2Rings
IB Groups, Rings and Modules
2.5 Factorization in polynomial rings
Since polynomial rings are a bit more special than general integral domains, we
can say a bit more about them.
Recall that for
F
a field, we know
F
[
X
] is a Euclidean domain, hence a
principal ideal domain, hence a unique factorization domain. Therefore we know
(i) If I C F [X], then I = (f) for some f ∈ F [X].
(ii) If f ∈ F [X], then f is irreducible if and only if f is prime.
(iii)
Let
f
be irreducible, and suppose (
f
)
⊆ J ⊆ F
[
X
]. Then
J
= (
g
) for some
g
. Since (
f
)
⊆
(
g
), we must have
f
=
gh
for some
h
. But
f
is irreducible.
So either
g
or
h
is a unit. If
g
is a unit, then (
g
) =
F
[
X
]. If
h
is a unit,
then (
f
) = (
g
). So (
f
) is a maximal ideal. Note that this argument is valid
for any PID, not just polynomial rings.
(iv)
Let (
f
) be a prime ideal. Then
f
is prime. So
f
is irreducible. So (
f
) is
maximal. But we also know in complete generality that maximal ideals are
prime. So in
F
[
X
], prime ideals are the same as maximal ideals. Again,
this is true for all PIDs in general.
(v) Thus f is irreducible if and only if F [X]/(f) is a field.
To use the last item, we can first show that
F
[
X
]
/
(
f
) is a field, and then use this
to deduce that
f
is irreducible. But we can also do something more interesting
— find an irreducible f, and then generate an interesting field F [X]/(f).
So we want to understand reducibility, i.e. we want to know whether we can
factorize a polynomial
f
. Firstly, we want to get rid of the trivial case where we
just factor out a scalar, e.g. 2
X
2
+ 2 = 2(
X
2
+ 1)
∈ Z
[
X
] is a boring factorization.
Definition (Content). Let
R
be a UFD and
f
=
a
0
+
a
1
X
+
···
+
a
n
X
n
∈ R
[
X
].
The content c(f) of f is
c(f) = gcd(a
0
, a
1
, ··· , a
n
) ∈ R.
Again, since the gcd is only defined up to a unit, so is the content.
Definition (Primitive polynomial). A polynomial is primitive if
c
(
f
) is a unit,
i.e. the a
i
are coprime.
Note that this is the best we can do. We cannot ask for
c
(
f
) to be exactly 1,
since the gcd is only well-defined up to a unit.
We now want to prove the following important lemma:
Lemma (Gauss’ lemma). Let
R
be a UFD, and
f ∈ R
[
X
] be a primitive
polynomial. Then
f
is reducible in
R
[
X
] if and only if
f
is reducible
F
[
X
], where
F is the field of fractions of R.
We can’t do this right away. We first need some preparation. Before that,
we do some examples.
Example. Consider
X
3
+
X
+ 1
∈ Z
[
X
]. This has content 1 so is primitive. We
show it is not reducible in Z[X], and hence not reducible in Q[X].
Suppose
f
is reducible in
Q
[
X
]. Then by Gauss’ lemma, this is reducible in
Z[X]. So we can write
X
3
+ X + 1 = gh,
for some polynomials
g, h ∈ Z
[
X
], with
g, h
not units. But if
g
and
h
are not
units, then they cannot be constant, since the coefficients of
X
3
+
X
+ 1 are all
1 or 0. So they have degree at least 1. Since the degrees add up to 3, we wlog
suppose g has degree 1 and h has degree 2. So suppose
g = b
0
+ b
1
X, h = c
0
+ c
1
X + c
2
X
2
.
Multiplying out and equating coefficients, we get
b
0
c
0
= 1
c
2
b
1
= 1
So
b
0
and
b
1
must be
±
1. So
g
is either 1 +
X,
1
− X, −
1 +
X
or
−
1
− X
, and
hence has
±
1 as a root. But this is a contradiction, since
±
1 is not a root of
X
3
+ X + 1. So f is not reducible in Q. In particular, f has no root in Q.
We see the advantage of using Gauss’ lemma — if we worked in
Q
instead,
we could have gotten to the step
b
0
c
0
= 1, and then we can do nothing, since
b
0
and c
0
can be many things if we live in Q.
Now we start working towards proving this.
