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
 A: Finite fields have finte non-zero characterisctic.
And characteristic of a field is a prime number.
Also characteristic of a field divides the order of the field.
Here order of the field is $4$, and the only prime that divides it is $2$, hence characteristic of this field is $2$, hence $1+1=0$ !!!
Alternative way:
Clearly, $ab=1$ as $ab$ can't equal $a$ or $b$(as then multiply by inverse to get a contradiction), also it can't equal $0$, as any field is also an integral domain and product of two non-zero elements is non-zero.
Since a field is also a group under addition, order of an element under addition divides the order of the field, now if order of the element $1$ is not $2$, then it's of course equal to $4$, hence $1$ is the generator of the field.
So $1+1+1+1=0$, W.L.O.G 
$a=1+1$, $b=1+1+1$
So $ab=1+1+1+1+1+1=0+1+1=1+1$
So $ab\neq1$ (a contradiction)
A: Trying to exclude $1+1=a$ and $1+1=b$ is very hard if you don't understand the following. All I am using here are field axioms, the fact that $0 \cdot x = 0$, and the notion of "order of an element in a group".
$(F , + )$ is a group of order 4, so there exists some element of order 2. Let $x \in F$ be such element. $x$ has order 2 means that $x \neq 0$ and
$$x+x=0$$
Hence
$$(1+1)= (x^{-1} \cdot x) (1+1) = x^{-1}(x \cdot (1+1)) = x^{-1} \cdot (x+x) = x^{-1} \cdot 0 = 0$$
A: By Lagrange's theorem applied to the additive group of the field we have
$$
1+1+1+1=0.
$$
Therefore, by distributivity,
$$
0=(1+1+1+1)=(1+1)(1+1)=(1+1)^2.
$$
So if $1+1\neq0$ we have found a zero divisor $1+1$. Fields don't have zero divisors, so this is a contradiction and we are left with $1+1=0$.
A: Consider {1,a,b} which is a group under multiplication. 1 is the identity under multiplication. ab is 1 under multiplication as argued in the above answer.
Now, consider {0,1,a,b} . Try to find the additive inverse of 1. 0 is not the additive inverse obviously. Suppose a or b is additive inverse. Say a is the inverse 
Then 1+a = 0
(Multiply with b )
1.b + a.b = 0.b
(Since, 1.b = b, 0.b = 0 and a.b = 1) 
b + 1 = 0
But additive inverse is unique. Hence a =b
But a$\neq$ b because they are distinct elements of the group .Contradiction
Similar contradictions if you start with b.So additive inverse of 1 is 1
Hence 1+1 = 0
A: 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.
.....
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 $1+1\ne b$.
