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Possible Duplicate:
Is there anything like GF(6)?

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

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    This is a [duplicate](http://math.stackexchange.com/q/53877/11619). If only other questions on finite fields would attract the same number of votes as these two :/2012-08-17

5 Answers 5

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$\def\x{\otimes}$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.

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    @MJD Thanks for your explanation!2018-10-06
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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.

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    Well, I think they are "very easy" once you have some experience, but they take some thought when you are first learning the subject. But you will learn abstract algebra (or any field) if you struggle with what you don't know and work at figuring it out. The exercises are broken down to manageable pieces, but the task of solving them is left to you, Andres. Good luck. :)2012-08-17
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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.

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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 (**).

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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.

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    I think Michael Joyce's answer was fairly elementary, but if it isn't elementary enough for you, you may prefer the new answer I just posted.2012-08-17