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: [math]\displaystyle{ 0\cdot a=a\cdot 0=0 }[/math]

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[math]\displaystyle{ 0\cdot a=(0+0)\cdot a=(0\cdot a)+(0\cdot a) }[/math]

By subtracting (i.e. adding the additive inverse of) [math]\displaystyle{ 0\cdot a }[/math] on both sides of the equation, we get the desired result. The proof that [math]\displaystyle{ a\cdot 0=0 }[/math] is similar.

Unique identity element per binary operation

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

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If there is another identity element [math]\displaystyle{ e' }[/math] for the binary operation, then [math]\displaystyle{ e'a = ae' = a }[/math], and when [math]\displaystyle{ a = e }[/math], [math]\displaystyle{ e'e = ee' = e = e' }[/math] where [math]\displaystyle{ ab }[/math] is the binary operation on ring elements [math]\displaystyle{ a }[/math] and [math]\displaystyle{ b }[/math].

Unique additive inverse element

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

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If there is another inverse element [math]\displaystyle{ - a' }[/math] for [math]\displaystyle{ a }[/math], then [math]\displaystyle{ - a = - a + 0 = - a + a - a' = 0 - a' = - a' }[/math].

Unique multiplicative inverse element

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

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If there is another inverse element [math]\displaystyle{ a^{-1'} }[/math] for [math]\displaystyle{ a }[/math], then [math]\displaystyle{ a^{-1} = a^{-1} \times 1 = a^{-1} \times a \times a^{-1'} = 1 \times a^{-1'} = a^{-1'} }[/math].

Zero ring

Theorem: A ring [math]\displaystyle{ (R, +, \cdot) }[/math] is the zero ring (that is, consists of precisely one element) if and only if [math]\displaystyle{ 0 = 1 }[/math].

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Suppose that [math]\displaystyle{ 1 = 0 }[/math]. Let [math]\displaystyle{ a }[/math] be any element in [math]\displaystyle{ R }[/math]; then [math]\displaystyle{ a = a \cdot 1 = a \cdot 0 = 0 }[/math]. Therefore, [math]\displaystyle{ (R, +, \cdot) }[/math] is the zero ring. Conversely, if [math]\displaystyle{ (R, +, \cdot) }[/math] is the zero ring, it must contain precisely one element by its definition. Therefore, [math]\displaystyle{ 0 }[/math] and [math]\displaystyle{ 1 }[/math] is the same element, i.e. [math]\displaystyle{ 0=1 }[/math].

Multiplication by negative one

Theorem: [math]\displaystyle{ (-1)a=-a }[/math]

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[math]\displaystyle{ (-1)\cdot a+a=(-1)\cdot a+1\cdot a=((-1)+1)\cdot a=0\cdot a=0 }[/math]

Therefore [math]\displaystyle{ (-1)\cdot a = (-1)\cdot a + 0 = (-1) \cdot a + (a + (-a))= ((-1) \cdot a + a) + (-a) = 0 + (-a) = (-a) }[/math].

Multiplication by additive inverse

Theorem: [math]\displaystyle{ (-a) \cdot b= a \cdot (-b) = -(ab) }[/math]

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To prove that the first expression equals the second one, [math]\displaystyle{ (-a)\cdot b = ((-1) \cdot a) \cdot b = (a \cdot (-1)) \cdot b = a\cdot ((-1) \cdot b) = a(-b). }[/math]

To prove that the first expression equals the third one, [math]\displaystyle{ (-a) \cdot b = ((-1) \cdot a) \cdot b = (-1) \cdot (a \cdot b). }[/math]

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 [math]\displaystyle{ (-a) \cdot b }[/math] is indeed the inverse of [math]\displaystyle{ (a \cdot b) }[/math] by showing that adding them up results in the additive identity element,

[math]\displaystyle{ (a \cdot b) + (-a) \cdot b = (a - a) \cdot b = 0 \cdot b = 0 }[/math].