2.4 Some Consequences of the Field Axioms.

2.59   Theorem (Cancellation laws.) Let $(F,+,\cdot)$ be a field, let $x,y,z$ be elements in $F$, and let $v\in F\backslash \{0\}$. Then

$$x+z=y+z \mbox{$\Longrightarrow$} x=y.$$ (2.60)

$$z+x=z+y \mbox{$\Longrightarrow$} x=y.$$ (2.61)

$$x\cdot v=y\cdot v \mbox{$\Longrightarrow$} x=y.$$ (2.62)

$$v\cdot x=v\cdot y \mbox{$\Longrightarrow$} x=y.$$ (2.63)

(2.60) and (2.61) are called cancellation laws for addition, and (2.62) and (2.63) are called cancellation laws for multiplication.

Proof: All of these results are special cases of the cancellation law for an associative operation (theorem 2.19). $\mid\!\mid\!\mid$

2.64   Theorem. In any field $(F,+,\cdot)$

$$-0=0 \mbox{ and } 1^{-1}=1.$$

Proof: These are special cases of the remark made earlier that an identity element is always invertible, and is its own inverse. $\mid\!\mid\!\mid$

2.65   Theorem (Double inverse theorem.) In any field $(F,+,\cdot)$,

$$\begin{eqnarray*} \mbox{ for all }\; x\in F \qquad\quad &\;&-(-x)=x, \\ \mbox{ for all }\; x\in F\backslash \{0\} \qquad\quad &\;&(x^{-1})^{-1}=x. \end{eqnarray*}$$

Proof: These are special cases of theorem 2.17. $\mid\!\mid\!\mid$

I will now start the practice of calling a field $F$. If I say `` let $F$ be a field" I assume that the operations are denoted by $+$ and $\cdot$.

2.66   Theorem. Let $F$ be a field. Then

$$\mbox{ for all } a\in F,\quad a\cdot 0=0.$$

Proof: We know that $0=0+0$, and hence

$$a\cdot 0=a\cdot(0+0)=a\cdot 0+a\cdot 0.$$

Also, $a\cdot 0+0=a\cdot 0$, so

$$a\cdot 0+0=a\cdot 0+a\cdot 0.$$

By the cancellation law for addition, $0=a\cdot 0$. $\mid\!\mid\!\mid$

2.67   Corollary. Let $F$ be a field. Then for all $a\in F$, $0\cdot a=0$.

2.68   Theorem. Let $F$ be a field. Then for all $x,y$ in $F$

$$\Big(x\cdot y=0 \Big)\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\Big(x=0 \mbox{ or } y=0\Big).$$ (2.69)

Proof:

Case 1:

Suppose $x=0$. Then (2.69) is true because every statement implies a true statement.

Case 2:

Suppose $x\neq 0$. By theorem 2.66, $x\cdot 0 = 0$, so

$$\begin{eqnarray*} x\cdot y=0 &\mbox{$\Longrightarrow$}& x\cdot y = x\cdot 0. \end{eqnarray*}$$

Since $x\neq 0$, we can use the cancellation law for multiplication to get

$$\Big( x\cdot y = x \cdot 0 \Big) \mbox{$\hspace{1ex}\Longrigh... ...ex}\Longrightarrow\hspace{1ex}$}\Big( x=0 \mbox{ or } y=0\Big),$$

and hence

$$\Big(x\cdot y=0 \Big)\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\Big(x=0 \mbox{ or } y=0\Big).$$

Thus (2.69) holds in all cases. $\mid\!\mid\!\mid$

2.70   Remark. We can combine theorem 2.66, corollary 2.67 and theorem 2.68 into the statement: In any field $F$,

$$\mbox{ for all } x,y\in F \qquad x\cdot y=0\hspace{1ex}\Longleftrightarrow\hspace{1ex}(x=0 \mbox{ or } y=0).$$

2.71   Exercise. Let $F$ be a field. Prove that $0$ has no multiplicative inverse in $F$.

2.72   Theorem (Commutativity of addition.) Let $F$ be any field. Then $+$ is a commutative operation on $F$.

Proof: Let $x,y$ be elements in $F$. Then since multiplication is commutative, we have

$$(1+x)\cdot (1+y)=(1+y)\cdot (1+x).$$

By the distributive law,

$$\left( (1+x)\cdot 1\right)+\left((1+x)\cdot y\right)=\left( (1+y)\cdot 1\right)+\left((1+y)\cdot x\right).$$

Since $1$ is the multiplicative identity,

$$\left( 1+x\right)+\left((1+x)\cdot y\right)= \left( 1+y\right)+\left((1+y)\cdot x\right),$$

