MATH 415 Lecture 15

« previous | Thursday, October 17, 2013 | next »

Rings

Ring Homomorphisms

Recall that $\varphi$ is a homomorphism of a ring $R$ if and only if $\varphi :R\to R^{\prime }$ satisfies the homomorphism property over both addition and multiplication.

For example, $\varphi :\mathbb{\mathbb{Z}}\to {\mathbb{\mathbb{Z}}}_n$ where $\varphi (m)=m(\mathrm{mod}n)$ is a homomorphism. $\varphi$ is obviously a homomorphism over addition, but multiplication can be trickier to see.

$\varphi (x\cdot y)=x\cdot y(\mathrm{mod}n)$. It is a property of modular arithmetic that the mod of a product is equivalent to the product of the mods, so $x\cdot y(\mathrm{mod}n)=\left(xmodn\right)\cdot \left(ymodn\right)=\varphi x\cdot \varphi y$.

Ring Isomorphisms

Recall the definition of an isomorphism:

$\varphi :R\to R^{\prime }$ is an isomorphism if it satisfies all of the following properties:

1. $\varphi$ is a homomorphism
2. $\varphi$ is one-to-one (injective)
3. $\varphi$ is onto (surjective)

Or, more simply stated, $\varphi$ is a bijective homomorphism.

Isomorphisms define an equivalence relation:

1. Reflexive. $R\simeq R$ via $\varphi (x)=x$ (identity function)
2. Symmetric. $R\simeq R^{\prime }$ via $\varphi :R\to R^{\prime }$. Then ${\varphi }^{-1}:R^{\prime }\to R$ is also an isomorphism, hence $R^{\prime }\simeq R$.
3. Transitive. $R\simeq R^{\prime }$ and $R^{\prime }\simeq R^{″}$. Define composition $\zeta =\psi \circ \varphi$

Warning

$⟨\mathbb{\mathbb{Z}},+⟩\simeq ⟨2\mathbb{\mathbb{Z}},+⟩$, but $⟨\mathbb{\mathbb{Z}},+,\cdot ⟩\simeq ̸⟨2\mathbb{\mathbb{Z}},+,\cdot ⟩$. This is because $⟨\mathbb{\mathbb{Z}},+,\cdot ⟩$ has one unity, but $⟨2\mathbb{\mathbb{Z}},+,\cdot ⟩$ has no unity.

(Unity is a structural property of rings)

Unity

The unity is the multiplicative identity element in a ring

For nontrivial groups, multiplicative unity 1 is defined as such: if $x\cdot 1=1\cdot x=x$ for all $x\in R$, where $1\ne 0$ (0 is additive identity).

In the degenerate case, $\left\{0\right\}$ is called the zero ring:

• 0 acts as the additive unity ($0+0=0$), and
• the multiplicative unity ($0\cdot 0=0$)

Definitions and Observations

• $R$ is commutative if $x\cdot y=y\cdot x$ for all $x,y\in R$.
• A ring with a multiplicative identity element is a ring with unity. (denoted $1$)

$(\underset{n}{\underbrace{1+\dots +1}})(\underset{m}{\underbrace{1+\dots +1}})=(\underset{n\cdot m}{\underbrace{1+\cdots +1}})$ because $(n\cdot 1)\cdot (m\cdot 1)=(m\cdot n)\cdot 1$.

Direct Product of Rings

If $\gcd (r,s)=1$, then the rings ${\mathbb{\mathbb{Z}}}_{rs}$ and ${\mathbb{\mathbb{Z}}}_r\times {\mathbb{\mathbb{Z}}}_s$ are isomorphic.

We already know from groups that ${\mathbb{\mathbb{Z}}}_r\times {\mathbb{\mathbb{Z}}}_s\simeq {\mathbb{\mathbb{Z}}}_{rs}$ if and only if $\gcd (r,s)=1$.

Unities $1\in {\mathbb{\mathbb{Z}}}_{rs}$ and $(1,1)\in {\mathbb{\mathbb{Z}}}_r\times {\mathbb{\mathbb{Z}}}_s$ are both generators. Therefore, $\varphi (x)=(x,x)$ is an isomorphism.

In general, The unity in $R_1\times \dots \times R_n$ is constructed as $(1_{R_1},\dots ,1_{R_n})$, where $1_{R_i}\in R_i$

Units

$b$ is called a multiplicative inverse of $a$ if $ab=ba=1$. We denote $b=a^{-1}$.

If $R$ is a ring with unity $1\ne 0$, an element $u\in R$ is a unit of $R$ if it has a multiplicative inverse.

Note: Unity is not the same as a unit

• If $u$ is a unit, then $u^{-1}$ is a unit.
• If $u_1$ and $u_2$ are units, then $u_1\cdot u_2$ is also a unit (and thus $(u_1\cdot u_2{)}^{-1}=u_2^{-1}\cdot u_1^{-1}$)

Collection of units $U$ forms a group that is a subgroup of the semigroup $⟨R,\cdot ⟩$.

