Elementary proof that all fields of four elements are isomorphic to each other A question in Rotman's Advanded Modern Algebra asks to prove the question in the title. I'm convinced of my proof, but a subquestion asked to prove that $1+1$ is zero, and for this I proceeded on a case by case basis; that is, assuming for the sake of contradiction that $1+1 \ne 0$, given that we now know that the field $F = \{0,1,1+1,a\}$, I proved that this structure cannot be a field. 
However, I think this solution is quite ugly and I was wondering if is a nicer but still elementary solution (avoiding the fact that char($F$) = $2$).
PS I'm not sure if this is a suitable question. If it's not, feel free to remove it.
 A: Since there is an odd number of non-zero elements in the field, at least one of them must be its own additive inverse (because those who aren't come in pairs). So let's call that element $a$.
Since $a\ne 0$, it has a multiplicative inverse $a^{-1}$. But then we have
$$1 + 1 = a a^{-1} + a a^{-1} = (a+a) a^{-1} = 0 a^{-1} = 0$$
Note that this in turn implies that all elements of the field are their own additive inverse, since
$$x+x = 1x + 1x = (1+1)x = 0x = 0$$
Or in short, any finite field with an even number of elements must be of characteristic $2$.
A: The underlying Abelian group $(F, +)$ has 4 elements, so the additive order of $1$ divides 4. If it were 4, the element $1+1$ would not be invertible. 
A: We need $0$ and $1$.  Call the other two elements $a$ and $b$.  Then $ab=1, a^2=b$ and $b^2=a$ (otherwise $a^2=1 \implies a=b$ etc...)  So the multiplication table is fixed. 
But so is the addition table:  $1+a=b$, necessarily.  (Also,  $b+1=a$.)  For if $1+a=a,0$ or $1$ we get a contradiction. For instance,  $1+a=0\implies a=-1\implies a=1$, since $1+1\not =1,a$ or $b$.  Again,  we would have a contradiction:   say $1+1=a$, then $1+b=b,0$ or $1$, in which case $1=0$ or $b=a$ or $0$.
So our hand winds up being forced. 
A: Another path is to consider the multiplicative group $F^\times$. Since it has order $3$, it has to be cyclic. Thus, the multiplication table is fixed. This will also fix the addition table.
