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.

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.
