Introduction to Extensions of \( \mathbb Q \)
What makes work with rational numbers and integers comfortable are
the essential properties they have, especially the unique
factorization property (the Main Theorem of Arithmetic). However,
the might of the arithmetic in \( \mathbb{Q} \) is bounded. Thus, some
polynomials, although they have zeros, cannot be factorized into
polynomials with rational coefficients. Nevertheless, such
polynomials can always be factorized in a wider field. For instance,
the polynomial \( x^2+1 \) is irreducible over \( \mathbb{Z} \) or
\( \mathbb{Q} \), but over the ring of the so called Gaussian
integers \( \mathbb{Z}[i]=\{a+bi\mid a,b\in\mathbb{Z}\} \) it can be
factorized as \( (x+i)(x-i) \). Sometimes the wider field retains many
properties of the rational numbers. In particular, it will turn out
that the Gaussian integers are a unique factorization domain, just
like the (rational) integers \( \mathbb{Z} \). We shall first discuss
some basics of higher algebra.
Definition
A number \( \alpha\in\mathbb{C} \) is
algebraic if there is a polynomial \( p(x)=a_nx^n+a_{n-1}
x^{n-1}+\dots+a_0 \) with integer coefficients such that
\( p(\alpha)=0 \). If \( a_n=1 \), then \( \alpha \) is an algebraic
integer.
Further, \( p(x) \) is the minimal polynomial of \( \alpha \) if it is
irreducible over \( \mathbb{Z}[x] \) (i.e. it cannot be written as a
product of nonconstant polynomials with integer coefficients).
Example 1
The number \( i \) is an algebraic integer, as
it is a root of the polynomial \( x^2+1 \) which is also its minimal
polynomial. Number \( \sqrt2+\sqrt3 \) is also an algebraic integer with
the minimal polynomial \( x^4-10x^2+1 \) (verify!).
Example 2
The minimal polynomial of a rational number
\( q=a/b \) (\( a\in\mathbb{Z} \), \( b\in\mathbb{N} \), \( (a,b)=1 \)) is \( bx-a \).
By the definition, \( q \) is an algebraic integer if and only if \( b=1 \),
i.e. if and only if \( q \) is an integer.
Definition
Let \( \alpha \) be an algebraic integer and
\( p(x)=x^n+a_{n-1}x^{n-1}+\dots+a_0 \) (\( a_i\in\mathbb{Z} \)) be its
minimal polynomial. The
extension of a ring \( A \) by the element
\( \alpha \) is the set \( A[\alpha] \) of all complex numbers of the form
\[ c_0+c_1\alpha+\dots+ c_{n-1}\alpha^{n-1}\;\;(c_i\in
A),\quad\quad\quad\quad\quad(\ast)\] with all the operations inherited from \( A \). The
degree of the extension is the degree \( n \) of the polynomial
\( p(x) \).
The theme of this text are extensions of the ring \( \mathbb{Z} \) of
degree 2, so called quadratic extensions. Thus, for example,
the polynomials \( x^2+1 \) and \( x^2+x+1 \) determine the extensions
\( \mathbb{Z}[i] \) and \( \mathbb{Z}[\omega] \), where
\( \omega=\frac{-1+i\sqrt3}2 \) (this notation will be used later).
All elements of a quadratic extension of \( \mathbb{Z} \) are algebraic
integers with the minimal polynomial of second degree. Two elements
having the same minimal polynomials are said to be conjugates.
Each nonrational element \( z \) of the quadratic extension has exactly
one conjugate, called the conjugate of \( z \) and denoted
\( \overline{z} \). For a rational integer \( z \) we define
\( \overline{z}=z \).
Definition
The
norm of an element \( z \) of a
quadratic extension of \( \mathbb{Z} \) is \( N(z)=z\overline{z} \).
The norm is always an integer. Roughly speaking, it is a kind of
equivalent of the absolute value in the set of integers
\( \mathbb{Z} \).
Example 3
If \( z\in\mathbb{Z}[\sqrt{d}] \),
\( z=a+b\sqrt{d} \) (\( a,b\in\mathbb{Z} \)), then
\( \overline{z}=a-b\sqrt{d} \) and \( N(z)=a^2-db^2 \). In particular, in
\( \mathbb{Z}[i] \) the norm of element \( a+bi \) (\( a,b\in\mathbb{N} \)) is
\( N(a+bi)=a^2+b^2 \).
If \( z=a+b\omega\in\mathbb{Z}[\omega] \) (\( a,b\in\mathbb{Z} \)), then
\( \overline{z}=a-b-b\omega \) and \( N(z)=a^2-ab+b^2 \).
