# Number Fields

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“God made the integers, all else is the work of man”.-Leopold Kronecker

The positive integers are the natural choice for counting, that is why they are also called natural numbers, and they form the foundation on which other kinds of numbers are built. Even, a fractional quantity can be, in a sense, expressed by an integer; “two-third cake” may be reckoned** **as **two** *thirdcakes*, where *thirdcake* is considered another kind of object different from a cake. However, the set of rational numbers extends the set of integers, every integer is a rational number with 1 as denominator.

## Rings and Fields

The set of integers, usually denoted by **Z**, is not closed under the arithmetic operation of division; except the unit, 1, the integers have no multiplicative inverses that are also integers (think about it). The set of integers form an algebraic structure called a **ring**, we’ve already discussed the properties of a **ring** in a previous article.

Now, a **field** is an algebraic structure which admits division, all the members (except 0) have multiplicative inverses. A field is a ring with the additional property of having multiplicative inverses. A field subsumes a ring just as the set of rationals subsumes the set of integers. Hence, when we mention the set of rationals, it covers both integers and fractions.

Fields are modeled after the arithmetic properties of the set of rational numbers. And the set of rational numbers, usually denoted by **Q** is the prototypical example of a field. We are interested in other examples of fields.

**Field Extensions; The X Factor**

“**X**” is perhaps the most loved (*tongue-in-cheek*) letter of the English alphabet; it has the unenviable job of representing unknown values in mathematical equations and expressions.

The most natural form of algebraic expression is the polynomial (with coefficients from the set of rationals). Consider the general first degree polynomial equation: a**x** + b = 0, where the coefficients **a** and **b** are rational numbers. Solving for **x** gives **x** = (-b/a) which is obviously a rational number. The root of a first degree polynomial equation will always be a rational number as long as the coeffients are rational. But solving for **x** in degree 2 and higher equations uncovers other kinds of numbers outside the familiar set of rational numbers.

Let’s solve a specific example of a second degree (quadratic) equation:

x² - 2 = 0 has the root x= **√**2 which is not a rational number, we can expand or extend our number system by *adjoining *this root to the set of rational numbers(**Q**)* *to get a system of numbers of the form **a + b√2, **where **a** and **b** belong to the set** Q**,** **we denote this new system of numbers by the symbol **Q(√2)**, reads Q adjoin √2. Examples of numbers in this new system include

3 + 5**√2**, 1/2 + 3**√2**, 2/7 + 1/3**√2**, 4 - 11**√2**, etc. We can see that Q(**√**2) subsumes Q, it is a generalization or *extension* of Q; every rational number can be written in the form **a + b√2**, with **b** set equal to 0. Also, very importantly, the elements of Q(**√**2) form a field just like the rationals, it is easy to check that they satisfy the field axioms, for example, every non-zero **a + b√2 **has a multiplicative inverse given by:

So, Q(**√**2) is an extension field of Q, and Q, a subfield of Q(**√**2).

**Degree of a Field Extension**:** **If** **the field Q(**√**2) is considered as a vector space over the rationals(Q), we find that the set {1, **√**2} with just **two** elements forms a basis for this vector space, that is, 1 and **√**2 are linearly independent (one is not a multiple of the other), and, every **a + b√2 **can be expressed as a linear combination of 1 and **√**2, for example, 3 + 5√2 = 3(1) + 5(√2). So the *degree* of the field extension **Q(√2)/Q** is two, which is the number of elements in the basis. Such field extensions of the rationals with a finite degree are called *algebraic number fields*, or *number fields* for short.

In a similar manner, we can adjoin other irrational quadratic roots to Q to get other quadratic field extensions; we can have Q(√3) = {a + b**√3**},

Q(√21) = {a + b**√21**}, etc., all of degree two. Generally, a quadratic field extension is of degree 2 and of the form **Q**(√m), where **m** is some integer, positive or negative, having no perfect square as a factor. If m > 0, we have a *real quadratic field*, if m < 0, we have an *imaginary quadratic field*. The special case **Q**(√-1) or **Q**(i) is a subfield of the complex numbers.

