Associativity of multiplication in a field I have the following definition of a field:

A field is a set $F$ together with two commutative binary operations $(+ 
 ,\times)$ and distinct elements $0$ and $1$ such that $F$ is a group under $+$ with identity $0$, $F\setminus\{0\}$ is a group under $\times$ with identity $1$ and the distributive law
$x \times (y + z) = x \times y + x \times z$
holds

My question is, since  $F\setminus\{0\}$ is a group under $\times$, associativity of $\times$ holds for $F\setminus\{0\}$, but does that mean that associativity of $\times$ does also holds for $F$ itself? Or do you have to prove that associativity for $\times$ in $F$ also holds in the case where $0$ is one of the elements you are multiplying by?
I saw an alternative definition of a field on wikipedia where $\times$ is defined to be associative. So is this property actually already part of the above definition?
 A: You can prove that $(F, \times)$ is associative using the field axioms you cite. Associativity states that for any $a,b,c \in F$, $(ab)c=a(bc)$. If none of $a$, $b$, or $c$ are zero then this equation is in $F^\times$, where it is true. Suppose then that at least one of $a$, $b$, or $c$ are zero. We'll need the following lemma.
Lemma. Let $a \in F$. Then $0a=0$.
Proof. $0a + 0a = (0+0)a =0a$ by distributivity. Then as $(F,+,0)$ is a group, $0a=0$ (subtract $0a$ from both sides).
Using this lemma, we can see that if any of $a,b,c$ were zero then $(ab)c=0$ and $a(bc)=0$. To be rigorous, we can do the 3 cases.

*

*$a=0$: $a(bc)=0(bc)=0$ and $(0b)c=0c-0$.


*$b=0$: $a(0c)=a0=0$ and $(a0)c=0c=0$.


*$c=0$: $a(b0)=a0=0$ and $(ab)0=0$.
A: Multiplicative associativity also holds if you include $0$, since $0$ is absorbing, i.e., for all $x\in F$,
$x\cdot 0 = 0$.
Absorption can be proved as follows:
$0=0+0$ and so by multiplying with $x$, $x0 = x0+x0$. Adding $-x0$ on both sides gives $0=x0$.
A: A better definition (imho) is this:
A ring is a tuple $(R,+,0,\cdot,1)$ with a set $R$ and binary operations $+,\cdot$ such that $(R,+,0)$ is a commutative group and $(R,\cdot,1)$ is a monoid (so a group without the existence of an inverse as an axiom). In addition, multiplication distributes over addition (from the left and from the right).
A ring is called commutative if its multiplicative monoid is commutative.
A field is a commutative ring where $(R\backslash\{0\},\cdot,1)$ is also a group.
So multiplication by $0$ is still associative by definition, since $(R,\cdot,1)$ is still a monoid, though not a group. But we could also deduce associativity, if we were so inclined, since multiplication by $0$ always yields $0$, so $(0a)b=0b=0=0(ab)$.
