{%hackmd @RintarouTW/About %}

# Ring Theory

The structures similar to the set of integers are called rings, and those similar to the set of real numbers are called fields.

## Ring

https://math.stackexchange.com/questions/61497/why-are-rings-called-rings

In German, "Ring" can also mean a (close) group of people with shared interests, an association. Maybe it is because of that: you've got some closely interacting elements but for all their interaction, they never leave the group. – Raphael Sep 2 '11 at 23:05

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A ring $[R;+,\cdot]$ is a set $R$ together with two binary operations, **addition** and **multiplication**, denoted by the symbols $+$ and $\cdot$ such that the following axioms are satisfied:
- $[R; +]$ is an abelian group. $\iff\forall a,b\in R,a+b=b+a$
- $\cdot$ is associative on $R\iff \forall a,b,c\in R, a(bc)=(ab)c$
- $\cdot$ is distributive over $+\iff\forall a,b,c\in R, \cases{
a(b+c)=ab+ac\\
(b+c)a=ba+ca
}$
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### The ring of integers (infinite ring)

$$
[\mathbb{Z};+,\cdot]\cases{
[\mathbb{Z};+]\in abelian\\
\forall a,b,c\in\mathbb{Z}, a(bc)=(ab)c\iff \cdot\ is\ associative\ on\ \mathbb{Z}.\\
\forall a,b,c\in\mathbb{Z}, a(b+c)=ab+ac
}
$$

### The ring of integers modulo $n$

$$
[\mathbb{Z}_n;+_n,\times_n]\cases{
[\mathbb{Z}_n;+_n]\in abelian\\
\forall a,b,c\in \mathbb{Z}_n, a\times_n(b\times_n c)=(a\times_n b)\times_n c\iff\times_n\ is\ associative\ on\ \mathbb{Z}_n\\
\forall a,b,c\in \mathbb{Z}_n,a\times_n(b+_n c)=a\times_n b+_n a\times_n c
}
$$

### Commutative Ring

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A ring in which **multiplication is a commutative operation** is called a commutative ring.

Abelian : commutative under $+$
Commutative : commutative under $\cdot$
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### Unity of a Ring

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A ring $[R;+,\cdot]$ that has a **multiplicative identity** is called **a ring with unity**.
The multiplicative identity itself is called **the unity of the ring**.
More formally, if there exists an element $1\in R$, such that $\forall x\in R, x\cdot 1 = 1\cdot x = x,$ then $R$ is called a ring with unity.
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### Types of Rings

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$$
\begin{array}{c|c|c}
[R;+,\cdot]\in Ring & Communtative\ (a\cdot b=b\cdot a) & Non-Commutative\\\hline
without\ unity\ (1\not\in R) & Commutative\ Ring & Ring\\
with\ unity\ (1\in R) & Commutative\ Ring\ with\ unity & Ring\ with\ unity\\
\end{array}
$$
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### Direct Products of Rings

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$$
1\le i\le n, [R_i;+_i,\cdot_i]\in Ring\\
\cases{
P=\mathop{\vcenter{\huge\times}}_\limits{i=1}^n R_i\\
a=(a_1,a_2,\ldots,a_n)\in P, a_i\in R_i\\
b=(b_1,b_2,\ldots,b_n)\in P, b_i\in R_i\\
}\\
a+b = (a_1+_1 b_1, a_2+_2 b_2,\ldots, a_n+_n b_n)\\
a\cdot b = (a_1\cdot_1 b_1, a_2\cdot_2 b_2,\ldots,a_n\cdot_n b_n)
$$
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### Units of Ring

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Let $[R; +, \cdot]$ be a ring with unity($1$).
$$
u\in R, \exists v\in R, u\cdot v=v\cdot u = 1\iff u\ has\ a\ multplicative\ inverse\ v=u^{-1}.
$$

A ring element that possesses a multiplicative inverse is **a unit of the ring**.

The set of **all units of a ring $R$** is denoted by
$$U(R)=\{u\in R\mid \exists u^{-1},u\cdot u^{-1}=u^{-1}\cdot u=1\}$$
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Example:

In the rings $[R; +, \cdot]$ and $[Q; +, \cdot]$ every nonzero element has a multiplicative inverse.
The only elements in $[Z; +, \cdot]$ that have multiplicative inverses are -1 and 1.
That is, $U(\mathbb{R}) = \mathbb{R}^∗, U(\mathbb{Q}) = \mathbb{Q}^∗$, and $U(\mathbb{Z}) = \{−1,1\}$.

