . . , an ) . . . ).
Actually computing irreducible or prime factorisations may, however, be difficult in practice (this difficulty is the basis for a standard method of securely(?)
encrypting electronic communications on the internet). But if R is a Euclidean
domain it is not necessary to compute prime factorisations in order to compute
greatest common divisors and least common multiples: the method based on
the Euclidean algorithm works just as well in this generality as it does for the
ring, Z, of integers (assuming that we are in a sufficiently explicit situation that
computations are possible).
The next result is another which generalises from the more familiar cases of
integers and polynomial rings.
Lemma 3.16. Let R be a unique factorisation domain and let a, b ∈ R be
irreducible and not associates. Then hai ∩ hbi = habi.
Proof. Since ab ∈ hai and ab ∈ hbi certainly habi ⊆ hai∩hbi (that part is true for
any commutative ring). For the converse, suppose that c ∈ hai∩hbi: say c = c′ a,
c = c′′ b. By unique factorisation, it follows that each of a and b occurs (up to
association) in a factorisation of c into irreducibles (which exists and is unique
since R is a UFD). Since a and b are not associates of each other, these are
different occurrences, so ab is a factor of c and hence c ∈ habi, as required.
3.3
The field of fractions of a commutative domain
The process of forming fractions, by which Q is obtained from Z, may be generalised with any commutative domain R in place of Z.
Let W = {(r, s) : r, s ∈ R, s 6= 0} be the set of pairs of elements of R with
the second element non-zero. The idea is that a pair (r, s) will represent the
fraction r/s, i.e. rs−1 (which is why s has to be non-zero). But a given fraction
has many different representations, e.g. 1/2 = 3/6 = −2/−4, so we introduce an
equivalence relation on W by defining (r, s) ∼ (t, u) iff ru = st (why? because
r/s = t/u iff ru = st).
It’s easy to check that this is indeed an equivalence relation and we define
Q to be the set of equivalence classes. Write (r, s)/ ∼ for the equivalence class
of (r, s).
Define an addition and multiplication on Q in the obvious way (“obvious”
when you think how it all works for ordinary fractions): define ((r, s)/ ∼) +
((t, u)/ ∼) = ((ru + st, su)/ ∼) and ((r, s)/ ∼) × ((t, u)/ ∼) = ((rt, su)/ ∼).
As usual, because we are defining operations on classes by reference to particular
14
representatives, we have to check that these operations are well-defined. This
is easily done, then one checks that the result is indeed a ring structure on the
set, Q, of equivalence classes, with (1, 1)/ ∼ being the 1 and (0, 1)/ ∼ being the
0. A bit tedious, but straightforward and left as an exercise.
What about the connection with the original ring R? Well, the map ι : R −→
S defined by r 7→ (r, 1)/ ∼ is easily checked to be an injective homomorphism.
So we can think of R as sitting inside Q. The latter is called the field of
fractions of R and also written Q(R). Another easy check shows that every
element of Q has the form ι(r)ι(s)−1 where r, s ∈ R, so Q really is built up as
fractions from R (at least, from the copy of R sitting inside it).
Examples 3.17. (1) Q(Z) = Q; (2) √
Q(K[X]) =√ K(X), the field of rational
functions, if K is a field; (3) Q(Z[ 2]) = Q[ 2]; (4) Q(Z[i]) = Q[i]; (5)
Q(Z[X]) = Q(X). In checking that these are correct one may make use of the
fact (which you can try as an exercise) that if R is a domain and if θ : R −→ L
is any embedding of R into a field then there is a unique factorisation of
θ : Q(R) → L through the embedding ι : R −→ Q(R) of R into its field of
fractions.
Our main example will be the case that R = K[X1 , . . . , Xt ], the ring of
polynomials in X1 , . . . , Xt with coefficients in a field K, in which case Q(R) is
the field of rational functions in X1 , . . . , Xt with coefficients in K, usually
denoted by K(X1 , . . . , Xt ). The typical element of K(X1 , . . . , Xt ) has the form
p(X1 ,...,Xt )
X 2 Y +XZ
q(X1 ,...,Xt ) where p, q ∈ K[X1 , . . . , Xt ] - for instance XY Z−Z 6 −2 ∈ Q(X, Y, Z).