Finding Units of a Ring

For example, find the units in ${\mathbb{\mathbb{Z}}}_{14}$

• $1$ is obviously a unit,
• If $1$ is a unit, then $-1\equiv 13(\mathrm{mod}14)$ is also a unit
• $(3)\cdot (5)=15\equiv 1(\mathrm{mod}14)$, so $3$ and $5$ are units.
• etc.

In general, units of ${\mathbb{\mathbb{Z}}}_n$ are elements that are coprime to $n$.

Division Rings

If every nonzero element is a unit, then $R$ is a division ring (or a skew field)

Note that $\mathbb{\mathbb{Z}}$ is not a division ring.

Subrings

If $R^{\prime }\subseteq R$ and $R^{\prime }$ is a ring, then $R^{\prime }$ is a subring of $R$.

Fields

A field is a commutative division ring.

Examples: $\mathbb{\mathbb{Q}}$, $\mathbb{ℝ}$, $\mathbb{ℂ}$.

Integral Domains

We start oddly with a quadratic equation $x^2-5x+6=(x-2)(x-3)=0$. This has solutions $x\in \{2,3\}$ in $\mathbb{ℝ}$, but what about in ${\mathbb{\mathbb{Z}}}_{12}$? $x\in \{2,3,6,11\}\subset {\mathbb{\mathbb{Z}}}_{12}$.

Zero Divisors

If $a$ and $b$ are two nonzero elements of a ring $R$ such that $ab=0$, then $a$ and $b$ are divisors of $0$ (or $0$ divisors)

Note that $4$ and $3$ are zero divisors in ${\mathbb{\mathbb{Z}}}_{12}$ since $4\cdot 3\equiv 0(\mathrm{mod}12)$.

Theorem 19.3

In ${\mathbb{\mathbb{Z}}}_n$, the divisors of 0 are precisely those nonzero elements that are not relatively prime to $n$.

Proof. Take $m\in {\mathbb{\mathbb{Z}}}_n$, where $m\ne 0$. Let $d=\gcd (m,n)>1$, so $m$ and $n$ are both divisible by $d$. Then $m\left(\frac{n}{d}\right)=\left(\frac{m}{d}\right)n\equiv 0(\mathrm{mod}n)$. Thus $m$ is a zero divisor.

Now suppose

$\gcd (m,n)=1$

and

$ms=0$

for some

$s\in {\mathbb{\mathbb{Z}}}_n$

. Then

$ms=0$

implies

$n\mid ms$

, so

$n\mid s$

since

$m$

and

$n$

are relatively prime. This means that

$s\equiv 0(\mathrm{mod}n)$

.

quod erat demonstrandum

Corollary. If $p$ is prime, then ${\mathbb{\mathbb{Z}}}_p$ has no zero-divisors. (because all integers are coprime to a prime number)

Cancellation Law

Recall that (left) cancellation law holds if $ab=ac$ implies $b=c$. (similarly right)

If $a,b,c\in R$, where $R$ is a ring, then $ab=ac$ can be rewritten as $ab-ac=a(b-c)=0$. Thus either $a$ or $b-c$ is a zero-divisor, and thus the cancellation law does not hold in general (cannot divide by zero). However,

Theorem 19.5

The left and right cancellation laws hold in $R$ if and only if $R$ has no zero divisors.

Definition of Integral Domain

An integral domain $D$ is a commutative ring with unity $1\ne 0$ containing no zero-divisors. (huh?)

Examples: $\mathbb{\mathbb{Q}},\mathbb{ℝ},\mathbb{ℂ},\mathbb{\mathbb{Z}}$. In general, any field is an integral domain: (See #Theorem 19.9→)

Note that $\mathbb{\mathbb{Z}}\times {\mathbb{\mathbb{Z}}}_p$ is not an integral domain even though $\mathbb{\mathbb{Z}}$ and ${\mathbb{\mathbb{Z}}}_p$ are both integral domains. This is because we can choose two elements $a=(1,0)$ and $b=(0,1)$ in the direct product such that $a\cdot b=(1,0)\cdot (0,1)=(1\cdot 0,0\cdot 1)=(0,0)=0\in \mathbb{\mathbb{Z}}\times {\mathbb{\mathbb{Z}}}_p$

Similarly, $D$ is an integral domain, but $M_n(D)$ ($n\times n$ matrices formed from elements of $D$) is not.

Theorem 19.9

Every field $F$ is an integral domain

If $R$ is a field, then $R\setminus \left\{0\right\}$ is a group of units, all of which have an inverse so the cancellation laws hold.

Proof. Assume $ab=0$, then if $a\ne 0$, we can find $a^{-1}\ne 0$. Thus $a^{-1}\cdot (a\cdot b)=0$ is equivalent to $(a^{-1}\cdot a)\cdot b=0$, thus $b=0$.