10.14 Fields

Fields

Definition and properties

Recall: \(\mathbb{N}\subseteq\mathbb{Z}\subseteq\mathbb{Q}\subseteq\mathbb{R}\subseteq\mathbb{C}\) with \(+,\times\)

Definition of fields

A field \(\mathbb{F}\) is a set with at least \(2\) different elements \(0\) and \(1\), and \(2\) (binary) operations

We denote as \((\mathbb{F},+,\cdot)\)

1. \(+:F\times F\rightarrow F\) sum

2. \(\times :F\times F\rightarrow F\) product

such that satisfies all the following axioms

Properties of fields

1. Associativity: \(x+(y+z)=(x+y)+z\) \(/~~x\cdot(y\cdot z)=\left(x\cdot y\right)\cdot z,\forall x,y,z\in \mathbb{F}\)

2. Commutativity: \(x+y=y+x~~/~~x\cdot y=y\cdot x\)

3. Existence of neutral (additive identity) and multiplicative identity: \(\forall x\in\mathbb{F},\exists0\in\mathbb{F},1\in\mathbb{F},x+0=0+x~,~x\cdot1=1\cdot x,\)

4. Existence of opposites (additive inverse) and multiplicative inverses: \(\forall x\in F,\exists\tilde x\in F\) such that \(x+\tilde x=0=\tilde x+x\)

   \(\forall x\in F,x\neq 0,\exists x^*\) such that \(x\cdot x^*=x^*\cdot x=1\)​

5. Distributivity: \(x\cdot (y+z)=xy+xz,\forall x,y,z\in \mathbb{F}\)​

Notation

\(x+y=+(x,y)/x\cdot y=x\times y=\times(x,y)\)​

Examples

\(\mathbb{Q},\mathbb{R},\mathbb{C}\) are fields

\(\N,\Z\) are non-examples

1. \(\mathbb{F}=\left\lbrace0,1\right\rbrace\) is a field

   \(+\)​	\(0\)​	\(1\)​
   \(0\)​	\(0\)​	\(1\)​
   \(1\)​	\(1\)​	\(0\)​

   \(\times\)​	\(0\)​	\(1\)​
   \(0\)​	\(0\)​	\(0\)​
   \(1\)​	\(0\)​	\(1\)​

   Notation: This is $\mathbb{F_2} $ or \(\Z_2\)(mod \(2\))

2. F={*,#} is a field

   Assume identity element is * and neutral element is #

   Then we have

   \(x+\#=\#+x~,~x\cdot *=*\cdot x\)

   Exists \(x+\tilde{x}=\#=\tilde{x}+x\) and \(x\cdot x^{*}=x^{*}\cdot x=*\)

   \(+\)​	\(*\)​	#
   \(*\)​	#	\(*\)​
   #	\(*\)​	#

   \(\times\)	\(*\)	#
   \(*\)	\(*\)	#
   #	#	#

   ‍

3. \(\mathbb{Z}_3=\left\lbrace0,1,2\right\rbrace\) is a field (modular)

   \(+\)​	\(0\)​	\(1\)​	\(2\)​
   \(0\)​	0	1	2
   \(1\)​	1	2	0
   \(2\)​	2	0	1

   \(\times\)​	\(0\)​	\(1\)​	\(2\)​
   \(0\)​	0	0	0
   \(1\)​	0	1	2
   \(2\)​	0	2	1

   It has the inverse even though it is a part of integers

4. \(\R\times \R\) with coordinatewise \(+\) and \(\times\) is not a field

   $(a,b)+(a',b')=(a+a',b+b') $​

   $(a,b)\times(a',b')=(aa',bb') $

   \((3,2)\times(\frac13,\frac12)=\left(1,1\right)\)​

   \((3,0)\times(?)=( ,\times)\)​

   No!