Lemma. Let R be a UFD. If f, g ∈ R[X] are primitive, then so is f g.
Proof. We let
f = a
0
+ a
1
X + ··· + a
n
X
n
,
g = b
0
+ b
1
X + ··· + b
m
X
m
,
where
a
n
, b
m
6
= 0, and
f, g
are primitive. We want to show that the content of
fg is a unit.
Now suppose
fg
is not primitive. Then
c
(
fg
) is not a unit. Since
R
is a
UFD, we can find an irreducible p which divides c(fg).
By assumption,
c
(
f
) and
c
(
g
) are units. So
p - c
(
f
) and
p - c
(
g
). So suppose
p | a
0
,
p | a
1
, . . . ,
p | a
k−1
but
p - a
k
. Note it is possible that
k
= 0. Similarly,
suppose p | b
0
, p | b
1
, ··· , p | b
`−1
, p - b
`
.
We look at the coefficient of X
k+`
in f g. It is given by
X
i+j=k+`
a
i
b
j
= a
k+`
b
0
+ ··· + a
k+1
b
`−1
+ a
k
b
`
+ a
k−1
b
`+1
+ ··· + a
0
b
`+k
.
By assumption, this is divisible by p. So
p |
X
i+j=k+`
a
i
b
j
.
However, the terms
a
k+`
b
0
+
···
+
a
k+1
b
`−1
, is divisible by
p
, as
p | b
j
for
j < `
.
Similarly,
a
k−1
b
`+1
+
···
+
a
0
b
`+k
is divisible by
p
. So we must have
p | a
k
b
`
.
As
p
is irreducible, and hence prime, we must have
p | a
k
or
p | b
`
. This is a
contradiction. So c(f g) must be a unit.
Corollary. Let
R
be a UFD. Then for
f, g ∈ R
[
X
], we have that
c
(
fg
) is an
associate of c(f)c(g).
Again, we cannot say they are equal, since content is only well-defined up to
a unit.
Proof.
We can write
f
=
c
(
f
)
f
1
and
g
=
c
(
g
)
g
1
, with
f
1
and
g
1
primitive. Then
fg = c(f)c(g)f
1
g
1
.
Since
f
1
g
1
is primitive, so
c
(
f
)
c
(
g
) is a gcd of the coefficients of
fg
, and so is
c(fg), by definition. So they are associates.
Finally, we can prove Gauss’ lemma.
Lemma (Gauss’ lemma). Let
R
be a UFD, and
f ∈ R
[
X
] be a primitive
polynomial. Then
f
is reducible in
R
[
X
] if and only if
f
is reducible
F
[
X
], where
F is the field of fractions of R.
Proof.
We will show that a primitive
f ∈ R
[
X
] is reducible in
R
[
X
] if and only
if f is reducible in F [X].
One direction is almost immediately obvious. Let
f
=
gh
be a product in
R
[
X
] with
g, h
not units. As
f
is primitive, so are
g
and
h
. So both have degree
> 0. So g, h are not units in F [X]. So f is reducible in F [X].
The other direction is less obvious. We let
f
=
gh
in
F
[
X
], with
g, h
not units.
So
g
and
h
have degree
>
0, since
F
is a field. So we can clear denominators
by finding
a, b ∈ R
such that (
ag
)
,
(
bh
)
∈ R
[
X
] (e.g. let
a
be the product of
denominators of coefficients of g). Then we get
abf = (ag)(bh),
and this is a factorization in
R
[
X
]. Here we have to be careful — (
ag
) is one
thing that lives in
R
[
X
], and is not necessarily a product in
R
[
X
], since
g
might
not be in R[X]. So we should just treat it as a single symbol.
We now write
(ag) = c(ag)g
1
,
(bh) = c(bh)h
1
,
where g
1
, h
1
are primitive. So we have
ab = c(abf) = c((ag)(bh)) = u ·c(ag)c(bh),
where u ∈ R is a unit, by the previous corollary. But also we have
abf = c(ag)c(gh)g
1
h
1
= u
−1
abg
1
h
1
.
So cancelling ab gives
f = u
−1
g
1
h
1
∈ R[X].
So f is reducible in R[X].
If this looks fancy and magical, you can try to do this explicitly in the case
where R = Z and F = Q. Then you will probably get enlightened.
We will do another proof performed in a similar manner.
