Can you construct a field with 6 elements? [duplicate]
Possible Duplicate:
Is there anything like GF(6)?
Could someone tell me if you can build a field with 6 elements.
Solution 1:
$\def\x{\otimes}$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.
Solution 2:
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.
Solution 3:
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.