3.4
Gauss’ Lemma
Suppose now that R is a UFD and that f ∈ R[X], f = an X n + · · · + a1 X +
a0 . Define the content, c(f ), of f to be greatest common divisor of all its
coefficients: c(f ) = gcd(an , . . . , a1 , a0 ). This is defined only up to association.
Clearly if f ∈ R[X] then f = c(f )f ′ where c(f ′ ) = 1. A polynomial f is said to
be primitive if c(f ) = 1 (more precisely, if c(f ) ∼ 1).
Example 3.18. Take R = Z and let f = 15X 4 − 21x2 + 6. Then c(f ) =
gcd(15, 21, 6) = 3 (we can ignore 0 coefficients) and f = c(f )f ′ = 3(5X 4 −
7X 2 + 2) where f ′ = 5X 4 − 7X 2 + 2 is a primitive polynomial.
Lemma 3.19. (Gauss’ Lemma) Let R be a unique factorisation domain and
let f, g ∈ R[X]. Then c(f g) = c(f )c(g) (up to multiplication by an invertible
element). In particular, any product of primitive polynomials is primitive.
Proof. Let f = c(f )f ′ , g = c(g)g ′ : so f ′ and g ′ are primitive. Then we have
f g = c(f )c(g)f ′ g ′ so, if we show that c(f ′ g ′ ) = 1 then we will have c(f g) =
c(f )c(g). Therefore, what we have to show is that the product of two primitive
polynomials is primitive.
So assume now that f and g are primitive. We have f g = c(f g)h, say, and
we want to show that c(f g) = 1. If not then, since R is a UFD, we can choose
some irreducible factor p, say, of c(f g): so p divides every coefficient of f g.
We will derive a contradiction from this. Let f = an X n + ·P
· · + a1 X + a0 and
n+m
m
k
g = bP
m X + · · · + b1 X + b0 with the ai , bj ∈ R. Then f g =
k=0 ck X where
ck = i+j=k ai bj .
Choose i such that p divides a0 , . . . , ai but not ai+1 and choose j such that
p divides b0 , . . . , bj but not bj+1 . Because f and g are primitive we have i < n,
15
j < m. Consider the coefficient ci+j+2 = ai+j+2 b0 + · · · + ai+2 bj + ai+1 bj+1 +
ai bj+2 + · · · + a0 bj+i+2 of X i+j+2 in f g (notice, e.g., that i + j + 2 could happen
to be greater than n - any “too-high-indexed” coefficients we just take to be 0).
Since p, by assumption, divides each coefficient of f g, p divides the left-hand
side of this equation and hence divides the right-hand side. By choice of i, j we
have that p divides every term of this sum except, possibly, ai+1 bj+1 so, since
p divides the whole sum it must also divide ai+1 bj+1 . But then, by 3.7(2), p
divides either ai+1 or bj+1 - contrary to choice of i or j.
This contradiction shows that c(f g) = 1, as required.
3.5
K[X] is a UFD
If R is a commutative domain then we can form its field, Q = Q(R), of fractions,
as in §3.3.
Proposition 3.20. Let R be a UFD and let Q be the field of fractions of R.
Suppose that f ∈ R[X]. Then f is irreducible in R[X] iff f is irreducible when
regarded as an element of Q[X].
Proof. Certainly if f has a proper factorisation in R[X] then it has a proper
factorisation in Q[X]. For the converse suppose that f = gh is a factorisation of f in Q[X] with deg(g), deg(h) > 1. Write g = d1 g ′ , h = 1e h′ where
d, e ∈ R and g ′ , h′ ∈ R[X] (for instance, take d to be the product (or the
least common multiple) of all denominators of coefficients of g, similarly for h).
′
′
Then write g ′ = c(g ′ )g ′′ , h′ = c(h′ )h′′ . So we have g = c(gd ) g ′′ , h = c(he ) h′′ ,
c(g ′′ ) = 1 = c(h′′ ). Note also that deg(g ′′ ) = deg(g) > 1, deg(h′′ ) = deg(h) > 1.
′
)c(h′ ) ′′ ′′
g h and hence (de)f = c(g ′ )c(h′ )g ′′ h′′ . We
We have f = gh = c(g de
want to divide now by de but we have to show that we can do this in R[X].
