2. Fields

Definition

A set K is a field if it is equipped with two two-argument operations + and \cdot called addition and multiplication and two selected elements 0,1\in K, satisfying the following conditions for every a,b,c\in K:

  1. addition is associative: (a+b)+c=a+(b+c),
  2. multiplication is associative: (a\cdot b)\cdot c=a\cdot (b\cdot c),
  3. addition is commutative: a+b=b+a,
  4. multiplication is commutative: a\cdot b=b\cdot a,
  5. 0 is the neutral element for addition: 0+a=a,
  6. 1 is the neutral element for multiplication: 1\cdot a=a,
  7. there are negations: there exists x\in K, such that a+x=0, such x is denoted by -a,
  8. there are reciprocals: ifa\neq 0, there exists x\in K, such that a\cdot x=1, such x is denoted by a^{-1} or 1/a,
  9. these operations are distributive: (a+b)\cdot c=a\cdot c+b\cdot c.

It is worth noticing that there exists exactly one negative for a i and exactly one reciprocal for a (if a\neq 0). Indeed, if a+b=a+c=0, then b=0+b=a+c+b=a+b+c=0+c=c.

Examples

Examples: real numbers \mathbb{R} with standard operations and standard 0 and 1, similarly rational numbers \mathbb{Q}.

Those examples are infinite fields, but there also exist finite fields. The simplest example of a finite field is \mathbb{Z}_p, for a prime number p. \mathbb{Z}_p=\{0,1,\ldots,p-1\}, and addition and multiplications are taken modulo p. E.g. in \mathbb{Z}_2 the operation work as follows:

    \[\begin{array}{c|cc} +&0&1\\ \hline 0& 0 &1\\ 1&1& 0\end{array}\]

    \[\begin{array}{c|cc} \cdot&0&1\\ \hline 0& 0 &0\\ 1&0& 1\end{array}\]

Subfields

L\subseteq K is a subfield of a field K, if for every a,b\in L, a+b, a\cdot b, -a \in L, and for a\neq 0, 1/a\in L, and moreover 0,1\in L. Obviously a subfield of a field is a field.

E.g., \mathbb{Q} is a subfield of \mathbb{R}.

Systems of linear equations over a given field

We can also solve systems of linear equations with coefficients and variables in a given field. We do this using the same method which was described for real numbers, i.e. transforming a matrix into reduced echelon form, but obviously calculating everything in the given field. E.g. to solve in \mathbb{Z}_{13} the system

    \[\begin{cases}x_{1}+3x_{2}+x_{3}+5x_{4}=2\\ 2x_{1}+7x_{2}+9x_{3}+2x_{4}=4\\ 4x_{1}+11x_{3}+12x_{4}=8\end{cases}\]

we get:

    \[\left[\begin{array}{cccc|c}1 & 3 & 1 & 5 & 2\\ 2 & 7 & 9 & 2 & 4\\ 4 & 0 & 11 & 12 & 8\end{array}\right]\]

    \[\underrightarrow{w_{2}-2w_{1}, w_{3}-4w_{1}} \left[\begin{array}{cccc|c}1 & 3 & 1 & 5 & 2\\ 0 & 1 & 7 & 5 & 0\\ 0 & 1 & 7 & 5 & 0\end{array}\right]\]

    \[\underrightarrow{w_{3}-w_{2}} \left[\begin{array}{cccc|c}1 & 3 & 1 & 5 & 2\\ 0 & 1 & 7 & 5 & 0\\ 0 & 0 & 0 & 0 & 0\end{array}\right]\]

    \[\underrightarrow{w_{1}-3w_{2}} \left[\begin{array}{cccc|c}1 & 0 & 6 & 3 & 2\\ 0 & 1 & 7 & 5 & 0\\ 0 & 0 & 0 & 0 & 0\end{array}\right],\]

So the general solution takes the following form:

    \[\begin{cases}x_{1}=2+7x_{3}+10 x_{4}\\ x_{2}=6x_{3}+8x_{4} \end{cases}\]

which gives \{(2+7x_{3}+10x_{4}, 6x_{3}+8x_{4}, x_{3},x_{4})\colon x_3,x_4\in\mathbb{Z}_{13}\}. Substituting, for example, x_3=1,x_4=0, we get and exemplary solution (9,6,1,0) — and it is easy to check that those numbers satisfy the equations from the system (again in \mathbb{Z}_{13}).

Extensions by an element

Given a subfield L of a field K and an element a\in K\setminus L, we will denote by L(a) the least field containing L\cup{a\}. Notice that if a is a root of a polynomial f\in L[x], then L(a)=L[a] (indeed we can assume that 0 is not a root of f, and then f=g(a-x) for some g\in L[a][x], so a\cdot \frac{g(0)}{f(0)}= 1, and f(0)\in L, so 1/a =g(0)/f(0)\in L[a]). In particular, if a^2\in L, then L(a)=\{xa+y\colon x,y\in L\}.