What are the fields with 4 elements? What are the  fields with 4 elements?

Let $K$ be your field. 
The additive group of $K$ is an abelian group with four elements. The order of $1$ in this group divides $4$, so it is either $2$ or $4$. Were it $4$, we would have $1+1\neq0$ and $(1+1)\cdot(1+1)=0$, which is absurd in a field. It follows that $1+1=0$ in $K$. But then for all $x\in K$ we have $x+x=x\cdot(1+1)=0$, and we see that all elements have order $2$. In particular, $-1=1$.
Let $a$ be an element in $K$ which is neither $0$ nor $1$. Then $a+1$ is neither $a$ nor $1$ and if we had $a+1=0$, then $a=-1=1$ which again is a not true. We conclude that the four elements of $K$ are $0$, $1$, $a$ and $a+1$. 
You should check that this knowledge complete determines the addition in $K$.
We have to determine the multiplication now. Since $0$ and $1$ are what they are, we need only see what $a\cdot a$, $a\cdot(a+1)$ and $(a+1)\cdot(a+1)$ are:


*

*We can't have $a^2=a$, for then $a(a-1)=0$ and we are supposing that $a\not\in\{0,1\}$; similarly, $a^2\neq0$. If $a^2=1$, then $(a-1)^2=a^2-1=0$ , which is also impossible. We must have thus $a^2=1+a$.

*Next, using this, $a\cdot(a+1)=a^2+a=1+a+a=1$.

*Finally, using that $1+1=0$, $(a+1)\cdot(a+1)=a^2+1=a$.
Multiplication is completely determined.
Now we have to check that with this operations we do have a field... You should have no trouble with that :)
