Math4302 Modern Algebra (Lecture 25)

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 $xx^{-1}=1$.

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

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.

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

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.

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

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Any field is a integral domain.

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

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).