Let $F$ be a field with four elements $F = \{0,1,a,b\}$. What is $a^2 + a$?

Question

Let $F$ be a field with four elements $F = \{0,1,a,b\}$. What is $a^2 + a$?

I proved that $a\cdot b = 1$ by eliminating other possibilities.

I used $a\cdot b = 1$ to prove that $a\cdot a = b$.

So $a^2 + a = b + a$

I tried prooving that $b + a = 1$ by elimination and I derived a contradiction for $b + a =a$ and $b+a = b$ but I don't know how to disprove $b+a=0$.

Also I am not sure how to prove other additions in the addition table such as $1 +1, a + 1, a+ a$

Answer 1

Assume that $b+a = 0$. Multiplication with $a$ gives $1+b = 0$. Multiplication with $b$ gives $a+1 = 0$. Hence, both $a$ and $b$ are the additive inverse of $1$. A contradiction.

Answer 2

Hint: what is the characteristic of $F$? What does that mean regarding $a+a$?

Answer 3

$F^\times = \{1,a,b\}$ has order $3$ and so is cyclic. Therefore, $b=a^2$ and $a^2 + a = b + a$.

Now, $a^2+a=0 \implies a^2=-a \implies a^4=a^2 \implies b=a^2=1$, which contradicts $b\ne1$.