Rings

Definition: Ring

A ring $(R,+,\cdot )$ is an algebraic structure whose domain $R$ is equipped with an addition $+:R\times R\to R$ and a multiplication $\cdot :R\times R\to R$ operations which satisfy the following properties.

1. $R$ is an abelian group under addition, i.e.:

• $a+b=b+a$ for all $a,b\in R$

• $(a+b)+c=a+(b+c)$ for all $a,b,c\in R$;

• There exists an additive identity $0_R\in R$ such that $a+0_R=0_R+a=0_R$ for all $a\in R$;

• For each $a\in R$ there exists $-a$ such that $a+(-a)=0_R$.

2. $R$ is a monoid under multiplication, i.e.:

• $(a\cdot b)\cdot c=a\cdot (b\cdot c)$ for all $a,b,c\in R$;
• There exists a multplicative identity $1_R\in R$ such that $a\cdot 1_R=1_R\cdot a=a$ for all $a\in R$.

3. Multiplication is distributive with respect to addition, i.e.

• $a\cdot (b+c)=a\cdot b+a\cdot c$ for all $a,b,c\in R$

• $(a+b)\cdot c=a\cdot c+b\cdot c$ for all $a,b,c\in R$

NOTE

The additive identity is often called “zero” and the multiplicative identity is often called “one”.

NOTATION

When it is clear from context which ring we are talking about, we can write simply $0$ and $1$ for the additive and multiplicative identities, respectively.

Multiplication is often denoted as $ab$ or $a\times b$ instead of $a\cdot b$.

Multiplying an element by itself $n$ times is denoted as

$$a^n=\underset{n\text{ times}}{\underbrace{a\times \cdots \times a}}$$

We also define $a^0=1_R$ for every $a\in R,a\ne 0$.

Properties

Theorem: Uniqueness of the Additive Identity

The additive identity $0_R$ of a ring $R$ is unique.

PROOF

Suppose there were two additive identities $0_R$ and $0_R^{\prime }$.

Since $0_R^{\prime }$ is an additive identity, we have

$$0_R+0_R^{\prime }=0_R$$

Similarly, since $0_R$ is an additive identity, we have

$$0_R^{\prime }+0_R=0_R^{\prime }$$

The commutativity of addition tells us that

$$0_R+0_R^{\prime }=0_R^{\prime }+0_R$$

and so $0_R=0_R^{\prime }$.

Theorem: Uniqueness of the Multiplicative Identity

The multiplicative identity $1_R$ of a ring $R$ is unique.

PROOF

Suppose there were two multiplicative identities $1_R$ and $1_R^{\prime }$.

Since $1_R^{\prime }$ is a multiplicative identity, we have

$$1_R\cdot 1_R^{\prime }=1_R^{\prime }\cdot 1_R=1_R$$

Similarly, since $1_R$ is a multiplicative identity, we have

$$1_R^{\prime }\cdot 1_R=1_R\cdot 1_R^{\prime }=1_R^{\prime }$$

From these two equations we conclude that $1_R=1_R^{\prime }$.

Theorem: Uniqueness of Additive Inverses

The additive inverse $-a$ of each element $a$ in a ring is unique.

PROOF

Let $a$ be a ring element and suppose it has two additive inverses $-a$ and $-a^{\prime }$.

Since $-a$ is an additive inverse, we have

$$a+(-a)=0$$

Since $-a^{\prime }$ is an additive inverse, we have

$$a+(-a^{\prime })=0$$

From the two equations we get

$$a+(-a)=a+(-a^{\prime })$$

Let’s add $-a$ to both sides.

$$a+(-a)+(-a)=a+(-a^{\prime })+(-a)$$

Using associativity on the left side and commutativity on the right side we get

$$0+(-a)=a+(-a)+(-a^{\prime })$$

We again apply associativity, this time to the right side.

$$0+(-a)=0+(-a^{\prime })$$

By the property of the additive identity we get

$$-a=-a^{\prime }$$

Theorem: Multiplication with the Additive Identity

Multiplying any element $a$ of a ring by its additive identity results in the additive identity.

$$a\cdot 0=0\cdot a=0$$

PROOF

The property of the additive identity tells us that $0=0+0$ and so

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

The distributivity of multiplication gives us

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

By the existence of additive inverses we obtain

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

Associativity entails

$$(2)(a\cdot 0+a\cdot 0)+(-(a\cdot 0))=a\cdot 0+(a\cdot 0+(-(a\cdot 0)))$$

Combining $(1)$ and $(2)$ we get

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

The property of the additive identity also tells us that

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

Therefore,

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

The proof for $0\cdot a$ is analogous.

Theorem: Multiplication with the Additive Inverse of the Multiplicative Identity

Multiplying any element $a$ of a ring by the additive inverse of the multiplicative identity results in the additive inverse of $a$.

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

PROOF

TODO