How do you prove that $a + 1 = b$ in a field with 4 elements?

Question

I have a Field with 4 distinct elements {$0, 1, a, b$}. And because $a \cdot b \neq a$, and vice versa, I have filled the $\cdot$ table as: $$ \begin{matrix} \mathbf\cdot & \mathbf0 & \mathbf1 & \mathbf{a} & \mathbf{b} \\ \mathbf0 & 0 & 0 & 0 & 0 \\ \mathbf1 & 0 & 1 & a & b \\ \mathbf{a} & 0 & a & b & 1\\ \mathbf{b} & 0 & b & 1 & a\\ \end{matrix} $$

Now to fill the $+$ table, I can't figure out how to prove by contradiction that $a + 1 = b$, and similarly, $b + 1 = a$.

Any pointers on how to proceed?

EDIT: I'm thinking that the distributive law fits in here somewhere?

Answer 1

If $a+1=a$, then we have $1=0$, impossible.

If $a+1=1$, then we have $a=0$, impossible.

If $a+1=0$, we multiply by $a$ on both sides, getting $a^2+a=0$ or $b+a=0$. So $b=1$, impossible.

Answer 2

One way to look at this problem is to use linear algebra. Let $F$ be the field $\mathbb{Z}/2\mathbb{Z}$. Then $F$ can be identified with a subfield of the field $K$ with four elements. Then $K$ is a vector space over $F$, where vector addition is the usual addition in $K$, and scalar multiplication is the restriction of the usual multiplication in $K$ to only allow the scalars to be in $F$.

A vector space over a field $F$ (or any field) is isomorphic to a direct sum

$$\bigoplus\limits_{i \in X} F$$

for some indexing set $X$. Since $K$ has four elements, it must be isomorphic as vector spaces, and in particular as abelian groups, to

$$\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$$

Now the three nonzero elements $a, b,1$ of $K$ correspond in some order to the elements $(1,0);(0,1);(1,1)$. Your conclusion follows from the observation that if you add any two of these elements together, you get the third.