Can you construct a field with 6 elements? 
Possible Duplicate:
Is there anything like GF(6)? 

Could someone tell me if you can build a field with 6 elements.

If such a field $F$ exists, then the multiplicative group $F^\times$ is cyclic of order 5. So let $a$ be a generator for this group and write
$F = \{ 0, 1, a, a^2, a^3, a^4\}$.
From $a(1 + a + a^2 + a^3 + a^4) = 1 + a + a^2 + a^3 + a^4$, it immediately follows that
$1 + a + a^2 + a^3 + a^4 = 0$. Let's call this (*).
Since $0$ is the additive inverse of itself and $F^\times$ has odd number of elements, at least one element of $F^\times$ is its own additive inverse. Since $F$ is a field, this implies $1 = -1$. So, in fact, every element of $F^\times$ is its own additive inverse (**).
Now, note that $1 + a$ is different from $0$, $1$ and $a$. So it is $a^i$, where i = 2, 3 or 4. Then, $1 + a - a^i = 1 + a + a^i = 0$. Hence, by $(*)$ one of $a^2 + a^3$, $a^2 + a^4$ and $a^3 + a^4$ must be $0$, a contradiction with (**).
