Is it possible to construct $GF(4)$ with two different multiplication operations mod n and mod m as "addition" and "multiplication"? I found that it's possible to construct multiplication and addition tables for $GF(2)$ and $GF(3)$ in that way, but I still can't find ones for $GF(4)$. Is it possible at all?
$GF(2)$ can be constructed with elements $\{21, 7\}$ and operations $* \bmod42$ and $* \bmod28$
$GF(2)$ multiplication:
\begin{array}{c|cc}
\ * \bmod42 & 21 & 7  \\ 
\hline
21 & 21 & 21  \\ 
7 & 21 & 7  \\ 
\end{array}
$GF(2)$ addition:
\begin{array}{c|cc}
\ * \bmod28 & 21 & 7  \\ 
\hline
21 & 21 & 7  \\ 
7 & 7 & 21  \\ 
\end{array}
$GF(3)$ can be constructed with elements $\{10, 16, 4\}$ and operations $* \bmod30$ and $* \bmod18$
$GF(3)$ multiplication:
\begin{array}{c|ccc}
\ * \bmod30 & 10 & 16 & 4    \\ 
\hline
10 & 10 & 10 & 10   \\ 
16 & 10 & 16 & 4\\
4  & 10 & 4 & 16\\ 
\end{array}
$GF(3)$ addition:
\begin{array}{c|ccc}
\ * \bmod 18 & 10 & 16 & 4    \\ 
\hline
10 & 10 & 16 & 4   \\ 
16 & 16 & 4 & 10\\
4  & 4 & 10 & 16\\ 
\end{array}

[Edit]
Adding the following for clarity because users well versed in algebra can make mistakes here.

Judging from the OP's examples and comments, mod should be the binary mod. In other words, $ab\bmod A$ is the remainder of the product $ab$ when divided by a positive integer $A$. This translates to the usual congruence only when both moduli are larger than all the elements of the "field", JL

[/Edit]
 A: Choosing $n$ distinct primes $q_1,\ldots,q_n$ such that $p |q_i-1$, letting $k = \prod_{i=1}^n q_i$, you can embed $(\Bbb{F}_{p^n},+)$ into $(\Bbb{Z}/k \Bbb{Z},\times)$.
Taking one more distinct prime $\ell\equiv 1\bmod p^n-1$ you can embed $(\Bbb{F}_{p^n},\times)$ into $(\Bbb{Z}/\ell\Bbb{Z},\times)$
This way you are mapping each element $a\in \Bbb{F}_{p^n}$ to a pair of residue classes $b\bmod k$ and $c\bmod \ell $. You can represent this pair of residue classes by an integer $f(a)<k\ell$ such that $f(a)\equiv b\bmod k, f(a)\equiv c\bmod \ell$.
You'll get that $f(a+a') \equiv f(a)f(a')\bmod k,f(aa') \equiv f(a)f(a')\bmod \ell$ which is (I think) what you are asking for.

For $p^n=4$ you can take $q_1=3,q_2=5, \ell = 7$,
$\Bbb{F}_4 = \Bbb{F}_2[x]/(x^2+x+1) = \{0,1,x,1+x\}$,
$f(0) \equiv 1\mod 15,f(0) \equiv 0\mod 7$,
$f(1) \equiv -1\mod 3,f(1) \equiv 1\mod 5, f(1) \equiv 1\mod 7$,
$f(x) \equiv 1\mod 3,f(x) \equiv -1\mod 5, f(x) \equiv 2\mod 7$,
$f(1+x) \equiv -1\mod 3,f(1+x) \equiv -1\mod 5, f(1+x) \equiv 4\mod 7$.
