MATH 415 Lecture 16

« previous | Tuesday, October 22, 2013 | next »

Review

(See MATH 415 Lecture 15→)

• Zero-Divisors: elements $a$ such that $\exists b$ where $a\,b=0$.
• Integral Domain $D$: commutative ring with multiplicative identity 1 different from additive identity 0 and containing no zero-divisors
• Division Ring: is a ring such that every nonzero element is a unit
• Field: is a commutative division ring
• Theorem 19.9: Every field $F$ is an integral domain.

Theorem 19.11

Theorem. Every finite integral domain is a field.

Proof. Let $D$ be an integral domain such that $\left|D\right|<\infty$. Then $D=\left\{0,1,a_{1},\ldots ,a_{n}\right\}$, where all elements are distinct. For any nonzero $a\in D$, there exists a $b$ such that $a\,b=1$.

Consider $aD=\left\{a\,b\mid b\in D\right\}=\left\{a\cdot 0,a\cdot 1,a\cdot a_{1},\ldots ,a\cdot a_{n}\right\}$.

We claim that all elements in $aD$ are distinct because $D$ has no nonzero divisors: consider $a\,a_{i}=a\,a_{j}$ for $i\neq j$. Then $a\,a_{i}-a\,a_{j}=a\,(a_{i}-a_{j})=0$. Since $a\neq 0$, it follows that $a_{i}-a_{j}=0$ and thus $a_{i}=a_{j}$.

Since $\left|D\right|=n+2$, then $\left|aD\right|=n+2$ by distinctiveness of elements. In particular, $\left|D\setminus \left\{0\right\}\right|=\left|aD\setminus \left\{0\right\}\right|=n+1$. Therefore, there exists some $i$ such that $a\,a_{i}=1$ in $aD\setminus \left\{0\right\}$.

quod erat demonstrandum

Corollary. $\mathbb {Z} _{n}$ is a field if and only if $n$ is prime.

Proof. if $n$ is not prime, then there exists $m\cdot k=n$, which would imply that $m$ and $n$ are zero divisors and thus $\mathbb {Z} _{n}$ would not be a field. However, if $n$ is prime, then $n$ necessarily has no divisors, so the set $\mathbb {Z} _{n}$ would not have any zero divisors and thus be a field.

quod erat demonstrandum

Characteristic of a Ring

Let $R$ be a ring, and let $n\in \mathbb {Z} ^{+}$. Then if $n\cdot a=0$ for all $a\in R$,

we call the least number $n$ with such property a characteristic of the ring $R$. Otherwise, R has characteristic zero (0).

For example, the rings $\mathbb {Z}$, $\mathbb {Q}$, $\mathbb {R}$, and $\mathbb {C}$ are all of characteristic $0$.

but $\mathbb {Z} _{n}$ has characteristic $n$ since $n\cdot a\equiv 0{\pmod {n}}$.

Theorem 19.15

Theorem. Let $R$ be a ring with unity $1$.

• If $n\cdot 1\neq 0$ for all $n\in \mathbb {Z} ^{+}$, then $R$ has characteristic 0.
• If $n\cdot 1=0$ for some $n\in \mathbb {Z} ^{+}$, then the smallest such integer $n$ is the characteristic of $R$.

Proof. If $n\cdot 1\neq 0$, then we cannot have $n\cdot a=0$ for all $a\in R$. Now suppose $n\cdot 1=0$, then take any $a\in R$, then $n\,a=\underbrace {a+\dots +a} _{n}=a\,(1+\dots +1)=a(n\cdot 1)=a\cdot 0=0$.

quod erat demonstrandum

Characteristic of a Field

Exercise 29: If $R=D$ is an integral domain, then the characteristic is either $0$ or $p$, a prime number. This holds for all fields in general

Section 20: Fermat's and Euler's Theorems

(Fermat's little and last)

Additive groups and quotient groups of integers mod $n$ are isomorphic:

$\mathbb {Z} _{n}\simeq \mathbb {Z} /n\mathbb {Z}$

We know that the cosets in $\mathbb {Z} /n\mathbb {Z}$ are exactly $a+n\mathbb {Z}$ for all $a\in n\mathbb {Z}$.

Let's define the product between $(a+n\mathbb {Z} )$ and $(b+n\mathbb {Z} )$ by $(a+n\mathbb {Z} )\cdot (b+n\mathbb {Z} )=(a\,b+n\mathbb {Z} )$

Consider $a_{1}=a+n\,r\equiv a{\pmod {n}}$ and $b_{1}=b+n\,k\equiv b{\pmod {n}}$. observe that their product belong to the same coset: $(a+n\,r)\cdot (b+n\,k)=a\,b+(a\,k+r\,b+r\,k\,n)\,n$. Thus multiplication is well-defined.

Fact: for any division ring, the nonzero elements form a group under multiplication:

• Division ring $R_{D}$: $R_{D}^{*}=R_{D}\setminus \left\{0\right\}$ is a group (over multiplication) consisting of units.
• Field $F$: $F^{*}=F\setminus \left\{0\right\}$ is an abelian group over multiplication

Fermat's Little Theorem

If we take $\mathbb {Z} _{p}^{*}$, this forms a group over multiplication with order $p-1$. Thus we have $b^{p-1}=1$ for any $b\in \mathbb {Z} _{p}^{*}$.

This is because $\left|G\right|=n$ implies $g^{n}=1$ for all $g\in G$.

Euler's Theorem

Theorem 20.6

The set $G_{n}$ of nonzero elements of $\mathbb {Z} _{n}$ that are not zero-divisors forms a group under multiplication modulo $n$.

Proof.

1. $G_{n}$ is closed under multiplication.
2. $1\in G_{n}$ (identity element)
3. multiplication is associative
4. Every element has an inverse modulo $n$ (see permutation argument under #Theorem 19.11).

Euler Function

We define $\phi (n)$ to be the number of positive integers less than $n$ that are relatively prime to $n$.

For example, $\phi (12)=4$ because:

$1,{\cancel {2}},{\cancel {3}},{\cancel {4}},5,{\cancel {6}},7,{\cancel {8}},{\cancel {9}},{\cancel {10}},11$