Arithmetic properties of fields

Remark

The 0 is the unique neutral element and 1 is the unique identity

\(0=0'+0=0'\)

\(1=1'\cdot 1=1'\)

1. The opposite of an \(x\) is unique

   Assume \(x+\tilde x=0\) and \(x+\tilde{\tilde x}=0\)

   So \(x+\tilde x=0=x+\tilde{\tilde x}\), then \(\tilde x+(x+\tilde x)=\tilde x+(x+\tilde{\tilde x})\)

   Then \(0+\tilde x=0+\tilde{\tilde x}\), thus \(\tilde x=\tilde{\tilde x}\)

2. The inverse of a \(x\neq 0\) is unique

   Assume \(x\cdot x^{**}=1\) and \(x\cdot x^{*}=1\)​

   So \(x\cdot x^{**}=\) \(x\cdot x^{*}=1\), then \(x^*\cdot (x\cdot x^{**})=x^*\cdot(x\cdot x^{*})\)​

   Then \(1\cdot x^{*}=1\cdot x^{**}\)​

Notation

The unique opposite of \(x\) is denoted \(-x\)

The unique inverse of \(x(x\neq 0)\) is \(x^{-1}=\frac{1}{x}\)

Proposition

1. Let \(\mathbb{F}\) be a field, the following propositions hold:

   1. \(-0=0,1^{-1}=1\)

      1. Proof

         \(0\left(x\right)+0\left(y\right)=0\Rightarrow y=-x\Rightarrow0=-0\)​ 2. Proof

         \(1\left(x\right)\cdot1\left(y\right)=1\Rightarrow x=y^{-1}\Rightarrow1^{-1}=1\)​ 2. \(x\cdot 0=0,\forall x\in \mathbb{F}\)

      Proof

      \(x\cdot 0=x\cdot (0+0)=x\cdot 0+x\cdot 0\)

      \(\tilde y+y=\tilde y+(y+y)\)

      \(0=y=x\cdot 0\) 3. If \(x+y=x+z\Rightarrow y=z\)

      \(-x+(x+y)=-x+(x+z)\Rightarrow0+y=0+z\Rightarrow y=z\) 4. For every \(x,y\in \mathbb{F}\), \(\exists !z\in \mathbb{F}\) such that \(x+z=y\)

      Proof

      1. Uniqueness

         Suppose exists \(z_1,z_2\) such that \(x+z_1=y,x+z_2=y\)​

         Thus \(x+z_1=x+z_2\), then \((-x)+x+z_1=(-x)+x+z_2\Rightarrow z_1=z_2\)​ 2. ==Existence==

         Let \(z=y-x\), then \(x+z=x+y-x=y\)​ 5. If \(x\cdot y=x\cdot z,x\neq 0\Rightarrow y=z\)

      Proof

      Since \(x\neq 0\) and \(x\cdot y=x\cdot z\), then \(x^{-1}\cdot x\cdot y=x^{-1}\cdot x\cdot z\Rightarrow1\cdot y=1\cdot z\Rightarrow y=z\)​ 6. For every \(x,y\in \mathbb{F},x\neq 0,\exists!z\in \mathbb{F}\) such that \(x\cdot z=y\)

      1. Existence

         Since we are in the field, then there exists \(x\cdot x^{-1}=1\)

         Thus let \(z=\frac{y}{x}\) and \(x\cdot z=x\cdot\frac{y}{x}=y\) 2. Uniqueness

         Suppose exists \(z_1,z_2\), then \(x\cdot z_1=y=x\cdot z_2\)​

         \(x^{-1}\cdot x\cdot z_1=x^{-1}\cdot x\cdot z_2\Rightarrow\left(x^{-1}\cdot x\right)\cdot z_1=\left(x^{-1}\cdot x\right)\cdot z_2\Rightarrow z_1=z_2\)​

2. Let \(\mathbb{F}\) be a field, the following propositions hold:

   1. \(-(-x)=x\)

      Proof

      \(-x+x=0\Rightarrow -(-x)=x\)​ 2. \(-1\cdot x=-x\)​

      \(-1\cdot x+x=x\cdot (-1+1)=x\cdot 0=0\)​

      Then \(-1\cdot x\) is the additive inverse of \(x\).

