Proofs of elementary ring properties

The following proofs of elementary ring properties use only the axioms that define a mathematical ring:

Basics

Multiplication by zero

Theorem: $0\cdot a=a\cdot 0=0$

(Click "show" at right to see the proof of this theorem or "hide" to hide it.)

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

By subtracting (i.e. adding the additive inverse of)

0\cdot a

on both sides of the equation, we get the desired result. The proof that

a\cdot 0=0

is similar.

Unique identity element per binary operation

Theorem: The identity element e for a binary operation (addition or multiplication) of a ring is unique.

(Click "show" at right to see the proof of this theorem or "hide" to hide it.)

If there is another identity element

e'

for the binary operation, then

e'a=ae'=a

, and when

a=e

,

e'e=ee'=e=e'

where

ab

is the binary operation on ring elements

a

and

b

.

Unique additive inverse element

Theorem: - a as the additive inverse element for a is unique.

(Click "show" at right to see the proof of this theorem or "hide" to hide it.)

If there is another inverse element

-a'

for

a

, then

-a=-a+0=-a+a-a'=0-a'=-a'

.

Unique multiplicative inverse element

Theorem: a⁻¹ as the multiplicative inverse element for a is unique.

(Click "show" at right to see the proof of this theorem or "hide" to hide it.)

If there is another inverse element

a^{-1'}

for

a

, then

a^{-1}=a^{-1}\times 1=a^{-1}\times a\times a^{-1'}=1\times a^{-1'}=a^{-1'}

.

Zero ring

Theorem: A ring $(R,+,\cdot )$ is the zero ring (that is, consists of precisely one element) if and only if $0=1$ .

(Click "show" at right to see the proof of this theorem or "hide" to hide it.)

Suppose that

1=0

. Let

a

be any element in

R

; then

a=a\cdot 1=a\cdot 0=0

. Therefore,

(R,+,\cdot )

is the zero ring. Conversely, if

(R,+,\cdot )

is the zero ring, it must contain precisely one element by its definition. Therefore,

0

and

1

is the same element, i.e.

0=1

.

Multiplication by negative one

Theorem: $(-1)a=-a$

(Click "show" at right to see the proof of this theorem or "hide" to hide it.)

$(-1)\cdot a+a=(-1)\cdot a+1\cdot a=((-1)+1)\cdot a=0\cdot a=0$

Therefore

(-1)\cdot a=(-1)\cdot a+0=(-1)\cdot a+(a+(-a))=((-1)\cdot a+a)+(-a)=0+(-a)=(-a)

.

Multiplication by additive inverse

Theorem: $(-a)\cdot b=a\cdot (-b)=-(ab)$

(Click "show" at right to see the proof of this theorem or "hide" to hide it.)

To prove that the first expression equals the second one, $(-a)\cdot b=((-1)\cdot a)\cdot b=(a\cdot (-1))\cdot b=a\cdot ((-1)\cdot b)=a(-b).$

To prove that the first expression equals the third one, $(-a)\cdot b=((-1)\cdot a)\cdot b=(-1)\cdot (a\cdot b).$

A pseudo-ring does not necessarily have a multiplicative identity element. To prove that the first expression equals the third one without assuming the existence of a multiplicative identity, we show that $(-a)\cdot b$ is indeed the inverse of $(a\cdot b)$ by showing that adding them up results in the additive identity element,

(a\cdot b)+(-a)\cdot b=(a-a)\cdot b=0\cdot b=0

.

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