Addition table for a 4 elements field

$\begingroup$ If you literally relabel: $0 \rightarrow a$ and $a \rightarrow 0$ then why would this addition table be less valid (other than the fact that it is a bit unintuitive because the identity element is labeled $a$ instead of $0$)? $\endgroup$

$\begingroup$ The change is true only inside the table, not in the title of each row\column. Any yet the relabel is not 100% because the changes are only in the 2nd and 4th rows\columns. $\endgroup$

2 Answers 2

The second table is perfectly fine. It is actually $\mathbb/4\mathbb$.

To see that there is no field with additive group $\mathbb/4\mathbb$ consider the following: $$aa =(1+1)(1+1) = 1\cdot 1+1\cdot 1+1\cdot 1+1\cdot 1=0.$$

$\begingroup$ While $\mathbb/4\mathbb$ is a perfectly fine abelian group, it is not the addition group of any field. $\endgroup$

$\begingroup$ I think you're over-killing it. I was supposed to build an addition table of a 4 elements field, I came up with the second table but I was told that the right answer is the first table. I want to know why is my table bad, cause I can't find any axiom that it is not enduring. $\endgroup$

If you have only one operation, it is difficult to speak about field. But, it is well-known that:

1) there exists exactly two groups (up to isomorphism) with 4 elements: one is $<\mathbb Z>/2<\mathbb Z>\times<\mathbb Z>/2<\mathbb Z>$ (the first table) and the other one is $<\mathbb Z>/4<\mathbb Z>$ (the second table)

2) there exists exaclty one field (up to isomorphism) with 4 elements, and it is isomorphic to $<\mathbb Z>/2<\mathbb Z>\times<\mathbb Z>/2<\mathbb Z>$

$\begingroup$ Actually I have two operations but the multiplication table was too easy to think of. $\endgroup$

$\begingroup$ @J.G: how do you define the Klein $4$-group? One possible definition is that it is $<\mathbb Z>/2<\mathbb Z>\times<\mathbb Z>/2<\mathbb Z>$. $\endgroup$