Ring/Linear algebra/Introduction/Section

Definition

A set ${}R$ is called a ring if there are two binary operations (called addition and multiplication)

$+:R\times R\longrightarrow R{\text{ and }}\cdot :R\times R\longrightarrow R$

and two elements ${}0,1\in R$ that fulfill the following properties.

1. Axioms for the addition:
   1. Associative law: ${}(a+b)+c=a+(b+c)$ holds for all ${}a,b,c\in R$.
   2. Commutative law: ${}a+b=b+a$ holds for all ${}a,b\in R$.
   3. ${}0$ is the neutral element of the addition, i.e., ${}a+0=a$ holds for all ${}a\in R$.
   4. Existence of the negative: For every ${}a\in R$, there exists an element ${}b\in R$ with ${}a+b=0$.
2. Axioms of the multiplication:
   1. Associative law: ${}(a\cdot b)\cdot c=a\cdot (b\cdot c)$ holds for all ${}a,b,c\in R$.
   2. ${}1$ is the neutral element for the multiplication, i.e., ${}a\cdot 1=a$ holds for all ${}a\in R$.
3. Distributive law: ${}a\cdot (b+c)=(a\cdot b)+(a\cdot c)$ holds for all ${}a,b,c\in R$.

Definition

A ring is called commutative

if its multiplication is commutative.

For us, the most important commutative rings are the set of integer numbers ${}\mathbb {Z}$, the rational numbers ${}\mathbb {Q}$, and the real numbers ${}\mathbb {R}$. The real numbers (and the rational numbers) with their natural operations fulfill all these axioms, as should be known from school. An axiomatic reasoning is possible, but we will not do this here. With its addition, a ring ${}(R,0,+)$ (forgetting the multiplicative structure) is in particular a commutative group.

In a ring, we use the convention that multiplication ties stronger than addition. Therefore, we write ${}a\cdot b+c\cdot d$ instead of ${}(a\cdot b)+(c\cdot d)$. To simplify further the notation, we omit the product symbol. The special elements ${}0$ and ${}1$ in a ring are called the null element and the unit. For ${}a\in R$, we call the (according to fact) uniquely determined element ${}y$ fulfilling ${}a+y=0$ the negative of ${}a$ and denote it by ${}-a$. We have ${}-(-a)=a$, since ${}a+(-a)=0$ shows that the element ${}a$ equals the uniquely determined negative of ${}-a$. We write ${}b-a$ instead of ${}b+(-a)$ and call this a difference. Hence, the difference is not a basic operation but is defined as the addition with the negative element.

The following properties are familiar for the real numbers; we prove them using only the axioms of a ring. So they hold for any ring.

Lemma

Let ${}R$ be a ring,

and let ${}a,b,c,a_{1},\ldots ,a_{r},b_{1},\ldots ,b_{s}$ denote elements from ${}R$. Then the following statements hold.

1. ${}a0=0$ (annihilation rule).

2. ${}(-a)b=-ab=a(-b)\,$

   (rules for sign).

3. ${}(-a)(-b)=ab\,.$

4. ${}a(b-c)=ab-ac\,.$

5. ${}{\left(\sum _{i=1}^{r}a_{i}\right)}{\left(\sum _{k=1}^{s}b_{k}\right)}=\sum _{1\leq i\leq r,\,1\leq k\leq s}a_{i}b_{k}$ (general law of distributivity).

Proof

In the noncommutative case, we only proof one half of the statements.

1. We have ${}a0=a(0+0)=a0+a0$. Subtracting ${}a0$ (meaning addition with the negative of ${}a0$) on both sides gives the claim.

2. ${}(-a)b+ab=(-a+a)b=0b=0\,$

   due to part (1). Therefore, ${}(-a)b$ is the (uniquely determined) negative of ${}ab$.

3. Due to (2), we have ${}(-a)(-b)=(-(-a))b$, and because of ${}-(-a)=a$ (which holds in every group), we get the claim.
4. This follows from the parts proved so far.
5. This follows with a double induction.

$\Box$