NoteNextra-origin/content/Math4302/Math4302_L25.md

Math4302 Modern Algebra (Lecture 25)

Midterm next, next Wednesday

Rings

Definitions

• commutative ring: elements a\cdot b=b\cdot a, \forall a,b\in R
• ring with unity: elements a\cdot 1=1\cdot a=a, \forall a\in R
• units: elements such that there is a\cdot b=1 for some b\in R.
• division ring: every element a\neq 0 has a multiplicative inverse a^{-1} such that a\cdot a^{-1}=1.
• field: division ring that is commutative

Examples of division ring that is not a field

Quaternions

Let i^2=-1, j^2=-1, k^2=-1, with ij=k, jk=i, ki=j.

R=\{a+bi+ci+dj\mid a,b,c,d\in \mathbb{R}\}

R is not commutative since ij\neq ji, but R is a division ring.

Let x=a+bi+cj+dk be none zero, then \bar{x}=a-bi-cj-dk, x^{-1}=\frac{\bar{x}}{a^2+b^2+c^2+d^2} is also non zero and xx^{-1}=1.

Recall from last time \mathbb{Z}_n is a field if and only if n is prime.

Units in \mathbb{Z}_n is coprime to n

More generally, [m]\in \mathbb{Z}_n is a unit if and only if \operatorname{gcd}(m,n)=1.

Proof

Let d=\operatorname{gcd}(m,n) and [m] is a unit, then \exists [x]\in \mathbb{Z}_n with [m][z]=[1], so mz\equiv 1\mod n. so mz-1=nt for some t\in \mathbb{Z}, but d|m, d|t, so d|1 implies d=1.

If d=1, so 1=mr+ns for some r,s\in \mathbb{Z}_n. If x=r\mod n, then [x] is the inverse of [m]. mr\equiv 1\mod n\implies [m][x]=[1].

Integral Domains

Definition of zero divisors

If a,b\in R with a,b\neq 0 and ab=0, then a,b are called zero divisors.

Example of zero divisors

Consider \mathbb{Z}_6, then 2\cdot 3=0, so 2 and 3 are zero divisors.

And 4\cdot 3=0, so 4 and 3 are zero divisors.

If a is a unit, then a is not a zero divisor.

ab=0\implies a^{-1}ab=0\implies 1b=0\implies b=0.

Note

If an element is not unit, it may not be a zero divisor.

Consider R=\mathbb{Z} and 2 is not a unit, but 2 is not a zero divisor.

Zero divisors in \mathbb{Z}_n

[m]\in \mathbb{Z}_n is a zero divisor if and only if \operatorname{gcd}(m,n)>1 (m is not a unit).

Proof

If d=\operatorname{gcd}(m,n)=1, then [m] is a unit, so [m] is not a zero divisor.

Therefore [m] is a zero divisor if \operatorname{gcd}(m,n)>1.

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If d=\operatorname{gcd}(m,n)>1, then n=n_1d,m=m_1d, 1\leq n_1<n.

Then mn_1=m_1dn_1=m_1n, n|mn_1 [m][n_1]=[0], n_1\neq 0, [m] is a zero divisor.

Definition of integral domain

A commutative ring with unity is called a integral domain (or just a domain) if it has no zero divisors.

Example of integral domain

\mathbb{Z} is a integral domain.

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Any field is a integral domain.

Corollaries of integral domain

If R is a integral domain, then we have cancellation property ab=ac,a\neq 0\implies b=c.

Units with multiplication forms a group

If R is a ring with unity, then the units in R forms a group under multiplication.

Proof

if a,b are units, then ab is a unit (ab)^{-1}=b^{-1}a^{-1}.

In particular, non-zero elements of any field form an abelian group under multiplication.

Example

Consider \mathbb{Z}_p field, then (\{1,2,\cdots,p-1\},\cdot) forms an abelian group of size p-1.

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Consider \mathbb{Z}_5, then we have a group of size 4 under multiplication.

• 1 has order 1
• 2 has order 4 2,4,3,1.
• 3 has order 4 3,4,2,1.
• 4 has order 2 4,1.

Therefore \mathbb{Z}_5\simeq \mathbb{Z}_4.

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Therefore in R=\mathbb{Z}_p, \mathbb{Z}_p^*=\{[1],[2],\cdots,[p-1]\} is a group of order p-1.

Therefore, for every a\in \mathbb{Z}_p, [a]^{p-1}=[1], then a^{p-1}\equiv 1\mod p (Fermat's little theorem).