Taking the content of each side of this equation and using 3.19 we obtain
(de)c(f ) = c(g ′ )c(h′ )c(g ′′ h′′ ) = c(g ′ )c(h′ )c(g ′′ )c(h′′ ) = c(g ′ )c(h′ ). Therefore
′
)c(h′ )
= c(f ) ∈ R and hence f = (c(f )g ′′ ) · h′′ is a proper factoriwe have c(g de
sation of f in R[X], showing that f is reducible in R[X] if it is reducible in
Q[X].
Corollary 3.21. Let f ∈ Z[X] be a polynomial with integer coefficients. If f
is irreducible in Z[X] then f is also irreducible in Q[X].
Proof. This is the special case of the previous result where R = Z.
So if a polynomial with integer coefficients cannot be split over Z then it
cannot be split over Q either.
Example 3.22. Let f = X 3 + 15X 2 − 3X − 6 ∈ Q[X]. By the above result this
is irreducible over Q iff it is irreducible over Z. If it were reducible over Z then,
being a cubic, it must have a linear factor, say f = (X − a)g for some integer
a. Any such a must be an integer divisor of 6 and hence is ±1, ±2, ±3, ±6. If
you substitute these values for X in f then in no case is the result equal to 0.
We conclude that f has no integer root and hence is irreducible over Z and so,
by the above result, is irreducible over Q.
Theorem 3.23. Suppose that R is a unique factorisation domain. Then R[X]
is a unique factorisation domain.
16
Proof. Suppose that f ∈ R[X] is non-zero. If f is invertible (that is, a nonzero constant polynomial) then there is nothing to prove. So assume also that
f is not invertible, hence deg(f ) ≥ 1. Write f = c(f )f ′ with f ′ a primitive
polynomial in R[X]. First we factorise c(f ) and f ′ separately and put their
factorisations together to obtain a factorisation of f .
Since R is a UFD there is a factorisation of c(f ) as c(f ) = c1 . . . cl with the
ci irreducible elements of R. As you would expect, the ci are also irreducible
elements of R[X] since if, say, ci = hh′ with h, h′ ∈ R[X] then, since 0 =
deg(ci ) = deg(h) + deg(h′ ), we have that h and h′ are of degree 0 and hence are
in R so, since ci is irreducible in R, one of h, h′ is invertible, as required.
If f ′ is already irreducible then fine; otherwise factorise it as f ′ = gh with
neither g, h invertible. That means that neither g nor h is a constant polynomial
because 1 = c(f ) = c(g)c(h). So deg(f ) > deg(g), deg(h). Since, note, both g
and h are primitive we can apply the same argument to each of them so, since
degrees can’t keep decreasing, eventually we obtain a factorisation f ′ = f1 . . . fn
of f ′ into irreducible polynomials in R[X] each of degree at least 1.
Putting these together we get a factorisation f = c1 . . . cl f1 . . . fn of f as a
product of irreducible elements of R[X]. Now we have to prove uniqueness.
So suppose also f = d1 . . . dk g1 . . . gm with the di irreducible elements of R
and the gj irreducible elements of R[X] of degree at least 1.
Notice that if g ∈ R[X] is irreducible then c(g) = 1: otherwise we would
have g = c(g)g ′ with neither c(g) nor g ′ invertible. So c(fi ) = 1 = c(gj ) for
all i, j. It follows that d1 . . . dk = c(f ) = c1 . . . cl and hence, since R is a UFD,
l = k and the ci and dj are, up to rearrangement, pairwise associated.
Therefore c1 . . . cl f1 . . . fn = f = d1 . . . dk g1 . . . gm = c1 . . . cl ug1 . . . gm for
some invertible element u. Since R[X] is a domain we can cancel c1 . . . cl from
each side to obtain f1 . . . fn = (ug1 )g2 . . . gm . Thus we have two ways of writing
f ′ as a product of irreducible elements. Rewrite ug1 as g1 (invertible factors of
R[X] can be ignored).
Let Q be the field of fractions of R and notice that R[X] is a subset (even
a subring) of Q[X]: so consider the two above factorisations of f ′ as being in
Q[X]. By 3.20 each fi and each gj is an irreducible element of Q[X].
Therefore f1 . . . fn and g1 . . . gm are two factorisations of f ′ into irreducibles
in Q[X]. Since Q is a field, Q[X] is a unique factorisation domain (3.5) and
hence m = n and, after renumbering if necessary, we have fi = ui gi where ui is
an invertible element of Q[X]. We can write each ui as haii with hi ∈ R[X] and
ai ∈ R, and hence hi ∈ R (since any invertible element of Q[X] is a constant).
