A ring is a structure $(R, +, \cdot)$ such that:
$(R, +)$ is an abelian group.
$(R, \cdot)$ is a monoid.
(Distributivity) For all $a$, $x$, $y\in R$,
\[\begin{align*} a(x + y) &= ax + ay, \\ (x + y)a &= xa + ya. \end{align*}\]
The neutral element of $(R, +)$, called the additive neutral element, is typically denoted by $0$.
The neutral element of $(R, \cdot)$, called the multiplicative neutral element, is typically denoted by $1$.
Examples.
- Trivial / zero ring: $\{0\}$ is the only ring where $0 = 1$.
A ring $(R, +, \cdot)$ is abelian if $(R, \cdot)$ is abelian.
A unit of a ring is a multiplicatively invertible element.
Let $R$ be a ring, $a \in R$ and $n \in \N$. $a^n$ is defined recursively:
- \[a^0 = 1.\]
- \[a^{n + 1} = a^n \cdot a.\]
For $n \in \N^+$, we have
\[a^n = \underbrace{a \cdot \cdots \cdot a}_{\text{$n$ times}}.\]
In a ring $R$, $0$ is an absorbing element with respect to multiplication: For all $a \in R$,
\[a \cdot 0 = 0 \cdot a = 0.\]
Proof.
\[\begin{align*} a \cdot 0 &= a \cdot (1 - 1) = a \cdot 1 - a \cdot 1 = 0, \\ 0 \cdot a &= (1 - 1) \cdot a = 1 \cdot a - 1 \cdot a = 0. \end{align*}\]
In a non-zero ring $R$, $0$ is not invertible.
Proof. Otherwise, there would be an $x \in R$ such that
\[0 = 0 \cdot x = 1.\]