and hence

$$1+(x+\left((1+x)\cdot y\right) )= 1+(y+\left((1+y)\cdot x\right)).$$

By the cancellation law for addition

$$x+\left((1+x)\cdot y\right)=y+\left((1+y)\cdot x\right).$$

By commutativity of multiplication and the distributive law,

$$x+\left(y\cdot (1+x)\right)=y+\left(x\cdot (1+y)\right)$$

and

$$x+\left((y\cdot 1)+(y\cdot x)\right)=y+\left((x\cdot 1)+(x\cdot y)\right).$$

Since $1$ is the multiplicative identity and addition is associative

$$x+\left(y+(y\cdot x)\right)=y+\left(x+(x\cdot y)\right)$$

and hence

$$(x+y)+(y\cdot x)=(y+x)+(x\cdot y).$$

Since multiplication is commutative

$$(x+y)+(x\cdot y)=(y+x)+(x\cdot y)$$

and by the cancellation law for addition,

$$x+y=y+x.$$

Hence, $+$ is commutative. $\mid\!\mid\!\mid$

2.73   Remark. Let $F$ be a field, and let $x,y\in F$. Then

$$\mbox{ To prove } x=-y, \mbox{ it is sufficient to prove } x+y=0.$$ (2.74)

$$\mbox{ To prove } x=y^{-1},\mbox{ it is sufficient to prove } x\cdot y=1.$$ (2.75)

Proof:

$$\begin{eqnarray*} x+y=0&\mbox{$\Longrightarrow$}&\left((x+y=0)\mbox{ and } (y+x=... ...arrow$}&x=y^{-1}\mbox{ and } y=x^{-1}.\mbox{ $\mid\!\mid\!\mid$} \end{eqnarray*}$$

2.76   Theorem. Let $F$ be a field. Then

$$\mbox{ for all } x,y\in F,\qquad x\cdot(-y)=-(x\cdot y).$$

Proof: Let $x,y\in F$. By (2.74) it is sufficient to prove

$$x\cdot (-y)+x\cdot y=0.$$

Well,

$$\begin{eqnarray*} x\cdot(-y)+x\cdot y &=& x\cdot\left((-y)+y\right) \\ &=&x\cdot 0 \\ &=&0.\mbox{ $\mid\!\mid\!\mid$} \end{eqnarray*}$$

2.77   Exercise. Let $F$ be a field, and let $a,b\in F$. Prove that

1. a) $(-a)\cdot b=-(a\cdot b).$
2. b) $a\cdot (-1) =-a.$
3. c) $(-a)\cdot (-b)=a\cdot b.$

2.78   Exercise. A Let $F$ be a field and let $b,d$ be non-zero elements in $F$. Prove that

$$b^{-1}\cdot d^{-1}=(b\cdot d)^{-1}.$$

2.79   Definition (Digits.) Let $F$ be a field. We define

$$\begin{array}{ll} 2=1+1, &\hspace{1in} 6=5+1, \\ 3=2+1, &\hs... ...1, & \hspace{1in}9=8+1, \\ \; & \hspace{1in}t=9+1. \end{array}$$

I'll call the set

$$\mbox{{\bf D}}_F=\{0,1,2,3,4,5,6,7,8,9\}$$

the set of digits in $F$. If $a,b,c$ are digits, I define

$$ab=t\cdot a+b,$$ (2.80)

and

$$abc=t\cdot (ab)+c.$$ (2.81)

Here $ab$ should not be confused with $a\cdot b$.

2.82   Example.

$$\begin{eqnarray*} 10&=&t\cdot 1+0=t. \\ 100&=&t\cdot 10+0=t\cdot 10=10\cdot 10. \\ 37&=&t\cdot 3+7=10\cdot 3+7. \end{eqnarray*}$$

In general, if $x\in F$, I define

$$x^2=x\cdot x.$$

Then for all digits $a,b,c$

$$\begin{eqnarray*} abc&=&t\cdot(ab)+c=t\cdot(t\cdot a+b)+c=\left(t\cdot(t\cdot a)... ...b\right)+c \\ &=&t^2\cdot a+t\cdot b+c=10^2\cdot a+10\cdot b+c, \end{eqnarray*}$$

so, for example

$$375=10^2\cdot 3+10\cdot 7+5.$$

2.83   Remark. The set $\mbox{{\bf D}}_F$ of digits in $F$ may contain fewer than ten elements. For example, in $(\mbox{{\bf Z}}_5,\oplus_5,\odot_5)$,

$$\begin{eqnarray*} 2&=&1\oplus_5 1=2. \\ 3&=&2\oplus_5 1=3.\\ 4&=&3\oplus_5 1=4. \\ 5&=&4\oplus_5 1=0. \end{eqnarray*}$$

and you can see that $\mbox{{\bf D}}_{\mathbf{Z}_5}=\{0,1,2,3,4\}$.

2.84   Theorem. In any field $F$, $2+2=4$ and $2 \cdot 2 = 4$.

Proof:

$$\begin{array}{lll} 2+2 & = 2+(1+1) & \mbox{ (by definition of... ... of } 3) \\ & = 4 & \mbox{ (by definition of } 4). \end{array}$$

Also,

$$2\cdot 2=2\cdot(1+1)=2\cdot 1+2\cdot 1=2+2=4.\mbox{ $\mid\!\mid\!\mid$}$$

2.85   Exercise. Prove that in any field $F$, $3+3=6$ and $3\cdot 2=6$.

2.86   Exercise. Prove that in any field $F$, $9+8=17$.

2.87   Remark. After doing the previous two exercises, you should believe that the multiplication and addition tables that you learned in elementary school are all theorems that hold in any field, and you should feel free to use them in any field.

2.88   Exercise. Let $F$ be a field and let $x\in F$. Prove that

$$x+x=2\cdot x.$$