Properties of Rings

We have listed 8 axioms in the definition of rings. These axioms are designed to be as concise and condense as possible. Many of facts we used to have in the elementary arithmetics still hold, and can be derived from these axioms.

We will first list some basic properties that can be derived solely from the axioms of rings. These properties are so fundamental that we will use them without mentioning to the proof in the future.

Basic Properties

Here is the list of properties we can prove from the axioms of rings:

1. Additive Identity is Unique: There is only one additive identity $0_R$ in a ring.
2. Additive Inverse is Unique: For all $a\in R$, the additive inverse $-a$ is unique.
3. Multiply by Zero: For all $a\in R$, $0_R\cdot a=a\cdot 0_R=0_R$.
4. Multiply by Negative: For all $a,b\in R$, $(-a)\cdot b=a\cdot (-b)=-(a\cdot b)$.
5. Double Negative: For all $a\in R$, $-(-a)=a$.
6. Negative of Sum: For all $a,b\in R$, $-(a+b)=-a+(-b)$.
7. Product of Negatives: For all $a,b\in R$, $(-a)\cdot (-b)=a\cdot b$.

We will prove each of these properties, before that, assuming the Uniqueness of Additive Inverse, we can define the Subtraction operation in a ring:

Definition:

For all $a,b\in R$, the subtraction $a-b$ is defined as $a+(-b)$.

Combined the Negative of Sum and the definition of subtraction, we have the following property:

Negative of Difference: For all $a,b\in R$, $-(a-b)=-a+b$.

We can restate the Negative of Sum as: $-(a+b)=-a-b$.

Proof of Basic Properties

Additive Identity is Unique

Suppose $0_R$ and $0_R^{\prime }$ are both additive identities. Then we have:

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

Additive Inverse is Unique

Suppose $a$ has two additive inverses $b$ and $b^{\prime }$. Then we have:

$$b=b+0_R=b+(a+b^{\prime })=(b+a)+b^{\prime }=0_R+b^{\prime }=b^{\prime }.$$

Multiply by Zero

We have:

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

Adding $-(0_R\cdot a)$ to both sides, we have:

$$0_R=0_R\cdot a.$$

Multiply by Negative

We have:

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

This implies $(-a)\cdot a=-(a\cdot a)$.

Double Negative

We have:

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

This implies $-(-a)=a$.

Negative of Sum

We have:

$$\begin{array}{rl}(a+b)+(-a+(-b))& =a+b+(-a)+(-b)\\ & =a+(-a)+b+(-b)\\ & =0_R+0_R=0_R.\end{array}$$

This implies $-(a+b)=-a+(-b)$.

Product of Negatives

We have:

$$\begin{array}{rl}-ab+(-a)(-b)=(-a)b+(-a)(-b)& =(-a)(b+(-b))\\ & =(-a)\cdot 0_R=0_R.\end{array}$$

Here we used the Multiply by Zero property. The identity we established above implies $(-a)(-b)=ab$.

Properties of Integral Domains

Integral domains are special rings, we list the following properties that are specific to integral domains:

Cancellation Law: For all $a,b,c\in R$, if $a\ne 0_R$ and $ab=ac$, then $b=c$.

Proof of Cancellation Law

Suppose $a\ne 0_R$ and $ab=ac$. Then we have:

$$0_R=ab-ac=a(b-c).$$

Since $a\ne 0_R$, we have $b-c=0_R$, which implies $b=c$.

Remark

One may expect cancellation law to hold when we have $a^{-1}$ available, which is the case in fields. The above proof tells us that cancellation law holds in integral domains even without the existence of multiplicative inverses.

Use the cancellation law, we can prove the classification of finite integral domains:

Theorem

If $R$ is a finite integral domain, then $R$ is a field.

Proof of Theorem

We need to show for all $a\in R$, $a\ne 0_R$, there exists a $x\in R$ such that $ax=1_R$. Since $R$ is finite, we can list the elements of $R$ as $a_1,a_2,\dots ,a_n$. Consider the set:

$$\{a\cdot a_1,a\cdot a_2,\dots ,a\cdot a_n\}.$$

Clearly it is a subset of $R$, we will show it is the entire $R$. Suppose there is an element $a\cdot a_i=a\cdot a_j$ for some $i\ne j$. Then we have $a_i=a_j$ by the cancellation law. This tells us the set contains $n$ distinct elements, so it must be the entire $R$. In particular, there exists a $a_i$ such that $a\cdot a_i=1_R$.