      Thus \(-1\cdot x=-x\) 3. \(-(x+y)=-x+(-y)\)

      Proof

      \(-(x+y)=-1\cdot (x+y)=-1\cdot x+(-1)\cdot y=-x+(-y)\)​ 4. \(-(x\cdot y)=(-x)\cdot y=x\cdot (-y)\)​

      Proof

      \(-(x\cdot y)=(-1)\cdot x \cdot y=(-x)\cdot y=x\cdot (-y)\)​ 5. \((-x)(-y)=xy\)​

      Proof

      \((-x)(-y)=(-1)\cdot x\cdot (-1)\cdot y=-(-1)\cdot x\cdot y=1\cdot x\cdot y=x\cdot y\)​ 6. \(\left(x^{-1}\right)^{-1}=x\left(x\neq0\right)\)​

      Proof

      Since \(x\cdot x^{-1}=1\), then \(x=\left(x^{-1}\right)^{-1}\) 7. \((xy)^{-1}=x^{-1}\cdot y^{-1}(x,y\neq 0)\)​

      Since \(x\cdot x^{-1}=1\) and \(y\cdot y^{-1}=1\), then \(x\cdot x^{-1}\cdot y\cdot y^{-1}=1\Rightarrow\left(x\cdot y\right)\cdot\left(x^{-1}\cdot y^{-1}\right)=1\Rightarrow\left(xy\right)^{-1}\left(xy\right)\left(x^{-1}\cdot y^{-1}\right)=\left(xy\right)^{-1}\Rightarrow1\cdot\left(x^{-1}\cdot y^{-1}\right)=\left(xy\right)^{-1}\Rightarrow x^{-1}\cdot y^{-1}=\left(xy\right)^{-1}\)​ 8. \(\left(-1\right)^{-1}=-1\)​

      Since we know \((-1)\cdot (-1)=1\), then \(-1\) is the inverse of \((-1)\)​

      Thus \(\left(-1\right)^{-1}=-1\) 9. \((-x)^{-1}=-x^{-1}\left(x\neq0\right)\)​

      Since 8, then \((-x)^{-1}=(-1)^{-1}\cdot x^{-1}\) and by 7 we have \((-1)^{-1}\cdot x^{-1}=-1\cdot x^{-1}=-x^{-1}\)

3. Let \(\mathbb{F}\) be a field and \(x,y\in\mathbb{F}\). Then \(x\cdot y=0\Leftrightarrow x=0\) or \(y=0\)

   \(\Rightarrow\))

   If \(x=0\), we have proved it

   If \(x\neq 0\)​

   Since \(x\cdot y=0\), ==then== \(x^{-1}\cdot x\cdot y=0\cdot x^{-1}\Rightarrow1\cdot y=0\Rightarrow y=0\)

   \(x\) is similar

   \(\Leftarrow\))

   Since \(x=0\), then \(x\cdot y=0\cdot y=0\)

   \(y\) is similar

   Equivalently, \(x,y\neq 0 \Leftrightarrow x\neq 0\) and \(y\neq 0\)​

The complex field

\(\mathbb{C}=\left\lbrace a+bi:a,b\in\mathbb{R}\right\rbrace\) \(\mathbb{R^2}=\left\lbrace\left(a,b\right):a,b\in\mathbb{R}\right\rbrace\)

The \(+\) is defined coordinatewise

Notation

\(a=a+0i\), \(bi=0+bi\), \(i=0+1\cdot i\)

The product is defined by extending the product in \(\R\) and by defining \(i^2=-1\)

\((a+bi)(a'+b'i)=a\cdot a'+a\cdot (b'\cdot i)+(b\cdot i)\cdot a'+(bi)(b'\cdot i)=(aa'-bb')+(ab'+ba')i\)

If \(a+bi\neq 0=0+0i\), \((a+bi)^{-1}=\frac{a-bi}{a^2+b^2}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i\)​

Notation: \(a-b=a+(-b)\)