Fix an algebraic closure
of
. For example,
could
be the subfield of the complex numbers
generated by all roots in
of all polynomials with coefficients in
.
Much of this course is about algebraic integers.
Definition 5.1.1 (Algebraic Integer)
An element
![$ \alpha\in\overline{\mathbf{Q}}$](img203.png)
is an
if it is a
root of some monic polynomial with coefficients in
![$ \mathbf{Z}$](img40.png)
.
Definition 5.1.2 (Minimal Polynomial)
The
of
![$ \alpha\in\mathbf{Q}$](img204.png)
is the monic polynomial
![$ f\in\mathbf{Q}[x]$](img205.png)
of least positive degree such that
![$ f(\alpha)=0$](img206.png)
.
The minimal polynomial of
divides any polynomial
such
that
, for the following reason. If
, use
the division algorithm to write
, where
. We have
,
so
is a root of
. However,
is the polynomial of least
positive degree with root
, so
.
Lemma 5.1.3
If
is an algebraic integer, then the minimal polynomial
of
has coefficients in
.
Proof.
Suppose
![$ f\in\mathbf{Q}[x]$](img205.png)
is the minimal polynomial of
![$ \alpha$](img11.png)
and
![$ g\in\mathbf{Z}[x]$](img212.png)
is a monic integral polynomial such that
![$ g(\alpha)=0$](img213.png)
.
As mentioned after the definition of minimal polynomial, we have
![$ g=fh$](img214.png)
, for some
![$ h\in\mathbf{Q}[x]$](img215.png)
. If
![$ f\not\in\mathbf{Z}[x]$](img216.png)
, then some prime
![$ p$](img4.png)
divides the denominator of some coefficient of
![$ f$](img162.png)
. Let
![$ p^i$](img217.png)
be the
largest power of
![$ p$](img4.png)
that divides some denominator of some
coefficient
![$ f$](img162.png)
, and likewise let
![$ p^j$](img218.png)
be the largest power of
![$ p$](img4.png)
that divides some denominator of a coefficient of
![$ g$](img135.png)
. Then
![$ p^{i+j}g
= (p^if)(p^j g)$](img219.png)
, and if we reduce both sides modulo
![$ p$](img4.png)
, then the
left hand side is
0 but the right hand side is a product of two
nonzero polynomials in
![$ \mathbf{F}_p[x]$](img220.png)
, hence nonzero, a contradiction.
Proposition 5.1.4
An element
is integral if and only if
is
finitely generated as a
-module.
Proof.
Suppose
![$ \alpha$](img11.png)
is integral and let
![$ f\in\mathbf{Z}[x]$](img222.png)
be the monic minimal polynomial
of
![$ \alpha$](img11.png)
(that
![$ f\in\mathbf{Z}[x]$](img222.png)
is Lemma
5.1.3). Then
![$ \mathbf{Z}[\alpha]$](img221.png)
is generated by
![$ 1,\alpha,\alpha^2,\ldots,\alpha^{d-1}$](img223.png)
, where
![$ d$](img179.png)
is
the degree of
![$ f$](img162.png)
. Conversely, suppose
![$ \alpha\in\overline{\mathbf{Q}}$](img203.png)
is such that
![$ \mathbf{Z}[\alpha]$](img221.png)
is finitely generated, say by elements
![$ f_1(\alpha), \ldots, f_n(\alpha)$](img224.png)
. Let
![$ d$](img179.png)
be any integer bigger
than the degree of any
![$ f_i$](img189.png)
. Then there exist integers
![$ a_i$](img115.png)
such
that
![$ \alpha^d = \sum a_i f_i(\alpha)$](img225.png)
, hence
![$ \alpha$](img11.png)
satisfies
the monic polynomial
![$ x^d - \sum a_i f_i(x) \in \mathbf{Z}[x]$](img226.png)
, so
![$ \alpha$](img11.png)
is integral.
The rational number
is not integral. Note that
is not a finitely generated
-module, since
is infinite
and
.
