MATH 415 Lecture 15

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Rings

Ring Homomorphisms

Recall that $\phi$ is a homomorphism of a ring $R$ if and only if $\phi :R\to R'$ satisfies the homomorphism property over both addition and multiplication.

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

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

Ring Isomorphisms

Recall the definition of an isomorphism:

$\phi :R\to R'$ is an isomorphism if it satisfies all of the following properties:

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

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

Isomorphisms define an equivalence relation:

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

Warning

$\left\langle \mathbb {Z} ,+\right\rangle \simeq \left\langle 2\mathbb {Z} ,+\right\rangle$, but $\left\langle \mathbb {Z} ,+,\cdot \right\rangle \not \simeq \left\langle 2\mathbb {Z} ,+,\cdot \right\rangle$. This is because $\left\langle \mathbb {Z} ,+,\cdot \right\rangle$ has one unity, but $\left\langle 2\mathbb {Z} ,+,\cdot \right\rangle$ 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 $1\neq 0$ (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 $1$)

$(\underbrace {1+\dots +1} _{n})(\underbrace {1+\dots +1} _{m})=(\underbrace {1+\cdots +1} _{n\cdot m})$ 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 $\mathbb {Z} _{r\,s}$ and $\mathbb {Z} _{r}\times \mathbb {Z} _{s}$ are isomorphic.

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

Unities $1\in \mathbb {Z} _{r\,s}$ and $(1,1)\in \mathbb {Z} _{r}\times \mathbb {Z} _{s}$ are both generators. Therefore, $\phi (x)=(x,x)$ is an isomorphism.

In general, The unity in $R_{1}\times \dots \times R_{n}$ is constructed as $\left(1_{R_{1}},\ldots ,1_{R_{n}}\right)$, where $1_{R_{i}}\in R_{i}$

Units

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

If $R$ is a ring with unity $1\neq 0$, 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 $\left\langle R,\cdot \right\rangle$.

Finding Units of a Ring

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

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

In general, units of $\mathbb {Z} _{n}$ 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 $\mathbb {Z}$ is not a division ring.

Subrings

If $R'\subseteq R$ and $R'$ is a ring, then $R'$ is a subring of $R$.

Fields

A field is a commutative division ring.

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

Integral Domains

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

Zero Divisors

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

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

Theorem 19.3

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

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

Now suppose $\gcd {(m,n)}=1$ and $m\,s=0$ for some $s\in \mathbb {Z} _{n}$. Then $m\,s=0$ implies $n\mid m\,s$, so $n\mid s$ since $m$ and $n$ are relatively prime. This means that $s\equiv 0{\pmod {n}}$.

quod erat demonstrandum

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

Cancellation Law

Recall that (left) cancellation law holds if $a\,b=a\,c$ implies $b=c$. (similarly right)

If $a,b,c\in R$, where $R$ is a ring, then $a\,b=a\,c$ can be rewritten as $a\,b-a\,c=a\,(b-c)=0$. 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 $1\neq 0$ containing no zero-divisors. (huh?)

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

Note that $\mathbb {Z} \times \mathbb {Z} _{p}$ is not an integral domain even though $\mathbb {Z}$ and $\mathbb {Z} _{p}$ 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 $a\cdot b=(1,0)\cdot (0,1)=(1\cdot 0,0\cdot 1)=(0,0)=0\in \mathbb {Z} \times \mathbb {Z} _{p}$

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