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) $1*(1+1)=0 \leftrightarrow (1*1)+(1*1)=0$
$\forall a,b,c\in M : a*(b+c)=(a*b)+(a*c)$

Could you please tell me if I have made a mistake?
And what about 2) the inverse element is that correct?(because in the "usual multiplication" the inverse element i is $i=a^{-1}$ defined?!)

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it's correct because false <-> false. –  user50336 Nov 29 '12 at 22:20
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
sure?! or are you joking?because if 1=true it will never implicate 0=false. –  phil Nov 29 '12 at 22:24
Existence of inverses if fine. Note that you only have to show left inverses (or, if you prefer, right inverses) because you prove (later) that the multiplication is commutative. For 2.6F, 2.7F, note that both sides are equal to 1, not to 0. –  Brett Frankel Nov 29 '12 at 22:24
@dtldarek ok, so it is still field, even when the 2. distributive law is not fullfilled? –  phil Nov 29 '12 at 22:26