F is a field that contains 4 elements {0,1,a,b} , given ab=1, prove that a^2 = b

Question

How do I prove that a^2 = b using the field axioms of closure, associativity, commutativity, multiplicative and additive identities, negatives and reciprocals, and distributivity?

My current attempt is:

a = a + 0 (additive identity)

a = a + (-b + b) (existence of negatives)

a * a = a * a + a*(-b) + a*(b) (distributivity of multiplication)

This is where I get stuck even if I sub in ab=1, there is still the a*a on the right side of the equation and no b. Does anyone see where I went wrong or an alternative way to begin this question?

Answer 1

HINT:

1. First, note that $a^2 \not = 1$, because $ab = 1$ and as $a \not = b$, it follows that $a^2 \not = ab$.

2. Next, $a^2 \not = a$, because $a \times 1 = a$.

3. Finally, $a^2 \not = 0$, because $F$ is a field and $a$ is a nonzero element.

What do you have left?