In every complex quadratic field the conjugation corresponds to the
complex conjugation.
The following two propositions follow directly from definition.
Theorem 1
The conjugation is multiplicative, i.e. for
arbitrary elements \( z_1,z_2 \) of a quadratic extension of
\( \mathbb{Z} \) it holds that
\( \overline{z_1z_2}=\overline{z_1}\overline{z_2} \). \( \Box \)
Theorem 2
The norm is multiplicative, i.e. for
arbitrary elements \( z_1,z_2 \) of a quadratic extension of
\( \mathbb{Z} \) it holds that \( N(z_1z_2)=N(z_1)N(z_2) \).
An element \( \epsilon\in\mathbb{Z}[\alpha] \) is called a unit if
there exists \( \epsilon^{\prime}\in\mathbb{Z}[\alpha] \) such that
\( \epsilon\epsilon^{\prime}=1 \). In that case
\( N(\epsilon)N(\epsilon^{\prime})=N(1)=1 \), so \( N(\epsilon)=\pm1 \). In fact,
\( \epsilon \) is a unit if and only if its norm is \( \pm1 \): indeed, if
\( N(\epsilon)=\pm1 \) then \( \epsilon\overline{\epsilon}=\pm1 \) by
definition.
Example 4
The only units in \( \mathbb{Z} \) are \( \pm1 \).
Let us find the units in \( \mathbb{Z}[i] \). If \( a+bi \)
(\( a,b\in\mathbb{Z} \)) is a unit, then \( N(a+bi)=a^2+b^2=\pm1 \), which
implies \( a+bi\in\{\pm1,\pm i\} \).
All units in \( \mathbb{Z}[\omega] \) are \( \pm1,\pm\omega,
\pm(1+\omega) \). Indeed, if \( a+b\omega \) is a unit then
\( a^2-ab+b^2=1 \), i.e. \( (2a-b)^2+3b^2=4 \) and he result follows. Note
that \( \omega^2 \) equals \( -(1+\omega) \).
Problem 1
Let \( p \) be a prime number and \( N=\prod
_{k=1}^{p-1}(k^2+1) \). Determine the remainder of \( N \) upon division
by \( p \).
Denote \( P(x)=(1+x)(2+x)\dots(p-1+x) \). We know
that \( P(x)=x^{p-1}-1+pQ(x) \) for some polynomial \( Q(x) \) with integer
coefficients.
Since \( k^2+1=(k+i)(k-i) \) for each \( k \), we immediately obtain that
\[
N=P(i)P(-i)=\left(i^{p-1}-1+pQ(i)\right)
\left((-i)^{p-1}-1+pQ(-i)\right)\] \[
\equiv\left\{\begin{array}{rl}
4,&\mbox{if }p\equiv3\; (\mbox{mod } 4); \\ 0,&\mbox{otherwise. }
\end{array}\right. \]
The divisibility and congruences in an extension \( K \) of the ring
\( \mathbb{Z} \) is defined in the usual way: \( x\in K \) is divisible by
\( y\in K \) (denoted \( y\mid x \)) if there exists \( z\in K \) such that
\( x=yz \), and \( x\equiv y \) (mod \( z \)) if \( z\mid x-y \).
Since every element of a quadratic ring is divisible by every unit,
the definition of the notion of a prime must be adjusted to the new
circumstances.
Definition
An element \( y \) of a quadratic ring \( K \) is
adjoint to element \( x \) (denoted \( y\sim x \)) if there exists a
unit \( \epsilon \) such that \( y=\epsilon x \).
Definition
A nonzero element \( x\in K \) which is not a
unit is
prime if it has no other divisors but the units and
elements adjoint to itself.
We have the following simple proposition.
Theorem 3
Let \( x\in K \). If \( N(x) \) is a prime, then \( x \)
is prime.
Suppose that \( x=yz \), \( y,z\in K \). Then \( N(x)=
N(y)N(z) \), so at least one of \( N(y),N(z) \) equals \( \pm1 \), i.e. either
\( y \) or \( z \) is a unit, while the other is (by definition) adjoint to
\( x \).
The converse does not hold, as \( 3 \) is a prime in \( \mathbb{Z}[i] \),
but \( N(3)=9 \) is composite.
Of course, the elements conjugate or adjoint to a prime are also
primes. Therefore the smallest positive rational integer divisible
by a prime \( z \) equals \( z\overline{z}=N(z) \).
