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.
 A: A finite field $F$ has a finite characteristic $p$  ($p \cdot 1=0$) which must be a prime since $F$ is a field. So $F$ is a vector space of some finite dimension, say $n$ over $Z/pZ$; thus $F$ has $p^n$ elements.
A: $\def\x{\otimes}$There is not. Suppose $\langle F, +, \x\rangle$ is a field where $F$ has six elements.  Then $\langle F, +\rangle$ is an abelian group; it must be $Z_6$, which is the only abelian group with six elements. So take $F=\{0,1,2,3,4,5\}$ and $+$ to be addition modulo 6. By Lagrange's theorem, every element of $\langle F, +\rangle$ has an order that divides 6, so any element $f$ of this group has the property that $f+f+f+f+f+f = 0$.
Now we consider multiplication. We don't know yet what $1\x1$ is—it might not be $1$—so let's call it $i$, and consider $2\x 3$:
$$\begin{eqnarray}
2\x 3 & = & (1+1)\x(1+1+1) \\
& = & 1\x 1 +1\x 1 +1\x 1 +1\x 1 +1\x 1 +1\x 1 \\
& = & i+i+i+i+i+i\\
& = & 0 
\end{eqnarray}$$
But this cannot happen in a field: $ab=0$ implies $a=0$ or $b=0$, and that fails here.  So there is no way to define $\x$ to make $\langle F, +, \x\rangle$ into a field.
A: 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 (**).
A: There is not.  The Wikipedia article on finite fields says: 

The order, or number of elements, of a finite field is of the form $p^n$, where $p$ is a prime number called the characteristic of the field, and $n$ is a positive integer.

A: No, I cannot and neither can you.  Here's the reason:
Suppose you have a field $\mathbb{F}$ with finitely many elements.  Take $1 \in \mathbb{F}$ and keep adding it to itself, giving you $2 = 1 + 1$, $3 = 1 + 1 + 1$, etc.  Because $\mathbb{F}$ is finite, there must be a smallest positive number, I'll call it $p$, such that $p \cdot 1 := 1 + 1 + \cdots + 1 \text{ ($p$ terms)} = 0$.  (Exercise: Prove this.)  In fact, $p$ must be prime. (Exercise: Prove this.)
The elements $\{0, 1, 2, \dots, p - 1\}$ are all distinct and form a subfield of $\mathbb{F}$ isomorphic to $\mathbb{F}_p := \mathbb{Z} / p \mathbb{Z}$.  By abuse of notation, I'll call this subfield simply $\mathbb{F}_p$.  Then $\mathbb{F}$ is a finite dimensional vector space over $\mathbb{F}_p$. (Exercise: Prove this.)  If $n$ is the dimension of this vector space, then $\mathbb{F}$ has $p^n$ elements.  (Exercise: Prove this.)
Conclusion: The number of elements in every finite field is a power of a prime number.  In particular, there is no finite field with six elements.