So ai fi = hi gi and hence ai = c(ai fi ) ∼ c(hi gi ) = hi (c(gi ) = 1 since gi is
irreducible in R[X]). Hence ai ∼ hi and so we can cancel ai to get fi = u′i gi for
some unit u′i ∈ R. Therefore fi and gi are associates, which is what we wanted
to show (that is, unique decomposition of f ′ and hence of f ).
Corollary 3.24. Let K be any field. Then K[X1 , . . . , Xt ] is a unique factorisation domain. That is, if f ∈ K[X1 , . . . , Xt ] is a non-constant polynomial and if
f = af1 . . . fn = bg1 . . . gm where a, b ∈ K and each fi , gj is a monic irreducible
polynomial then a = b, m = n and there is a permutation σ of {1, . . . , n} such
that fi = gσ(i) for each i = 1, . . . , n.
Proof. That K[X1 , . . . , Xt ] is a UFD follows from 3.1.2 and induction. The
second statement follows directly.
17
That is, every polynomial f may be written as a scalar times a product of
monic irreducible polynomials and, except for re-arranging the terms, there is
just one way of writing f in this form.
Of course, we could as well assume in 3.24 that K is any UFD and obtain
the same result by the same argument.
Example 3.25. You can check that X 4 Y 4 −X 2 Y 2 Z 2 +X 2 Y 2 Z −X 2 Y 2 −Z 3 +Z 2
is equal to (XY − Z)(X 2 Y 2 + Z − 1)(XY + Z) and, with a bit of work, you
can check that these three factors are irreducible. By 3.24 there is no other
essentially different way of factoring this polynomial as a product of irreducible
polynomials.
Let f ∈ K[X1 , . . . , Xt ]. Define the zero-set, V (f ), of f to be the set of
all t-tuples of elements of K which, when sustituted into f , give 0: V (f ) =
{(a1 , . . . , at ) ∈ K t : f (a1 , . . . , at ) = 0}.
Example 3.26. Let K = R, f = X 4 + X 2 Y 2 − X 2 ∈ R[X, Y ]. Notice that
f = X.X.(X 2 + Y 2 − 1) = ggh, say. We saw in 3.1 that X 2 + Y 2 − 1 is
irreducible, so this is the irreducible decomposition of f . Suppose that (a, b) ∈
V (f ). Then f (a, b) = 0 so, g(a, b)g(a, b)h(a, b) = 0. Hence either g(a, b) = 0 or
h(a, b) = 0 so either (a, b) ∈ V (g) or (a, b) ∈ V (h). The converse is also true:
if (a, b) ∈ V (g) or (a, b) ∈ V (h) then f (a, b) = 0 so (a, b) ∈ V (f ). The same
argument shows that this is true in general: the zero-set of any polynomial is
the union of the zero-sets of its irreducible factors. In this case, we have that
V (X 4 + X 2 Y 2 − X 2 ) = V (X) ∪ V (X 2 + Y 2 − 1) = the union of the Y -axis with
the circle of radius 1, centred at the origin (0, 0).
3.6
Irreducible polynomials in K[X]
Most of what is in this section is review of things you have seen before.
1. Suppose f 6= 0, f ∈ K[X]. Then V (f ) is a finite set - any non-zero
polynomial in just one indeterminate has only finitely many roots.
2. Suppose that f 6= 0, f ∈ K[X] and let a ∈ K. Then f (a) = 0 iff (X − a)
is a factor of f .
3. Recall that if f ∈ R[X] is a polynomial with real roots then f factorises
as a product of linear and quadratic polynomials. That is:
Proposition 3.27. If f ∈ R[X] is irreducible then deg(f ) ≤ 2.
Recall also that the proof of this uses the fact that we can extend R to a
field (C) over which every polynomial splits as a product of linear polynomials.
4. There is a formula for finding the roots (and hence the linear factors) of
any quadratic polynomial. This was known (at least special instances) to the
Babylonians around 2000BC. At the end of the 1400’s Luca Pacioli (at Venice)
judged that a solution to cubic equations was impossible by the methods known
at the time, yet in 1545 Girolamo Cardano in Milan published his book Ars
Magna which contained a procedure for finding the solution for any cubic (this
procedure was also found earlier, at least in part, by Scipione del Ferro at
Bologna and by Niccolo Tartaglia at Venice).
