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-trivial ring $R$, $0$ is not invertible.
Proof.Otherwise, there would be $x\in R$ such that
\[0 = 0\cdot x = 1.\]