Proving a result in a given field My thinking:
Due to the axiom describing closure of a field we know that there are four possibilities for the result.
Thanks

The elements $0,1,a,b $ form an additive group.  Either it is equivalent to $\mathbb Z_4$ and it is equivalent to $\mathbb Z_2\times \mathbb Z_2$. If it is equivalent to $\mathbb Z_2\times \mathbb Z_2$ then $1+1=0$
So we can assume it is equivalent to $Z_4$ and we can relable $0,1,a,b$ as $0,1,2,3$ and $1+1=2,2+1=3,$ etc.
But we can't make a field of this.
$2*2=2 (1+1)=2+2=0$.  But that would mean $1=\frac 12 *2*\frac 12*2=\frac 12*\frac 12 *2*2=\frac 12 *\frac 12 *0=0$.  A contradiction.
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Oh, I guess I should point out that when we relable $0,1,a,b$ as $0,1,2,3$ we don't have to map $1$ to $1$.  We could map $a$ or $b $ to $1$ and map $1$ to $2$ or $3$.  But $2+2=0$ so if we map $1$ to $2$ we have $1+1=0$.  If we map $1$ to $3$, we still have $3+3=2$ and a proof will still hold.  
If $0,1,a,b $ is a field and $1+1=x\ne 0$ then $x+x=0$ as that is the only additive group possibility. And $0=x+x=x (1+1)=x*x $ which is impossible.
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If $1+1=a $ then $1+b\ne 1,a,b$ so $1+b=0$ and $1+a=b $
$b+a\ne 0$ because $b+1=0$.  $b+a\ne b $. $b+a\ne a $. So $ b+a=1$.
$a+a\ne a $, $a+a\ne 1$ because $a+b=1$, $a+a\ne b $ because $a+1=b $.  So $a+a=0$
So $0=a+a=a (1+1)=a*a $.
So $0*\frac 1a =a*a*\frac 1a=a $
But $0*x+0*x=(0+0)x=0x $.  So $0x=0x+0x-0x=0x-0x=0$.
$a=0*\frac 1a =0$ which is a contradiction.
So $1+1\ne a $.
The exact same argument holds for $1+1\ne b$.
