Prove $a+a=1$ using field table

Question

Suppose a field $F=\{0,1,a\}$, what would be the table for addition . I know that for addition table, to prove $a+a=1$

Answer 1

$(F,+)$ is an Abelian group and there is only one group with three elements. Specifically, if $x \in F$ and $x \ne 0$ then the subgroup generated by $x$ must have an order dividing $3$ by Lagrange's theorem. Therefore $(F,+)$ is a cyclic group of order $3$.

Answer 2

As you stated the first row and first column of the multiplication table are zero entries. The row/column belonging to $1 \in F$ are the identity row/column. This leaves $a*a=1$ as $F$ is a field and $a$ must have an inverse which certainly is not $1$ nor $0$. Now, as $F$ has three elements and is a group under addition, $(F,+) \cong C_3$ the cyclic group of order 3. By looking at the table of $C_3$ we deduce that $1+1 \neq a+a$. Further, we have $a +a \neq 0$ else we would have \begin{equation}a+a = 0 \implies a(a+a) = a(0) \implies 1+1=0 \end{equation} which contradicts our aforementioned deduction. Finally, as $C_3$ can have no idempotent elements other than the identity, this leaves us with $a +a =1$.

Answer 3

What is $a+1$?

If $a+1=a $ then $a+1-a=a-a $ and $1=0$. That's impossible.

If $a+1=1$ then $a+1-1=1-1$ and $a=0$. That's impossible.

So $a+1=0$.

So what is $a+a $?

If $a+a=0$ then $a=a+0=a+a+1=0+1=1$. That is impossible.

If $a+a=a $ then $a+a+1=a+1$ so $a+0=a=0$. That is impossible.

The only option left is $a+a=1$.