# Thread: Field with three elements

1. ## Field with three elements

Show that there exists one and only one field with three elements.

If I had to guess, it would be the cyclic subgroup $\displaystyle Z_{3}$. Am I anywhere near on target? I figure this since each element would have inverses in the field. I am aware that $\displaystyle Z_{p}$ whereby p is prime is a field of p elements. Obviously, 3 is prime. Should I solve this generally for p and then simply state that 3 is an instance of a prime number? There has to be something I am either forgetting or simply not getting. The part I am confused about is HOW to shoe that there is ONE and ONLY ONE field with 3 elements. Thanks.

Barton

2. Originally Posted by Barton
Show that there exists one and only one field with three elements.

If I had to guess, it would be the cyclic subgroup $\displaystyle Z_{3}$. Am I anywhere near on target? I figure this since each element would have inverses in the field. I am aware that $\displaystyle Z_{p}$ whereby p is prime is a field of p elements. Obviously, 3 is prime. Should I solve this generally for p and then simply state that 3 is an instance of a prime number? There has to be something I am either forgetting or simply not getting. The part I am confused about is HOW to shoe that there is ONE and ONLY ONE field with 3 elements. Thanks.
Let $\displaystyle \{x,y,z\}$ be the field. One element needs to be $\displaystyle 0$ so say $\displaystyle x=0$. One element needs to be $\displaystyle 1$ (and $\displaystyle 1\not = 0$) say $\displaystyle y=1$. This leaves us with $\displaystyle \{0,1,z\}$. Now $\displaystyle z$ has an additive inverse. It cannot be zero because $\displaystyle z+0=z\not = 0$. And if cannot be $\displaystyle z$ because $\displaystyle z+z=0\implies z(1+1)=0\implies z=0$ a contradiction. Thus, $\displaystyle -z = 1$. Using a similar argument we can show that $\displaystyle z^{-1} = z$. Now define a mapping from $\displaystyle \{0,1,z\}$ to $\displaystyle \{[0],[1],[2]\} = \mathbb{Z}_3$ by $\displaystyle 0\mapsto 0$, $\displaystyle 1\mapsto [1]$, $\displaystyle z\mapsto [2]$ and show it is an isomorphism.

3. ## Re: Field with three elements

I don't see why z+z=0\implies z(1+1)=0\implies z=0 is a contradiction. Why can't it be true that 1+1=0?

4. ## Re: Field with three elements

Remember in any field, the set of all elements form a group with operation +. In this case, we have an additive group of order 3. In such a group we can have no element of order 2; in particular 1 does not have order 2; i.e. 1+1 is not 0. (Actually, of course, 1 has order 3 or 1+1+1 is 0; i.e. the characteristic of the field is 3)

5. ## Re: Field with three elements

even without lagrange, we can see if 1+1 = 0, (that is, z = -z), we have no possible choice for 1+z:

1+z = 0 contradicts -1 = 1 (that is, 1 + 1 - 0).

1+z = 1 contradicts z ≠ 0 (as we see by adding 1 to both sides).

1+z = z contradicts 1 ≠ 0 (as we see by adding z to both sides).

so if 1+1 = 0, so that z+z = 0, then {0,1,z} is not closed under addition.

since 0 and 1 are thus ruled out as possible inverses for 1, it must be the case that 1+z = 0. that is: z = -1.

therefore:

1+1 ≠ 0
1+1 ≠ 1

and so 1+1 = z, so that: 1+1+1 = 1+z = 0.

thus F = {0,1,1+1} <---(F,+) is a cyclic group generated by 1 of order 3.

it really doesn't matter if we name 1+1 "-1", or "2", in either case we get a group isomorphism h:

0 <--> [0]
1 <--> [1]
1+1 <--> [2] of (F,+) with (Z3,+).

we can then use this group isomorphism to define a ring multiplication on F, by (for a,b in F):

a*b = h-1(h(a)h(b)).

then h(a*b) = h(h-1(h(a)h(b)) = h(a)h(b), which shows that h is now a ring-isomorphism. since Z3 is a field (both [1] and [2] are units), F is, as well.

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### the field with three elements

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