# 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

<details>
<summary>Examples of division ring that is not a field</summary>

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 $xx^{-1}=1$.

</details>

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 $\operatorname{gcd}(m,n)=1$.

<details>
<summary>Proof</summary>

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 $[m]$. $mr\equiv 1\mod n\implies [m][x]=[1]$.

</details>

### 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.

<details>
<summary>Example of zero divisors</summary>

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.

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

</details>

> [!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).

<details>
<summary>Proof</summary>

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.

</details>

#### Definition of integral domain

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

<details>
<summary>Example of integral domain</summary>

$\mathbb{Z}$ is a integral domain.

---

Any field is a integral domain.
</details>

#### 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.

<details>
<summary>Proof</summary>

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

</details>

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

<details>
<summary>Example</summary>

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).
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