Mathematical Analysis - problems regarding field axioms

Question

Let $F = \{0,1,2\}$. Prove that there is exactly one way to define addition and multiplication so that $F$ is a field if $0$ and $1$ have their usual meaning of ($A4$) and ($M4$).

$(A4)$: There exists an element $0 \in F$ such that $0+x = x; \, \forall x \in F$.

$(M4)$: There exists an element $1 \in F$ (and $16 = 0$) such that $1x = x; \, \forall x \in F$.

My main concern is that I don't really understand the question so I have no idea how to start this. Do I have to prove that $A4$ and $M4$ applies to $F$ or prove the other field axioms applies to $F$ so $A4$ and $M4$ are correct?

Answer 1

Hint: Filling in the tables, using that:

• $0+x=x+0=x$ for all $x\in F$
• $1\cdot x=x\cdot 1=x$ for all $x\in F$

You can immediately write out the tables as:

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

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

Now, how many possibilities are there for the white squares?