Reviewing for a term test and just confused about how these tables were derived. While I'm pretty sure I know what a field is (a set that satisfies those specific field axioms), that table for deriving values of a field (ex. 1 + 1 = x, in field F = {0, 1,x)) is something that I don't recall.
How would you come up with that table on your own for any arbitrary amount of elements, or even just for the 3,4 here? Clarification on 'fields' in general would also be very helpful. Thanks.
Note that the additive table of $\mathbb{F}_3$ consists of the 3 possible cyclyc permutations, starting with $01x$. This follows from $x+1=1+x=0$ and $1+1=x$ and from noting that $0$ is the additive identity. Now to show these 2 identities, suppose $1+1=1$, then $1=0$ contradiction. Similarly we have that $x+x \neq x$. By commutativity of addition, we have that $1+x=x+1$. Suppose $x+1=1$ then contadiction again: $x=0$. Similarly we show that $x+1 \neq x$. For the additive table in $\mathbb{F}_4$ similar reasonign can be used.
First note that in both multiplcative tables the $0$-th row will be all $0$'s. The $1$-th row will leave each element on top the same, since $1$ is the identity. For the $x$-th row in $\mathbb{F}_3$ it is clear that $x0=0$ and $x1=x$ (axioms of a field). As for $xx$, recall that every nonzero element in a field has a multiplicative inverse, so there must be a 1 in each non-$0$-th row, thus $xx=1$.
Now all we have left are the $a$-th and $b$-th rows in the multiplicative table of $\mathbb{F}_4$. Clearly, $a0=b0=0$, $a1=a$ and $b1=b$. As said before, every nonzero-th row has a 1 ( in the spot corresponding to the inverse of that row). However, it cannot be the case that $ab=ba= a$ or $b$. Simply apply the cancellation rule. This gives us $ab=ba=1$. Finally note that $a^2=b$ and $b^2=a$. In fact, $a^2 \neq 1$ $a^2 \neq a$ (same for b). The former violates the fact that nonzero elements in the multiplicative group of 3 elements $\mathbb{F}^{x}_4$ have order 3. The former implies that $a=1$ (or $b=1$). By closure under multiplication, $a^2 \in \mathbb{F}^{x}_4$, but we showed it is neither $1$ or $a$, therefore $a^2=b$. With similar reasoning, we deduce $b^2=a$.