### Ring Isomorphisms and Subrings

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Ring Isomorphism

$$
[R_1;+_1,\cdot_1],[R_2;+_2,\cdot_2]\in Ring\iff
\cases{
\exists f: R_1\to R_2, f\in bijection\\
\forall a,b\in R,f(a+_1 b)=f(a)+_2 f(b)\\
\forall a,b\in R,f(a\cdot_1 b)=f(a)\cdot_2 f(b)\\
}
$$
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Subring

$$
S\ is\ a\ subring\ of\ R\iff
\cases{
[R;+,\cdot]\in Ring\\
[S;+]\ is\ a\ subgroup\ of\ [R;+]\\
S\ is\ closed\ under\ \cdot, \forall a,b\in S, a\cdot b\in S
}
$$
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Example:

$$
[\mathbb{2Z};+,\cdot]\ is\ a\ subring\ of\ [\mathbb{Z};+,\cdot]\\
\cases{
[\mathbb{2Z};+]\ is\ a\ subgroup\ of [\mathbb{Z};+]\\
\forall a=2m,b=2n\in \mathbb{2Z}, m,n\in\mathbb{Z}, \\
a\cdot b=(2m)\cdot(2n)=2(2mn)\in \mathbb{2Z}\implies \mathbb{2Z}\ is\ closed\ under\ \cdot
}
$$

### Basic Properties of Ring

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Since Ring's multiplicative operation may not have inverse. we can only use distributive to prove.

$$
[R;+,\cdot]\in Ring\implies\cases{
\forall a\in R, a\cdot 0=0\cdot a=0\\\hline
\because
\begin{array}l
a\cdot 0 = a\cdot(0+0) &=a\cdot 0+a\cdot 0\\
(a\cdot 0) + -(a\cdot 0) &= (a\cdot 0 + a\cdot 0) + -(a\cdot 0)\\
0 &= a\cdot 0
\end{array}\\\hline
a\cdot (-b)=(-a)\cdot b=-(a\cdot b)\\\hline
\because
\begin{array}l
a\cdot (-b) + a\cdot b &= a\cdot (-b+b)\\
&= a\cdot 0\\
&= 0\\
\therefore -(a\cdot b) = a\cdot (-b)
\end{array}\\\hline
(-a)\cdot(-b)=a\cdot b\\\hline
\because
\begin{array}l
a\cdot(-b)=-(a\cdot b)\\
(-a)\cdot(-b)+ a\cdot(-b) &= (-a+a)\cdot(-b)\\
&= 0\\
\therefore (-a)\cdot (-b) &= -(-(a\cdot b))\\
&= a\cdot b
\end{array}
}
$$
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### Integral Domains and Zero Divisors

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Zero Divisor
Let $[R; +,\cdot]$ be a ring.

$$
\exists a,b\in R, a\ne 0, b\ne 0, a\cdot b=0\iff a,b\ are\ called\ zero\ divisors.
$$
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Multiplicative Cancellation
The multiplicative cancellation laws hold in a ring $[R;+,\cdot]$ if and only if $R$ has **no zero divisors**.
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Integral Domain ($\mathfrak{D}$)
A **commutative** ring **with unity** containing **no zero divisors** is called an integral domain.
$$
R\ is\ an\ integral\ domain\iff
\cases{
[R;+,\cdot]\in Ring\\
1\in R\\
\forall a,b\in R,a\cdot b=b\cdot a\\
\not\exists a,b\in R,a\ne 0, b\ne 0, a\cdot b=0\iff \forall a,b\in R,a\cdot b=0\implies a=0\lor b=0
}
$$
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### Zero divisors of $\mathbb{Z_n}$

If $m \in \mathbb{Z_n}, m\ne 0$, then $m$ is a **zero divisor** if and only if $m$ and $n$ are **not relatively prime**; i.e., $gcd(m, n) > 1$.

This also implies $[\mathbb{Z_p};+_p,\times_p], p\in prime$ has no zero divisors.

### Examples of Integral Domain ($\mathfrak{D}$)

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$$
\mathfrak{D}\cases{
[\mathbb{Z}; +, \cdot]\\
[\mathbb{Z_p};+_p,\times_p] with\ p\in prime\\
[\mathbb{Q};+,\cdot]\\
[\mathbb{R};+,\cdot]\\
[\mathbb{C};+,\cdot]\\
}
$$
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The product of two algebraic systems may not be an algebraic system of the same type.

ex:
$$
[\mathbb{Z_2};+_2,\times_2],
[\mathbb{Z_3};+_3,\times_3]\in\mathfrak{D},\\
P=\{(a,b)\mid a\in \mathbb{Z_2}, b\in\mathbb{Z_3}\}\\
but\\
\because (1,0)\cdot(0,2)=(0,0)\\
\therefore P\not\in\mathfrak{D}
$$

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Boolean Rings
$$
U\ is\ a\ nonempety\ set\\
[\mathcal{P}(U);\oplus,\cap]\in commutative\ ring
$$
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###### tags: `math`