Arithmetic in Extensions of \(\mathbb Q\) (Table of contents)
# Introduction to Extensions of \(\mathbb Q\)

**Definition** 1.
**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** 2. 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
\(p(x)\).
**Definition** 3. The * norm* of an element \(z\) of a
quadratic extension of \(\mathbb{Z}\) is \(N(z)=z\overline{z}\).
**Definition** 4.
**Theorem** 1.
**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 \(N(z_1z_2)=N(z_1)N(z_2)\).
**Example** 3.
**Problem** 1.
**Definition** 5. 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** 6. 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.
**Theorem** 3.
Let \(x\in K\). If \(N(x)\) is a prime, then \(x\)
is prime.
**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 \(z\).
**Definition** 7. 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.
**Definition** 8. DWR means:
For each \(a,b\in K\) with \(b\neq0\) there exist
\(p,q\in K\) such that \(a=pb+q\) and \(N(q) < N(b)\).
**Theorem** 5. If the division with remainder in a
quadratic ring \(K\) is always possible, then FTA holds in \(K\).
**Example** 4. 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** 5. 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 \(\mathbb{Z}[\omega]\).

What makes work with rational numbers and integers comfortable are
the essential properties they have, especially the unique
factorization property (the Fundamental 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* \(\mathbb Z [i]=\{a+b i \mid a,b\in\mathbb Z \}\) 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.

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).

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
\(\overline{z}=z\).

The norm is always an integer. Roughly speaking, it is a kind of equivalent of the absolute value in the set of integers \(\mathbb{Z}\).

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.

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.

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.

The only units in \(\mathbb{Z}\) are \(\pm1\).

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 \(-(1+\omega)\).

Let \(p\) be a prime number and \( N=\prod _{k=1}^{p-1}(k^2+1)\). Determine the remainder of \(N\) upon division by \(p\).

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.

We have the following simple proposition.

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. Therefore, we have

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.

The division with remainder in a quadratic extension \(K\) of \(\mathbb{Z}\) can be formulated as follows:

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:

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.