Is a set that is an abelian group under addition and a group under multipliation a field? I suspect the answer to my question is yes, but I'm just checking my understanding. If we have a set which is an Abelian group under addition and a group under multiplication is it then defined as a field? 
Cheers Matt
 A: In general, no, because the distributive laws may not hold.
With that and a minor modification, yes. The multiplicative group has to leave out $0$, because $0$ won't be invertible.
So if you have a ring (with identity) such that the nonzero elements are an abelian group, then yes, you have a field.
If the nonzero elements are just a group then you have a division ring.
A: Take for example $(\Bbb Z_4,+,0,-)$ which is an Abelian group. On $\{1,2,3\}=\Bbb Z_4-\{0\}$ define $a*b=((a-1)+(b-1)\bmod 3)+1$ (so that this set becomes $\Bbb Z_3$). But it fails to be a field as $2*2=3$ but $2*(1+1)$ had to give $2*1+2*1=2+2=0$. That means that $*$ does not distribute over $+$, so it is not a field.
More generally, there cannot be a field where you have to add $1$ to itself at least $4$ times before you get $0$ as the sum, since the characteristic $p$, the minimal number such that $\underbrace{1+1+\cdots+1}_{p\text{ times}}=0$, must always be a prime.
On the other hand there is a unique field with $4$ elements, the factor ring $\Bbb Z_2[X]/\langle X^2+X+1\rangle$. It's additive group is the Klein four-group and its multiplicative group is $\Bbb Z_3=\langle X\rangle$
