MATH 415 Lecture 16

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Review

(See MATH 415 Lecture 15→)

• Zero-Divisors: elements $a$ such that $\mathrm{\exists }b$ where $ab=0$.
• 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 $\left|D\right|<\mathrm{\infty }$. Then $D=\{0,1,a_1,\dots ,a_n\}$, where all elements are distinct. For any nonzero $a\in D$, there exists a $b$ such that $ab=1$.

Consider $aD=\{ab\mid b\in D\}=\{a\cdot 0,a\cdot 1,a\cdot a_1,\dots ,a\cdot a_n\}$.

We claim that all elements in $aD$ are distinct because $D$ has no nonzero divisors: consider $a{a}_i=a{a}_j$ for $i\ne j$. Then $a{a}_i-a{a}_j=a(a_i-a_j)=0$. Since $a\ne 0$, 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,

$|D\setminus \left\{0\right\}|=|aD\setminus \left\{0\right\}|=n+1$

. Therefore, there exists some

$i$

such that

$a{a}_i=1$

in

$aD\setminus \left\{0\right\}$

.

quod erat demonstrandum

Corollary. ${\mathbb{\mathbb{Z}}}_n$ 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

${\mathbb{\mathbb{Z}}}_n$

would not be a field. However, if

$n$

is prime, then

$n$

necessarily has no divisors, so the set

${\mathbb{\mathbb{Z}}}_n$

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 $n\in {\mathbb{\mathbb{Z}}}^{+}$. 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 $\mathbb{\mathbb{Z}}$, $\mathbb{\mathbb{Q}}$, $\mathbb{ℝ}$, and $\mathbb{ℂ}$ are all of characteristic $0$.

but ${\mathbb{\mathbb{Z}}}_n$ has characteristic $n$ since $n\cdot a\equiv 0(\mathrm{mod}n)$.

Theorem 19.15

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

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

Proof. If

$n\cdot 1\ne 0$

, 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

$na=\underset{n}{\underbrace{a+\dots +a}}=a(1+\dots +1)=a(n\cdot 1)=a\cdot 0=0$

.

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:

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

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

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

Consider $a_1=a+nr\equiv a(\mathrm{mod}n)$ and $b_1=b+nk\equiv b(\mathrm{mod}n)$. observe that their product belong to the same coset: $(a+nr)\cdot (b+nk)=ab+(ak+rb+rkn)n$. 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 ${\mathbb{\mathbb{Z}}}_p^{*}$, this forms a group over multiplication with order $p-1$. Thus we have $b^{p-1}=1$ for any $b\in {\mathbb{\mathbb{Z}}}_p^{*}$.

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 ${\mathbb{\mathbb{Z}}}_n$ 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 $\varphi (n)$ to be the number of positive integers less than $n$ that are relatively prime to $n$.

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

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