Math4302 Modern Algebra (Lecture 28)

Rings

Let $R$ be an integral domain ( $R$ has unity and commutative with no zero divisors).

Consider the pair $S=\{(a,b)|a,b\in R, b\neq 0\}$ .

And define the equivalence relation on $S$ as follows:

$(a,b)\sim (c,d)$ if and only if $ad=bc$ .

We denote $[(a,b)]$ as set of all elements in $S$ equivalent to $(a,b)$ .

Let $F$ be the set of all equivalent classes. We define addition and multiplication on $F$ as follows:

[(a,b)]+[(c,d)]=[(ad+bc,bd)]

[(a,b)]\cdot[(c,d)]=[(ac,bd)]

Addition:

If $(a,b)\sim (a',b')$ , and $(c,d)\sim (c',d')$ , then we want to show that $(ad+bc,bd)\sim (a'd+c'd,b'd)$ .

Since $(a,b)\sim (a',b')$ , then $ab'=a'b$ ; $(c,d)\sim (c',d')$ , then $cd'=dc'$ ,

So $ab'dd'=a'bdd'$ , and $cd'bb'=dc'bb'$ .

$adb'd'+bcb'd'=a'd'bd+b'c'bd$ , therefore $(ad+bc,bd)\sim (a'd+c'd,b'd)$ .

Multiplication:

If $(a,b)\sim (a',b')$ , and $(c,d)\sim (c',d')$ , then we want to show that $(ac,bd)\sim (a'c',b'd')$ .

Since $(a,b)\sim (a',b')$ , then $ab'=a'b$ ; $(c,d)\sim (c',d')$ , then $cd'=dc'$ , so $(ac,bd)\sim (a'c',b'd')$

additive identity: $(0,1)\in F$
additive inverse: $(a,b)\in F$ , then $(-a,b)\in F$ and $(-a,b)+(a,b)=(0,1)\in F$
additive associativity: bit long.
multiplicative identity: $(1,1)\in F$
multiplicative inverse: $[(a,b)]$ is non zero if and only if $a\neq 0$ , then $a^{-1}=[(b,a)]\in F$ .
multiplicative associativity: bit long
distributivity: skip, too long.

Such field is called a quotient field of $R$ .

And $F$ contains $R$ by $\phi:R\to F$ , $\phi(a)=[(a,1)]$ .

This is a ring homomorphism.

and $\phi$ is injective.

If $\phi(a)=\phi(b)$ , then $a=b$ .

Let $D\subset \mathbb R$ and

\mathbb Z \subset D\coloneqq \{a+b\sqrt{2}:a,b\in \mathbb Z\}

Then $D$ is a subring of $\mathbb R$ , and integral domain, with usual addition and multiplication.

(a+b\sqrt{2})(c+d\sqrt{2})=(ac+2bd)+(ad+bc)\sqrt{2}

-(a+b\sqrt{2})=(-a)+(-b)\sqrt{2})

…

$D$ is a integral domain since $\mathbb R$ has no zero divisors, therefore $D$ has no zero divisors.

Consider the field of quotients of $D$ . $[(a+b\sqrt{2},c+d\sqrt{2})]$ . This is isomorphic to $\mathbb Q(\sqrt2)=\{r+s\sqrt{2}:r,s\in \mathbb Q\}$

m+n\sqrt{2}=\frac{m}{n}+\frac{m'}{n'}\sqrt{2}\mapsto [(mn'+nm'\sqrt{2},nn')]

And use rationalization on the forward direction.

Let $R$ be a ring, a polynomial with coefficients in $R$ is a sum

a_0+a_1x+\cdots+a_nx^n

where $a_i\in R$ . $x$ is indeterminate, $a_0,a_1,\cdots,a_n$ are called coefficients. $a_0$ is the constant term.

If $f$ is a non-zero polynomial, then the degree of $f$ is defined as the largest $n$ such that $a_n\neq 0$ .

Let $f=1+2x+0x^2-1x^3+0x^4$ , then $deg f=3$

If $R$ has a unity $1$ , then we write $x^m$ instead of $1x^m$ .

Let $R[x]$ denote the set of all polynomials with coefficients in $R$ .

We define multiplication and addition on $R[x]$ .

$f:a_0+a_1x+\cdots+a_nx^n$

$g:b_0+b_1x+\cdots+b_mx^m$

Define,

f+g=a_0+b_0+a_1x+b_1x+\cdots+a_nx^n+b_mx^m

fg=(a_0b_0)+(a_1b_0)x+\cdots+(a_nb_m)x^m

In general, the coefficient of $x^m=\sum_{i=0}^{m}a_ix^{m-i}$ .

Caution

The field $R$ may not be commutative, follow the order of computation matters.

We will show that this is a ring and explore additional properties.