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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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    $\begingroup$ No, all finite fields have order $p^n$ for some prime $p$ and positive integer $n$. $\endgroup$ – William Aug 17 '12 at 2:13
  • $\begingroup$ William and and how they could justify my exercise only with elementary properties of fields? $\endgroup$ – Roiner Segura Cubero Aug 17 '12 at 2:49
  • $\begingroup$ This is a duplicate. If only other questions on finite fields would attract the same number of votes as these two :/ $\endgroup$ – Jyrki Lahtonen Aug 17 '12 at 19:02
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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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    $\begingroup$ This is a particularly nice solution for a student who has studied a fair amount of group theory in his first abstract algebra course, but is just beginning field theory, which is not an uncommon occurrence. It might be worth explicitly mentioning that the reason you use $i = 1 \cdot 1$ is that we don't know a priori whether the element $1$ in the abelian group $F$ will play the role of the multiplicative identity, but whatever $1 \cdot 1$ turns out to be, we still conclude that $2 \cdot 3 = 0$. $\endgroup$ – Michael Joyce Aug 17 '12 at 3:18
  • $\begingroup$ Yes, thanks. An earlier draft had "... because we don't know what it is." I will put that back. $\endgroup$ – MJD Aug 17 '12 at 3:19
  • $\begingroup$ Why must $\langle F, +\rangle$ be a cyclic group? $\endgroup$ – miracle173 Nov 26 '13 at 7:29
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    $\begingroup$ $\langle F,+\rangle$ must be an abelian group by the definition of a field. The only two groups of order 6 are $\Bbb Z_6$ and $S_3$, but $S_3$ is nonabelian. $\endgroup$ – MJD Nov 26 '13 at 15:30
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    $\begingroup$ But even if you don't already know what the groups of order 6 are, it's easy to show there is no other abelian group of order 6, without using the full classification theorem. It must have at least two generators (or else it would be cyclic). The generators must have order 2 or 3, by Lagrange's theorem. (6 is also possible, but then it would be cyclic.) There must be at least one of each. Say that $a$ and $b$ are generators of order 2 and 3 respectively. We know $ab=ba$ (because it is abelian). Now a quick calculation shows that $ab$ has order 6, so it is $\Bbb Z_6$ after all. $\endgroup$ – MJD Oct 6 '18 at 11:35
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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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  • $\begingroup$ Michael and and how they could justify my exercise only with elementary properties of fields? $\endgroup$ – Roiner Segura Cubero Aug 17 '12 at 2:45
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    $\begingroup$ @Andres All of Michael Joyce exercises are very easy. The first two exercises use "elementary properties of fields". The last one is a simple exercise about vector spaces over finite fields. $\endgroup$ – William Aug 17 '12 at 2:57
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    $\begingroup$ 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. :) $\endgroup$ – Michael Joyce Aug 17 '12 at 3:05
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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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  • $\begingroup$ Mark and and how they could justify my exercise only with elementary properties of fields? $\endgroup$ – Roiner Segura Cubero Aug 17 '12 at 2:49
  • $\begingroup$ 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. $\endgroup$ – MJD Aug 17 '12 at 3:14

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