'Multiplying' by 0 in a field, field axiom proofs

Question

The question says:

The solution set was posted and there are a few things I don't quite understand from it.

For the first one, I'm not entirely sure what's happening. It appears to be using the case where a² = 0, but I'm not sure where the (a · a) comes from. If that's supposed to represent a², I don't recall a field axiom that would make that work as well as have it end up being 'a' in the end.

The second one seems to use the multiplicative inverse but I'm unsure as to what a² · a^-1 would return. The multiplicative inverse would seem to work only for a · a^-1, not for a² unless I'm misunderstanding the field axiom.

The last one seems the result already proven but also uses the multiplicative inverse axiom in a way I don't quite understand with a² · a^-1 = 1, apparently.

Any enlightenment on this question would be very appreciated.

Answer 1

$a^2$ is an informal abbreviation for $a\cdot a$.

As to your later question about $a^2\cdot a^{-1}$, it means $(a\cdot a)\cdot a^{-1}$. By associativity this is $a\cdot (a\cdot a^{-1})$, which is $a\cdot 1$, which is $a$.

Answer 2

There appears to be a typo in the last bit. Here's what I would write:

If $a^2=1$, then - since we know that $ab=1$ - we have $a^2=ab$. So multiplying both sides by $a^{-1}$ we get $a=b$, a contradiction.

I don't see how the author deduces $b=1$ here; I think "$1$" should be "$a$".