Proposition. Let
R
be a UFD, and
F
be its field of fractions. Let
g ∈ R
[
X
] be
primitive. We let
J = (g) C R[X], I = (g) C F [X].
Then
J = I ∩ R[X].
In other words, if
f ∈ R
[
X
] and we can write it as
f
=
gh
, with
h ∈ F
[
X
], then
in fact h ∈ R[X].
Proof.
The strategy is the same — we clear denominators in the equation
f
=
gh
,
and then use contents to get that down in R[X].
We certainly have J ⊆ I ∩ R[X]. Now let f ∈ I ∩ R[X]. So we can write
f = gh,
with h ∈ F [X]. So we can choose b ∈ R such that bh ∈ R[X]. Then we know
bf = g(bh) ∈ R[X].
We let
(bh) = c(bh)h
1
,
for h
1
∈ R[X] primitive. Thus
bf = c(bh)gh
1
.
Since
g
is primitive, so is
gh
1
. So
c
(
bh
) =
uc
(
bf
) for
u
a unit. But
bf
is really a
product in R[X]. So we have
c(bf) = c(b)c(f) = bc(f).
So we have
bf = ubc(f)gh
1
.
Cancelling b gives
f = g(uc(f)h
1
).
So g | f in R[X]. So f ∈ J.
From this we can get ourselves a large class of UFDs.
Theorem. If R is a UFD, then R[X] is a UFD.
In particular, if R is a UFD, then R[X
1
, ··· , X
n
] is also a UFD.
Proof.
We know
R
[
X
] has a notion of degree. So we will combine this with the
fact that R is a UFD.
Let
f ∈ R
[
X
]. We can write
f
=
c
(
f
)
f
1
, with
f
1
primitive. Firstly, as
R
is a
UFD, we may factor
c(f) = p
1
p
2
···p
n
,
for
p
i
∈ R
irreducible (and also irreducible in
R
[
X
]). Now we want to deal with
f
1
.
If f
1
is not irreducible, then we can write
f
1
= f
2
f
3
,
with
f
2
, f
3
both not units. Since
f
1
is primitive,
f
2
, f
3
also cannot be constants.
So we must have
deg f
2
, deg f
3
>
0. Also, since
deg f
2
+
deg f
3
=
deg f
1
, we must
have
deg f
2
, deg f
3
< deg f
1
. If
f
2
, f
3
are irreducible, then done. Otherwise, keep
on going. We will eventually stop since the degrees have to keep on decreasing.
So we can write it as
f
1
= q
1
···q
m
,
with q
i
irreducible. So we can write
f = p
1
p
2
···p
n
q
1
q
2
···q
m
,
a product of irreducibles.
For uniqueness, we first deal with the p’s. We note that
c(f) = p
1
p
2
···p
n
is a unique factorization of the content, up to reordering and associates, as
R
is
a UFD. So cancelling the content, we only have to show that primitives can be
factored uniquely.
Suppose we have two factorizations
f
1
= q
1
q
2
···q
m
= r
1
r
2
···r
`
.
Note that each
q
i
and each
r
i
is a factor of the primitive polynomial
f
1
, so are
also primitive. Now we do (maybe) the unexpected thing. We let
F
be the
field of fractions of
R
, and consider
q
i
, r
i
∈ F
[
X
]. Since
F
is a field,
F
[
X
] is
a Euclidean domain, hence principal ideal domain, hence unique factorization
domain.
By Gauss’ lemma, since the
q
i
and
r
i
are irreducible in
R
[
X
], they are also
irreducible in
F
[
X
]. As
F
[
X
] is a UFD, we find that
`
=
m
, and after reordering,
r
i
and q
i
are associates, say
r
i
= u
i
q
i
,
with
u
i
∈ F
[
X
] a unit. What we want to say is that
r
i
is a unit times
q
i
in
R
[
X
].
Firstly, note that u
i
∈ F as it is a unit. Clearing denominators, we can write
a
i
r
i
= b
i
q
i
∈ R[X].
Taking contents, since
r
i
, q
i
are primitives, we know
a
i
and
b
i
are associates, say
b
i
= v
i
a
i
,
with
v
i
∈ R
a unit. Cancelling
a
i
on both sides, we know
r
i
=
v
i
q
i
as required.