Consider an arbitrary nonzero and nonunit element \( x\in K \). If \( x \)
is not prime then there are nonunit elements \( y,z\in K \) such that
\( yz=x \). Hereby \( N(y)N(z)=N(x) \) and \( N(y),N(z)> 1 \). Hence
\( N(y),N(z)< N(x) \). Continuing this procedure we end up with a
factorization \( x=x_1x_2\cdots x_k \) in which all elements are prime.
This shows that:
Theorem 4
Every nonzero and nonunit \( x\in K \) can be
factorized into primes.
Problem 2
Given a nonzero and nonunit element \( z\in
K \), find the number of equivalence classes in \( K \) modulo \( z \).
Let \( K=\mathbb{Z}[\alpha] \), where \( \alpha^2=
p\alpha+q \), \( p,q\in\mathbb{Z} \), and let \( z=a+b\alpha \) (\( a,b\in
\mathbb{Z} \)). If \( b=0 \) then \( a_1+b_1\alpha\equiv a_2+b_2\alpha \) (mod
\( z \)) if and only if \( a_1\equiv a_2 \) and \( b_1\equiv b_2 \) (mod \( z \)).
Thus there are \( N(z)=z^2 \) equivalence classes.
Now suppose that \( b\neq0 \) and that \( (a,b)=d \). Then \( \alpha z=
(a+pb)\alpha+qb \). Since \( (a+pb,b)=d \), the coefficient at \( \alpha \) in
\( xz \) (\( x\in K \)) can be any integer divisible by \( d \) and no other
integer. Moreover, the smallest natural number divisible by \( z \) is
\( |(a+b\alpha)(\overline{a+b\alpha})|/d=|N(z)|/d \). We conclude that
for every \( x\in K \) there is a unique \( X=A+B\alpha\in K \) with
\( A,B\in\mathbb{Z} \), \( 0\leq A< |N(z)|/d \), \( 0\leq B< d \) such that
\( x\equiv X \) (mod \( z \)). Therefore the required number of equivalence
classes equals \( |N(z)| \).
Naturally, we would like to know when the factorization into primes
is unique, i.e. when the Fundamental Theorem of Arithmetic holds.
But let us first note that, by the above definition, the primes of
\( \mathbb{Z} \) are \( \pm2,\pm3,\pm5 \), etc, so the factorization into
primes is not exactly unique, as e.g. \( 2\cdot3=(-2)(-3) \). Actually,
in this case the uniqueness of factorization is true in the
following wording.
Definition
FTA, or
"The Fundamental Theorem of
Arithmetic" means: Each nonzero and nonunit element of
\( \mathbb{Z} \) or of
its quadratic extension \( K \) can be written as a product of primes.
This factorization is unique up to the order of the factors and
adjoining between corresponding factors.
The division with remainder in a quadratic extension \( K \) of
\( \mathbb{Z} \) can be formulated as follows:
Definition
DWR means:
For each \( a,b\in K \) with \( b\neq0 \) there exist
\( p,q\in K \) such that \( a=pb+q \) and \( N(q)< N(b) \).
Obviously, such a division, if it exists, is not necessarily unique
- it is not so even in \( \mathbb{Z} \) itself. Moreover, it does not
exist in some quadratic extensions, as we shall see later. The
significance of the existence of a division with remainder, however,
lies in the fact that it implies the uniqueness of factorization:
Theorem 5
If the division with remainder in a
quadratic ring \( K \) is always possible, then FTA holds in \( K \).
If the division with remainder is possible in
\( K \), then the Euclidean algorithm ends in a finite number of steps.
A simple implication of the Euclidean algorithm is that if \( p \) is a
prime, \( a,b\in K \) and \( p\mid ab \), then \( p\mid a \) or \( p\mid b \). The
uniqueness of factorization into primes (FTA) now easily follows.
There are quadratic rings in which FTA holds despite the
nonexistence of a division with remainder. However, FTA is an
exception rather than a rule.
Example 5
FTA is false in \( \mathbb{Z}[\sqrt{-5}] \), as
9 has two factorizations into primes:
\( 9=3\cdot3=(2+\sqrt{-5})(2-\sqrt{-5}) \), which are not equivalent
since \( 2\pm\sqrt{-5}\not\sim3 \).
Example 6
The factorizations of the element \( 4-\omega \)
in \( \mathbb{Z}[\omega] \) as
\( (1-\omega)(3+\omega)=(-2-3\omega)(1+2\omega) \) are considered the
same, since \( 1+2\omega=\omega(1-\omega)\sim 1-\omega \) and
\( -2-3\omega=-(1+\omega)(3+\omega)\sim 3+\omega \). We shall show later
that FTA is true in \( \mathbb{Z}[\omega] \).
