Field with four elements

The addition: $\;a+b=0\implies a=-b=b\;$ , since $\;\text{char}\, F=2\;$ . It also can't be $\;a+b=a\;,\;\;a+b=b\;$ , else $\;b=0\;$ or $\;a=0\;$ . Thus it must be $$\;a+b=1\implies b=1-a=1+a\;$$

Generalize the above and get the addition table.

As for multiplication $\;ab\neq 0,a,b\;$ , else $\;a=0\;$ or $\;b=0\;$ , or $\;a,b=1\;$ and none of this is true, thus it must be $\;ab=1\iff b=a^{-1}\;$ . Generalize now for the table.

A field has characteristic a prime number $p$ or $0$, (consider the homomorphism $\mathbf Z\rightarrow F, \enspace n\mapsto n\cdot 1$.

Any finite field $\mathbf F_{p^n}$ is an extension of degree $n$ of the prime field $\mathbf F_p$ and it is a simple extension of $\mathbf F_p$ generated by a root of any irreducible polynomial of degree $n$ in $\mathbf F_p[x]$. In the present case the only irreducible polynomial of degree $2$ is $x^2+x+1$. Let $\omega$ be one of its roots; the other root is its inverse, $\omega+1$.

Here is its multiplication table: $$\begin{array}{c|cccc} &0&1 &\omega&\omega+1\\\hline 0&0&0&0&0\\\hline 1&0&1 &\omega&\omega+1\\\hline \omega&0&\omega&\omega+1&1\\\hline \omega+1&0&\omega+1&1&\omega \end{array}$$