Suppose a field $F=\{0,1,a\}$, what would be the table for addition . I know that for addition table, to prove $a+a=1$
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1You should be able to fill in more of the multiplication table, e.g. the second row and second column. – hardmath May 31 '17 at 00:36
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1A finite field is not an ordered field. – Dilip Sarwate May 31 '17 at 00:39
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Possible duplicate of Finite Fields of Order 3 – Chris Culter May 31 '17 at 00:40
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@hardmath 11 =1 1x =x, x1 =x xx=1(because u cant have multiple values?) – Bob May 31 '17 at 00:41
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Okay, you seem to have switched from $a$ to $x$ for the nonzero, non-identity element. But that fills out the multiplication table. How about addition? – hardmath May 31 '17 at 00:46
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first row is 0 1 a and second row is 0 1 and x, idk about the rest – Bob May 31 '17 at 00:52
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Both the addition and multiplication tables are "symmetric" about their diagonals (because addition and multiplication are both commutative operations in a field). – hardmath May 31 '17 at 00:57
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wow thanks for that info lol, how would i prove x+x=1 – Bob May 31 '17 at 01:13
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@Bob What exactly do you mean by $x$? Your field only has the elements $0,1$, and $a$. – Théophile May 31 '17 at 01:18
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sorry im used to using x, its a though – Bob May 31 '17 at 01:18
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You shouldn't need multiplication; the fact that $(F,+)$ is a group is enough. Note that since $a,1\neq0$, $a+1$ can't equal $1$ or $a$, so $a+1=0$. Since $a\neq1$, $a+a$ can't equal $0$ or $a$, so $a+a=1$. – stewbasic May 31 '17 at 01:21
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@stewbasic if u dont mind can u explain a bit, i dont get a,1=!0 and why a!=1 and a+a!=0 – Bob May 31 '17 at 01:50
3 Answers
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$(F,+)$ 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 $3$.
Trevor Gunn
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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 $a +a =1$.
Birch Bryant
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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 $a+a=1$.
fleablood
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To figure out what a+a is, is simply a matter of figuring out what a+1 isn't. If a+a = 0 then a+1 = 1 which is impossible. – fleablood Jun 01 '17 at 02:19