The key idea is to use Gauss’ lemma to say the reducibility in
R
[
X
] is the
same as reducibility in
F
[
X
], as long as we are primitive. The first part about
contents is just to turn everything into primitives.
Note that the last part of the proof is just our previous proposition. We
could have applied it, but we decide to spell it out in full for clarity.
Example. We know
Z
[
X
] is a UFD, and if
R
is a UFD, then
R
[
X
1
, ··· , X
n
] is
also a UFD.
This is a useful thing to know. In particular, it gives us examples of UFDs
that are not PIDs. However, in such rings, we would also like to have an easy to
determine whether something is reducible. Fortunately, we have the following
criterion:
Proposition (Eisenstein’s criterion). Let R be a UFD, and let
f = a
0
+ a
1
X + ··· + a
n
X
n
∈ R[X]
be primitive with a
n
6= 0. Let p ∈ R be irreducible (hence prime) be such that
(i) p - a
n
;
(ii) p | a
i
for all 0 ≤ i < n;
(iii) p
2
- a
0
.
Then
f
is irreducible in
R
[
X
], and hence in
F
[
X
] (where
F
is the field of fractions
of R).
It is important that we work in
R
[
X
] all the time, until the end where we
apply Gauss’ lemma. Otherwise, we cannot possibly apply Eisenstein’s criterion
since there are no primes in F .
Proof. Suppose we have a factorization f = gh with
g = r
0
+ r
1
X + ··· + r
k
X
k
h = s
0
+ s
1
X + ··· + s
`
X
`
,
for r
k
, s
`
6= 0.
We know
r
k
s
`
=
a
n
. Since
p - a
n
, so
p - r
k
and
p - s
`
. We can also look at
bottom coefficients. We know
r
0
s
0
=
a
0
. We know
p | a
0
and
p
2
- a
0
. So
p
divides exactly one of r
0
and s
0
. wlog, p | r
0
and p - s
0
.
Now let j be such that
p | r
0
, p | r
1
, ··· , p | r
j−1
, p - r
j
.
We now look at a
j
. This is, by definition,
a
j
= r
0
s
j
+ r
1
s
j−1
+ ··· + r
j−1
s
1
+ r
j
s
0
.
We know r
0
, ··· , r
j−1
are all divisible by p. So
p | r
0
s
j
+ r
1
s
j−1
+ ··· + r
j−1
s
1
.
Also, since
p - r
j
and
p - s
0
, we know
p - r
j
s
0
, using the fact that
p
is prime. So
p - a
j
. So we must have j = n.
We also know that
j ≤ k ≤ n
. So we must have
j
=
k
=
n
. So
deg g
=
n
.
Hence
`
=
n − h
= 0. So
h
is a constant. But we also know
f
is primitive. So
h
must be a unit. So this is not a proper factorization.
Example. Consider the polynomial
X
n
− p ∈ Z
[
X
] for
p
a prime. Apply
Eisenstein’s criterion with
p
, and observe all the conditions hold. This is
certainly primitive, since this is monic. So
X
n
− p
is irreducible in
Z
[
X
], hence
in
Q
[
X
]. In particular,
X
n
− p
has no rational roots, i.e.
n
√
p
is irrational (for
n > 1).
Example. Consider a polynomial
f = X
p−1
+ X
p−2
+ ··· + X
2
+ X + 1 ∈ Z[X],
where
p
is a prime number. If we look at this, we notice Eisenstein’s criteria
does not apply. What should we do? We observe that
f =
X
p
− 1
X −1
.
So it might be a good idea to let Y = X −1. Then we get a new polynomial
ˆ
f =
ˆ
f(Y ) =
(Y + 1)
p
− 1
Y
= Y
p−1
+
p
1
Y
p−2
+
p
2
Y
p−3
+ ··· +
p
p − 1
.
When we look at it hard enough, we notice Eisenstein’s criteria can be applied —
we know
p |
p
i
for 1
≤ i ≤ p −
1, but
p
2
-
p
p−1
=
p
. So
ˆ
f
is irreducible in
Z
[
Y
].
Now if we had a factorization
f(X) = g(X)h(X) ∈ Z[X],
then we get
ˆ
f(Y ) = g(Y + 1)h(Y + 1)
in Z[Y ]. So f is irreducible.
Hence none of the roots of
f
are rational (but we already know that — they
are not even real!).