Proposition 5.1.5
The set
of all algebraic integers is a ring, i.e., the sum and
product of two algebraic integers is again an algebraic integer.
Proof.
Suppose
![$ \alpha, \beta\in \mathbf{Z}$](img231.png)
, and let
![$ m, n$](img232.png)
be the degrees of the
minimal polynomials of
![$ \alpha, \beta$](img233.png)
, respectively. Then
![$ 1,\alpha,\ldots,\alpha^{m-1}$](img234.png)
span
![$ \mathbf{Z}[\alpha]$](img221.png)
and
![$ 1,\beta,\ldots,\beta^{n-1}$](img235.png)
span
![$ \mathbf{Z}[\beta]$](img236.png)
as
![$ \mathbf{Z}$](img40.png)
-module. Thus
the elements
![$ \alpha^i\beta^j$](img237.png)
for
![$ i \leq m, j\leq n$](img238.png)
span
![$ \mathbf{Z}[\alpha, \beta]$](img239.png)
. Since
![$ \mathbf{Z}[\alpha + \beta]$](img240.png)
is a submodule of the
finitely-generated module
![$ \mathbf{Z}[\alpha, \beta]$](img239.png)
, it is finitely
generated, so
![$ \alpha+\beta$](img241.png)
is integral. Likewise,
![$ \mathbf{Z}[\alpha\beta]$](img242.png)
is a submodule of
![$ \mathbf{Z}[\alpha, \beta]$](img239.png)
, so it is also finitely
generated and
![$ \alpha\beta$](img243.png)
is integral.
Recall that a is a subfield
of
such
that the degree
is finite.
Definition 5.1.6 (Ring of Integers)
The
of a number field
![$ K$](img9.png)
is the ring
![$\displaystyle \O _K = K \cap \overline{\mathbf{Z}}= \{x \in K :$](img245.png)
$x$ is an algebraic integer
The field
of rational numbers is a number field of degree
,
and the ring of integers of
is
. The field
of
Gaussian integers has degree
and
. The field
has ring of integers
.
Note that the Golden ratio
satisfies
.
According to , the ring of integers of
is
, where
.
Definition 5.1.7 (Order)
An
in
![$ \O _K$](img200.png)
is any subring
![$ R$](img91.png)
of
![$ \O _K$](img200.png)
such that the
quotient
![$ \O _K/R$](img256.png)
of abelian groups is finite.
(Note that
![$ R$](img91.png)
must contain
![$ 1$](img147.png)
because it is a ring, and for us
every ring has a
![$ 1$](img147.png)
.)
As noted above,
is the ring of integers of
. For every
nonzero integer
, the subring
of
is an order.
The subring
of
is not an order, because
does not
have finite index in
. Also the subgroup
of
is not an order because it is not a ring.
We will frequently consider orders in practice because they are often
much easier to write down explicitly than
. For example, if
and
is an algebraic integer, then
is an order in
, but frequently
.
Lemma 5.1.8
Let
be the ring of integers of a number field. Then
and
.
Proof.
Suppose
![$ \alpha\in \O _K\cap\mathbf{Q}$](img265.png)
with
![$ \alpha=a/b$](img266.png)
in lowest terms and
![$ b>0$](img267.png)
. The monic minimal polynomial of
![$ \alpha$](img11.png)
is
![$ bx-a\in\mathbf{Z}[x]$](img268.png)
, so
if
![$ b\neq 1$](img269.png)
then Lemma
5.1.3 implies that
![$ \alpha$](img11.png)
is
not an algebraic integer, a contradiction.
To prove that
, suppose
, and let
be the minimal monic polynomial of
. For any
positive integer
, the minimal monic polynomial of
is
, i.e., the polynomial obtained from
by
multiplying the coefficient of
by
, multiplying the
coefficient of
by
, multiplying the coefficient of
by
, etc. If
is the least common multiple of
the denominators of the coefficients of
, then the minimal monic
polynomial of
has integer coefficients, so
is
integral and
. This proves that
.
In the next two sections we will develop some basic properties of
norms and traces, and deduce further properties of rings of integers.
William Stein
2004-05-06