NoteNextra-origin/content/Math4302/Math4302_L28.md

Math4302 Modern Algebra (Lecture 28)

Rings

Field of quotients

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)]

The multiplication and addition is well defined

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).

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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')

Claim (F,+,*) is a field

• 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.

• \phi(a+b)=[(a+b,1)]=[(a,1)][(b,1)]\phi(a)+\phi(b)
• \phi(ab)=[(ab,1)]=[(a,1)][(b,1)]\phi(a)\phi(b)

and \phi is injective.

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

Example

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.

Polynomial rings

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.

Example

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.