More than one root can be adjoined, like **Q**(√2, √3), this is like a composite adjunction; adjoin √2 to Q to get Q(√2), then adjoin √3 to Q(√2) to get **Q**(√2, √3), or the other way round. So, **Q**(√2, √3) = *a* + *b*√3, where *a* and b belongs to Q(√2), that is, *a* = *d*+ *e*√2, *b* = *f*+ *g*√2 (where *d, e, f, g* are some rational numbers). Substituting these values of *a* and *b* in **Q**(√2, √3) = *a* + *b*√3, we have, **Q**(√2, √3) = *d*+ *e*√2 + *f*√3 + g√6.

This is the smallest extension field of Q that contains both √2 and √3. It has the basis {1, √2, √3, √6} and is therefore a degree four extention over Q. It is an example of a *biquadratic* field.

**EXERCISE** : Find the general representation of numbers in the field extension **Q**(√2, √3, √5). What is the basis? Try to verify that the field axioms are satisfied.

We’ve discussed mainly quadratic field extensions, we can have higher degree field extensions as well, by adjoining roots of higher degree polynomials to Q; cubic fields, quartic fields, quintic fields, sextic fields etc. Generally, if **α** is the root of an irreducible polynomial(cannot be factored into lower degree polynomials) of degree ** n** with coefficients from the rationals, then Q adjoin

**α**is given by:

Specific example - a cubic field extension; adjoin the cube root of 7 to Q, we have Q(∛7) = {a + b∛7 + c∛49}. Where, as usual, a, b, c, are rational numbers. The 49 comes about by raising 7 to the power of 2, i.e 3–1.

Number fields provide a general setting for studying the properties of numbers and make it easier to solve certain number-theoretical problems like diophantine equations. The polynomial, **x² + 1**, ordinarily, is irreducible but working in the extension field Q(√-2), we can factorize it as follows:

x² + 1 = (x + √-2)(x - √-2).

Also, 13 is known as a prime number, but not anymore, if we adjoin √3 to Q. It now has the following factorization; 13 = (4 + √3)(4 - √3).

## Algebraic Integers

Consider again the linear equation **ax + b = 0**, where **a** and **b** are both *integers*, the solution, **x = -b/a**,** **will** **be an integer *only* if **a** divides **b**. We don’t know the definite values of **a** and **b**,** **but if we set **a = 1**, then certainly **x** will be an integer(**-b**). Looking at it another way, if **a** divides **b**, irrespective of the value of **a**, then the original equation can be simplified by dividing all the coefficients by **a** to get **x + c = 0** where **c = b/a**(an integer). So, whichever way we look at it, we arrive at the same result; **x** is *sure* to be an integer *only* if the leading coefficient(**a**) of the equation is 1. This fact is used to define integers in higher degree polynomials as well. *An algebraic integer is the root of a polynomial equation with integer coefficients whose leading coefficient is 1*. Such polynomials are called *monic* polynomials.

The root of the equation **x³ +5x² - 9x - 47 = 0**, whatever it is, is an algebraic integer because the coefficient of the highest power of **x** is 1.

The root of the equation **6x² - 7x + 11 = 0** is not an algebraic integer because the coefficient of the highest power of **x** is 6, not 1.

Is **1 - 5√3** an algebraic integer? To answer that, we’ll construct a polynomial equation with **1 - 5√3 **as a root, and see if the polynomial is monic.

`Set `**x = 1 - 5√3**; simplify as follows:

**x - 1 = –5√3**.

To remove the square root sign from √3, square both sides of the equation:

(x - 1)**² = (**–5√3)**² = 25 * 3**

x² - 2x + 1 = 75.

Subtract 75 from both sides to have 0 on the RHS:

x² - 2x - 74 = 0*(a monic polynomial equation).*

x = 1 – 5√3 is a root of this monic polynomial, it is therefore an algebraic integer.

