# 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 $$p$$.

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.

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 $$z$$.

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 $$K$$.

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 $$\mathbb{Z}[\omega]$$.

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