# Prove that (M,+,*) is a field

Prove that the multiplication $*:M \times M \to M$ defined by this table:
* | 0 1
--------
0 | 0 0
1 | 0 1
together with the commutative group (M,+), is a field (M,+,*).

Group axioms:

1) Closure:
$0*0=0 \in M$
$0*1=0 \in M$
$1*0=0 \in M$
$1*1=1 \in M$

2) Inverse element:
$1*1=1$
$1*1=1$
$\forall i,a,e \in M : a*i=i*a=e$
here the inverse element is i=e=1.

3) Identity element:
$0*1=0$
$1*0=0$
$1*1=1$
$\forall e,a\in M: e*a=a*e=a$ with $e=1$

4) Associativity:
4.1) $0*(0*0)=0 \leftrightarrow (0*0)*0=0$
4.2) $0*(0*1)=0 \leftrightarrow (0*0)*1=0$
4.3) $0*(1*0)=0 \leftrightarrow (0*1)*0=0$
4.4) $0*(1*1)=0 \leftrightarrow (0*1)*1=0$
4.5) $1*(0*0)=0 \leftrightarrow (1*0)*0=0$
4.6) $1*(0*1)=0 \leftrightarrow (1*0)*1=0$
4.7) $1*(1*0)=0 \leftrightarrow (1*1)*0=0$
4.8) $1*(1*1)=1 \leftrightarrow (1*1)*1=1$
$\forall a,b,c\in M : a*(b*c)=(a*b)*c$

the addition $+:M \times M \to M$ defined by:
$+ | 0$ $1$
------------
$0 | 0$ $1$
$1 | 1$ $0$

Field condition:

1F) Commutativity:
$0*0=0=0*0$
$1*0=0=0*1$
$1*1=1=1*1$

2F) Distributivity:

2.1F) $0*(0+0)=0 \leftrightarrow (0*0)+(0*0)=0$
2.2F) $0*(0+1)=0 \leftrightarrow (0*0)+(0*1)=0$
2.3F) $0*(1+0)=0 \leftrightarrow (0*1)*(0*0)=0$
2.4F) $0*(1+1)=0 \leftrightarrow (0*1)+(0*1)=0$
2.5F) $1*(0+0)=0 \leftrightarrow (1*0)+(1*0)=0$
2.6F) $1*(0+1)=1 \leftrightarrow (1*0)+(1*1)=1$
2.7F) $1*(1+0)=1 \leftrightarrow (1*1)+(1*0)=1$
2.8F) $1*(1+1)=0 \leftrightarrow (1*1)+(1*1)=0$
$\forall a,b,c\in M : a*(b+c)=(a*b)+(a*c)$

And what about 2) the inverse element is that correct?(because in the "usual multiplication" the inverse element i is $i=a^{-1}$ defined?!)
Only $a\cdot(b+c) = a\cdot b + a\cdot c$ is true, $a + (b\cdot c) \neq (a+b)\cdot(a+c)$. Moreover it is a field because $\mathbb{Z}_p$ is a field for any prime $p$. – dtldarek Nov 29 '12 at 22:23