18
The same book contained a procedure for finding a solution to the general
quartic (=degree 4) polynomial, due to Cardano’s pupil Ludovico Ferrari.
In all these cases, the roots of the polynomial could be expressed by a formula
involving the ordinary arithmetic operations, addition, subtraction, multiplication and division, together with the “extraction” of roots (think, for instance,
of the formula for the solution of a quadratic). The question then was whether
there exists such a type of formula for the roots of a general polynomial of degree
5 (=a quintic) and similarly for all higher degrees.
Gradually it came to be suspected that there is no such formula and this
was proved in the early 1800s by Galois and Abel. In fact, Galois gave the
exact conditions for a given polynomial to be “solvable by radicals”. This proof
arose out of consideration of symmetries among the roots of a polynomial or, as
we would now express it, of consideration of permutation groups acting on the
roots of a polynomial. Indeed, this work is often regarded as the birth of group
theory.
For the history and a good deal of the mathematics, see Tignol’s book.
5. Recall the “fundamental theorem of algebra”.
Theorem 3.28. Let f ∈ R[X] or C[X] be a non-zero polynomial of degree d.
Then, regarded as an element of C[X], f splits as a product of d linear factors.
In particular, a polynomial f of degree d has exactly d roots, where we count
repeated roots with their multiplicity (=the number of times the corresponding
linear factor appears in the irreducible decomposition of f ).
3.7
Tests for irreducibility
Proposition 3.29. (Eisenstein’s Criterion for Irreducibility) Let f ∈ Z[X], say
f = an X n + · · · + a1 X + a0 with an 6= 0. Suppose that there is a prime p ∈ Z
such that p does not divide an , p divides every other coefficient ai and p2 does
not divide a0 . Then f is irreducible (in Z[X], and hence, 3.21, in Q[X]).
Proof. Suppose, on the contrary, that f = gh with deg(g) > 0, deg(h) > 0, say
g = bm X m + · · · + b1 X + b0 and h = ck X k + · · · + c1 X + c0 with bm 6= 0, ck 6= 0,
m ≥ 1, k ≥ 1. Notice that bm ck X m+k is the leading term of gh = f and hence
bm ck = an . Since p2 does not divide a0 but p does divide a0 exactly one of b0 ,
c0 is divisible by p. Say p | b0 and p ∤ c0 . Let i be minimal such that p does not
divide bi - since bm ck = an is not divisible by p neither is bm , so there is such
a value i ≤ m < n. Consider ai = bi c0 + bi−1 c1 + · · · + b0 ci : on the right-hand
side each of bi−1 , . . . , b0 is divisible by p and also, by assumption, ai is divisible
by p. Hence the remaining term bi c0 is divisible by p. But p does not divide c0
hence p divides bi - contrary to choice of i. This contradiction shows that f is
irreducible.
Examples 3.30. (1) Let f = 2X 9 − 25X 5 + 10X 3 − 15X − 30 ∈ Z[X]. Taking
p = 5 we have that p does not divide the leading coefficient, 2, of f , p divides
every other coefficient of f and p2 = 25 does not divide the constant term of f .
Hence, by 3.29, f is irreducible in the ring Z[X] and hence, by 3.21, f also is
irreducible as a member of the ring Q[X]. It is not, of course, irreducible as a
member of R[X] or of C[X].
19
(2) Let p be a prime integer and consider Φ(X) = X p−1 +X p−2 +· · ·+X +1 ∈
Q[X]. We claim that Φ(X) is an irreducible polynomial in Q[X]. Notice that
p
−1
X +1 for X to get Φ(X +1) = (X+1)
=
Φ(X) = (X p−1)/(X −1). Substitute
X
p
p
p
p−1
p−2
X
+ p−1 X
+ · · · + 2 X + 1 . The leading coefficient, 1, is not divisible
by p, each binomial coefficient pi is divisible by p and the constant coefficient,
p, is not divisible by p2 . Hence, by Eisenstein’s Criterion, 3.29, Φ(X + 1) is
irreducible. But this means that Φ(X) is irreducible (because, if Φ(X) factorised
then so would Φ(X + 1) - think about it).
(3) Eisenstein’s criterion is a test which can be used to show that some
polynomials are irreducible: it cannot be used to show that a polynomial is
reducible. For instance, let f = X 3 − 9 ∈ Q[X]. Eisenstein’s criterion cannot
be applied to this polynomial, nevertheless, it is irreducible because the (real)
cube root of 9 is irrational. (We will give a proof of this in the example which
follows the next result.)
