Let $s$ be a formal symbol.
Define addition and multiplication operations on the set $S := {a+bs : a,b ∈ R}$ (with curly brackets) by the rules
$(a+bs)+(c+ds) := (a+c)+(b+d)s$,
$(a+bs)(c+ds) := (ac+2bd)+(ad+bc)s$.
a) Find $α, β ∈ R$ such that $s^ 2 = α+βs$.
b) By considering the product $( √ 2+s)( √ 2−s)$ inside $S$ or otherwise, prove that $S$ is not a field. [8] [You may assume that multiplication in $S$ is associative.]
For a), I took $(0 + 1s)(0 + 1s) = α + βs$, which gave me $α = 2, β = 0$, so $s^2 = 2$... is this correct?
For b), the product is $0$ (which is logical since $(√2−s) = 0$), but what ring axiom does this fail? The only real candidate is the multiplicative inverse law, but I can't yet see how.
a) Your answer is correct: $s^2=2$.
b) In a field, if the product of two elements is $0$, then (at least) one of them is $0$, which is not the case here.