# Field with three elements

• Aug 12th 2008, 10:28 PM
Barton
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
• Aug 13th 2008, 09:29 AM
ThePerfectHacker
Quote:

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.
• Jan 31st 2013, 05:31 PM
Teeps
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?
• Jan 31st 2013, 09:17 PM
johng
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)
• Feb 1st 2013, 06:54 AM
Deveno
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.