Lemma 3.31. Suppose that f ∈ Z[X] is of degree d and suppose that n ≥ 2 is
an integer such that the reduct of f modulo n is irreducible and has degree d.
Then f is irreducible (in Z[X], hence, by 3.20, in Q[X]).
Proof. Suppose, for a contradiction, that f = gh with deg(g) = k, deg(h) = l
k, l < d = deg(f ). Write [f ]n for the reduct of f modulo n, that is, the
polynomial in Zn [X] which is obtained by replacing each coefficient in f by its
congruence class modulo n, and similarly for g, h. Then we have [f ]n = [g]n [h]n .
Clearly, deg([g]n ) ≤ k, deg([h]n ) ≤ l. By assumption, deg([f ]n ) = d so, since
deg([g]n )+deg([h]n ) = deg([f ]n ) = d we must have deg([g]n ) = k, deg([h]n ) = l.
But that means that [f ]n is not irreducible, contrary to hypothesis. Hence f
must be irreducible, as claimed.
Notice that the condition that the reduct of f modulo n also has degree d is
equivalent to the condition that n should not divide the leading coefficient of f .
Examples 3.32. (1) In Example 3.30(3) we claimed that the cube root of 9 is
irrational. We will prove this now by showing directly that X 3 − 9 is irreducible
as an element of Q[X]. So consider the reduction of this polynomial modulo
the prime 7: you can check that each value 0, 1, . . . , 6 gives a non-zero value for
this polynomial in Z7 so X 3 − 9 has no root in Z7 and hence no linear factor.
Therefore, being of degree 3, the reduction of X 3 − 9 modulo 7 is irreducible
as an element of Z7 [X]. Therefore, by 3.31, X 3 − 9 ∈ Q[X] is irreducible. It
follows that X 3 − 9 has no rational root and we conclude that the real cube root
of 9 is, indeed, irrational.
(2) Consider X 3 + X 2 + X − 1. Consider the reduction of this modulo 3 and
set X = 0, 1, 2 in turn. The values obtained in Z3 for the polynomial are 2,2,1,
none of which is 0. Hence, since the polynomial is of degree 3, it has no linear
factor, hence is irreducible and hence X 3 + X 2 + X − 1 ∈ Q[X] is irreducible.
(3) Consider X 5 + X 2 + X − 1. Consider the reduction of this modulo 3 and
set X = 0, 1, 2 in turn. The values obtained in Z3 for the polynomial are 2,2,1,
none of which is 0. It follows that X 5 +X 2 +X −1 does not have any linear factor
but we cannot conclude from this that the polynomial is irreducible: there is
still the possibility that it factorises as gh where deg(g) = 3, deg(h) = 2. We can
proceed as follows: we can take g, h to have the forms g = X 3 + aX 2 + bX + c,
h = X 2 + dX + e. Multiply these together and compare coefficients in the
20
equation gh = f . We get the following system of equations: a+d = 0, ad+b = 0,
ae + c + bd = 1, cd + be = 1, ce = −1. The first two equations allow use to get
rid of d and b and the last one lets us write e in terms of c. This leaves two
equations in a and c to be solved. But, since we can work in Z3 we know that
a = 0, 1 or −1 so, trying these in turn we find, in each case, that no value of c
will solve the equations (exercise: do these computations). We conclude that
X 5 + X 2 + X + 1 is, indeed, irreducible in Z3 [X] and hence, by the result above,
is irreducible in Q[X].
Example 3.33. This example shows why, in 3.31, we need the hypothesis that
deg([f ]n ) = deg(f ). Let f = 3X 3 − X 2 − 6X + 2. Is f irreducible?
If we reduce modulo 3, we have [f ]3 = −X 2 + 2. Now this is irreducible:
[f ]3 (0) = 2 6= 0, [f ]3 (1) = −1 + 2 6= 0, [f ]3 (2) = −4 + 2 6= 0 so [f ]3 has no linear
factor and hence (since it has degree just 2) is irreducible. We cannot, however,
conclude from this that f is irreducible.
In fact f = (3X − 1)(X 2 − 2) = gh, say. Reducing modulo 3 we have
[g]3 = −1 - a scalar. So the reducibility of f is “hidden” when we take reducts
modulo 3.
21