We are interested in integers because when we talk of factorization or divisibility, it is with respect to integers. **1 - 5√3 **is not an integer in the usual sense, but it is an integer in some quadratic field, it is a *quadratic integer*. Determining the set of integers in various number fields is a major preoccupation of number theorists. This set of integers forms a ring, so we usually talk of the ring of integers of a number field.

For the basic field of rational numbers(**Q**), the ring of integers is just **Z**, the usual integers, also called the *rational integers*. Every extension field of **Q** and its ring of integers will have **Z** already included.

For quadratic fields Q(**√d**), where **d** is a rational integer not divisible by a perfect square, the ring of integers is described as follows:

(i) If **d** leaves a remainder of 2 or 3 when divided by 4, then the ring of integers in Q(**√d**) is **a + b√d**, where **a** and **b** are any rational integers.

E.g., 2 + 7**√**10 is an integer in the extension field Q(**√10**).

(ii) If **d** leaves a remainder of 1 when divided by 4, then the ring of integers in Q(**√d**) is (**a + b√d)/2**, where **a** and **b** are rational integers that are either both odd or both even. E.g., (9 + 13**√**5)/2 is an integer in the extension field Q(**√5**).

## Cyclotomic Fields

**Roots of Unity**. What is the square root of 1? What is the cube root? What is the fourth root? In general, what is the **n**th root? That’s equivalent to asking what values of **x** satisfy the equation **x**ⁿ** **= 1 (or **x**ⁿ - 1 = 0), given n. Obviously, 1 is always a solution. We expect **n** solutions (including complex roots) for an **n**th degree polynomial equation, according to the fundamental theorem of algebra. What are the other roots apart from 1? The theory of complex numbers can help in obtaining all the roots. We start with Euler’s formula,

The sum on the right will become **1** when **x** = 2**π***, *or in general when **x** is a multiple of 2π, i.e., when **x= k2π**,** **where** k** is the multiplying integer, since *cos* and *sin* are both periodic functions with period 2π. So, with **x= 2kπ**, we have;

The **n**th root of **1** is **1** raised to the power of ** 1/n, **so we

**raise both sides of the above equation to the power of**

*1/n;*There we have it, the **n***th* rooth of unity. But, there are supposed to be **n** of them. Yes, use **n** values of **k** from **k**= 0, 1, 2, 3…,to **k**= **n-1** in the formula to get **n** different roots. Anything after that gives a repeated root.

Example. The five fifth roots of unity are(using k = 0, 1, 2, 3, 4):

These are just complex numbers written in exponential form, they could be written in other forms as well. Observe that if we represent the root corresponding to **k = 1** with the greek letter **ζ**, then we can write the five roots as **1**, **ζ**, **ζ²**, **ζ³**, **ζ⁴**. It applies generally, the set of **n** roots of unity for any **n** can be written as

**Cylotomic Fields**. A cyclotomic field is a number field obtained by adjoining a root of unity to the set of rational numbers(Q). That’s why we had to discuss roots of unity first. The word *cyclotomic* is derived from two Greek words “*cyclo*” (**circle**) and “*tomos*” (**cutting**) and refers to how the *n* roots of unity cuts or divides a circle into *n* equal parts when plotted on the complex plane.

If **ζ** is an *n**th* root of unity(with k = 1), **Q** adjoin **ζ** is given by:

That’s the general form of a cyclotomic field extension. The coefficients(**a**₀, **a**₁, **a**₂, etc.) are rational numbers, when they are strictly integers we have the cyclotomic integers.

The theory of cyclotomic fields was developed in the attempt to prove the notorious Fermat’s Last Theorem and it did help in providing a partial proof and in enriching number theory as a whole. We won’t go into the details of the proof here but just state that the factorization of *x*ⁿ* + y*ⁿ *=*** zⁿ **for odd

**n**which is not possible in

**Q**is possible in

**Q(ζ)**:

That is the starting point.

**Further Study**Marcus, Daniel A. (1977).

*Number Fields*. Berlin, New York: Springer-Verlag.