Prove or disprove this table is a field

Question

Any help will be appreciated on how to approach this and get started.

Answer 1

A field has no non-zero divisors of zero.

Answer 2

Note from the multiplication table, the one for the $\cdot$ operation, that

$2 \cdot 2 = 0; \tag 1$

thus $2$ is nilpotent; but a field $F$ can have no non-zero nilpotents (note $2 \ne 0$ since $3 + 2 = 1 \ne 3$).

If $F$ is a field and $a \in F$ were nilpotent, that is, if

$a^k = 0, \; 2 \le k \in \Bbb N, \tag 2$

then

$a = a^{1 - k} a^k = a^{1 - k} = 0; \tag 3$

which shows a field has no non-zero nilpotent elements.