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Let $S=\left \{ 0,1,2 \right \}$. How can someone prove that there is unique way of defining addition and multiplication such that $S$ is a field if $0$ of the set $S$ has the meaning (for any element $a$ in $S$: $0+a=a$), and $1$ in $S$ has the meaning (for any element $a$ in $S$: $1.a=a$)? Also, can $S$ be an ordered field?

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up vote 5 down vote accepted

The axioms for a ring already imply that $0\cdot a=0$ for all $a\in S$, so you have almost all of the multiplication table:

$$\begin{array}{c|cc} \cdot&0&1&2\\ \hline 0&0&0&0\\ 1&0&1&2\\ 2&0&2 \end{array}$$

And $2$ has to have a multiplicative inverse, so we must have $2\cdot 2=1$.

For addition we automatically have this much:

$$\begin{array}{c|cc} +&0&1&2\\ \hline 0&0&1&2\\ 1&1&&\\ 2&2& \end{array}$$

Now $1$ must have an additive inverse, so either $1+1=0$, or $1+2=0$. Suppose that $1+1=0$; then what is $1+2$? If $1+2=0$, then $1=1+0=1+(1+2)=(1+1)+2=0+2=2$, which is absurd. If $1+2=1$, then $0=1+1=1+(1+2)=(1+1)+2=0+2=2$, which is also absurd. Thus, $1+2=2$. But we know that $2$ has an additive inverse $-2$, even if we don’t yet know what it is, so $1=1+\big(1+(-2)\big)=(1+2)+(-2)=2+(-2)=0$, which is also impossible. Thus, $1+1$ cannot be $0$. It also can’t be $1$ (why not?), so it must be $2$, and we have

$$\begin{array}{c|cc} +&0&1&2\\ \hline 0&0&1&2\\ 1&1&2&\\ 2&2& \end{array}$$

Clearly $2$ must be $-1$, giving us

$$\begin{array}{c|cc} +&0&1&2\\ \hline 0&0&1&2\\ 1&1&2&0\\ 2&2&0 \end{array}$$

and it’s not hard to check that $2+2$ can now only be $1$.

Added: By the way, that last bit can be shortened to practically nothing if you recall that up to isomorphism there is only one group of order $3$, $\Bbb Z/3\Bbb Z$: that gives you the addition table right away.

How to see that $S$ cannot be an ordered field depends on how you’ve defined ordered field. If you’ve defined it in terms of a positive cone, note that $1+1=2=-1$, so $1$ can’t be in the positive cone: it’s closed under addition and never contains a non-zero element and its additive inverse. Similarly, $(-1)+(-1)=2+2=1$, so $-1$ can’t be in the positive cone, either. But this is also impossible: one of them has to be in it.

If you’ve defined it in terms of an order relation, you can get essentially the same contradictions. If $0<1$, then $1=0+1<1+1=2$, $2=1+1<2+1=0$, so by transitivity $0<0$; OOPS! A similar problem arises if $1<0$.

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Did you avoid the uniqueness of finite fields because you felt it was too advanced than what the OP may understand ? – Belgi Oct 10 '12 at 4:24
@Belgi: I wanted to demonstrate that the problem could be done quite easily with very elementary tools. At most I might have used the uniqueness of a group of order $3$. – Brian M. Scott Oct 10 '12 at 4:29

Hint #1: The additive group has to be abelian for a field, and there is only one additive group of order $3$. So, what ways can the multiplication be defined so that $1\not=0$? (required for a field)?

Hint #2: (Order) There are only a finite number of elements in your additive group. You'll get your contradiction from $1 \leq 2 \Rightarrow 1+1 = 2 \leq 2+1=0$, and similar arguments.

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