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A \emph{field} is a set $F$ together with two binary operations on $F$, called addition and multiplication, and denoted $+$ and $\cdot$, satisfying the following properties, for all $a,b,c \in F$: A \emph{field} is a set $F$ together with two binary operations on $F$, called addition and multiplication, and denoted $+$ and $\cdot$, satisfying the following properties, for all $a,b,c \in F$:
\begin{enumerate} \begin{enumerate}
\item $a + (b+c) = (a+b)+ c$ (associativity of addition) \item $a + (b+c) = (a+b)+ c$ (associativity of addition)
\item $a+b = b+a$ (commutativity of addition) \item $a+b = b+a$ (commutativity of addition)
\item $a+0 = a$ for some element $0 \in F$ (existence of zero element) \item $a+0 = a$ for some element $0 \in F$ (existence of zero element)
\item $a+(-a) = 0$ for some element $-a \in F$ (existence of additive inverses) \item $a+(-a) = 0$ for some element $-a \in F$ (existence of additive inverses)
\item $a\cdot (b\cdot c) = (a\cdot b)\cdot c$ (associativity of multiplication) \item $a\cdot (b\cdot c) = (a\cdot b)\cdot c$ (associativity of multiplication)
\item $a\cdot b = b\cdot a$ (commutativity of multiplication) \item $a\cdot b = b\cdot a$ (commutativity of multiplication)
\item $a\cdot 1 = a$ for some element $1 \in F$, with $1 \neq 0$ (existence of unity element) \item $a\cdot 1 = a$ for some element $1 \in F$, with $1 \neq 0$ (existence of unit element)
\item If $a \neq 0$, then $a \cdot a^{-1} = 1$ for some element $a^{-1} \in F$ (existence of multiplicative inverses) \item If $a \neq 0$, then $a \cdot a^{-1} = 1$ for some element $a^{-1} \in F$ (existence of multiplicative inverses)
\item $a\cdot (b+c) = (a\cdot b) + (a\cdot c)$ (distributive property) \item $a\cdot (b+c) = (a\cdot b) + (a\cdot c)$ (distributive property)
\end{enumerate} \end{enumerate}
Equivalently, a field is a commutative ring $F$ with identity such that: Equivalently, a field is a commutative ring $F$ with identity such that:
\begin{itemize} \begin{itemize}
\item $1 \neq 0$ \item $1 \neq 0$
\item If $a \in F$, and $a \neq 0$, then there exists $b \in F$ with $a \cdot b = 1$. \item If $a \in F$, and $a \neq 0$, then there exists $b \in F$ with $a \cdot b = 1$.
\end{itemize} \end{itemize}