2Rings
IB Groups, Rings and Modules
2.7 Algebraic integers
We generalize the idea of Gaussian integers to algebraic integers.
Definition (Algebraic integer). An
α ∈ C
is called an algebraic integer if it is
a root of a monic polynomial in
Z
[
X
], i.e. there is a monic
f ∈ Z
[
X
] such that
f(α) = 0.
We can immediately check that this is a sensible definition — not all complex
numbers are algebraic integers, since there are only countably many polynomials
with integer coefficients, hence only countably many algebraic integers, but there
are uncountably many complex numbers.
Notation. For
α
an algebraic integer, we write
Z
[
α
]
≤ C
for the smallest subring
containing α.
This can also be defined for arbitrary complex numbers, but it is less inter-
esting.
We can also construct
Z
[
α
] by taking it as the image of the map
φ
:
Z
[
X
]
→ C
given by g 7→ g(α). So we can also write
Z[α] =
Z[X]
I
, I = ker φ.
Note that
I
is non-empty, since, say,
f ∈ I
, by definition of an algebraic integer.
Proposition. Let α ∈ C be an algebraic integer. Then the ideal
I = ker(φ : Z[X] → C, f 7→ f(α))
is principal, and equal to (f
α
) for some irreducible monic f
α
.
This is a non-trivial theorem, since
Z
[
X
] is not a principal ideal domain. So
there is no immediate guarantee that I is generated by one polynomial.
Definition (Minimal polynomial). Let
α ∈ C
be an algebraic integer. Then
the minimal polynomial is a polynomial
f
α
is the irreducible monic such that
I = ker(φ) = (f
α
).
Proof.
By definition, there is a monic
f ∈ Z
[
X
] such that
f
(
a
) = 0. So
f ∈ I
.
So
I 6
= 0. Now let
f
α
∈ I
be such a polynomial of minimal degree. We may
suppose that
f
α
is primitive. We want to show that
I
= (
f
α
), and that
f
α
is
irreducible.
Let
h ∈ I
. We pretend we are living in
Q
[
X
]. Then we have the Euclidean
algorithm. So we can write
h = f
α
q + r,
with
r
= 0 or
deg r < deg f
α
. This was done over
Q
[
X
], not
Z
[
X
]. We now clear
denominators. We multiply by some a ∈ Z to get
ah = f
α
(aq) + (ar),
where now (
aq
)
,
(
ar
)
∈ Z
[
X
]. We now evaluate these polynomials at
α
. Then
we have
ah(α) = f
α
(α)aq(α) + ar(α).
We know
f
α
(
α
) =
h
(
α
) = 0, since
f
α
and
h
are both in
I
. So
ar
(
α
) = 0. So
(
ar
)
∈ I
. As
f
α
∈ I
has minimal degree, we cannot have
deg
(
r
) =
deg
(
ar
)
<
deg(f
a
). So we must have r = 0.
Hence we know
ah = f
α
· (aq)
is a factorization in
Z
[
X
]. This is almost right, but we want to factor
h
, not
ah
.
Again, taking contents of everything, we get
ac(h) = c(ah) = c(f
α
(aq)) = c(aq),
as
f
α
is primitive. In particular,
a | c
(
aq
). This, by definition of content, means
(
aq
) can be written as
a¯q
, where
¯q ∈ Z
[
X
]. Cancelling, we get
q
=
¯q ∈ Z
[
X
]. So
we know
h = f
α
q ∈ (f
α
).
So we know I = (f
α
).
To show f
α
is irreducible, note that
Z[X]
(f
α
)
∼
=
Z[X]
ker φ
∼
=
im(φ) = Z[α] ≤ C.
Since
C
is an integral domain, so is
im
(
φ
). So we know
Z
[
X
]
/
(
f
α
) is an integral
domain. So (f
α
) is prime. So f
α
is prime, hence irreducible.
If this final line looks magical, we can unravel this proof as follows: suppose
f
α
=
pq
for some non-units
pq
. Then since
f
α
(
α
) = 0, we know
p
(
α
)
q
(
α
) = 0.
Since
p
(
α
)
, q
(
α
)
∈ C
, which is an integral domain, we must have, say,
p
(
α
) = 0.
But then deg p < deg f
α
, so p 6∈ I = (f
α
). Contradiction.
Example.
(i) We know α = i is an algebraic integer with f
α
= X
2
+ 1.
(ii) Also, α =
√
2 is an algebraic integer with f
α
= X
2
− 2.
(iii)
More interestingly,
α
=
1
2
(1 +
√
−3
) is an algebraic integer with
f
α
=
X
2
− X − 1.
(iv)
The polynomial
X
5
− X
+
d ∈ Z
[
X
] with
d ∈ Z
≥0
has precisely one real
root
α
, which is an algebraic integer. It is a theorem, which will be proved
in IID Galois Theory, that this
α
cannot be constructed from integers
via +
, −, ×, ÷,
n
√
·
. It is also a theorem, found in IID Galois Theory, that
degree 5 polynomials are the smallest degree for which this can happen (the
prove involves writing down formulas analogous to the quadratic formula
for degree 3 and 4 polynomials).
Lemma. Let α ∈ Q be an algebraic integer. Then α ∈ Z.
Proof.
Let
f
α
∈ Z
[
X
] be the minimal polynomial, which is irreducible. In
Q
[
X
],
the polynomial
X − α
must divide
f
α
. However, by Gauss’ lemma, we know
f ∈ Q
[
X
] is irreducible. So we must have
f
α
=
X − α ∈ Z
[
X
]. So
α
is an
integer.
It turns out the collection of all algebraic integers form a subring of
C
. This
is not at all obvious — given
f, g ∈ Z
[
X
] monic such that
f
(
α
) =
g
(
α
) = 0, there
is no easy way to find a new monic
h
such that
h
(
α
+
β
) = 0. We will prove this
much later on in the course.