>>9654963

Let (K,+,x) be a commutative ring with unit 1 and L:= K^2 (the cartesian product).

Is (a,b) and (c,d) are in K, define (a,b)+(c,d):= (a+c,b+d) and (a,b)*(c,d):= (ac-bd, ad+bc). Define also i:= (0,1). It turns out that (L,+,*) is also a commutative ring with unit = (1,0), the map f:x->(x,0) from K to L is an injective ring homomorphism, and for every x,y in K we have (x,y)= f(x)+ i * f(y) (since f i s injective, if we identify t and f(t) for every t, we get the more familiar notation (x,y) = x+i*y). Wed also have i*i = -1 from the definition.

In addition if K is a field and for every x in K, x^2 is different from -1, then L is also a field (for every a,b in K, we cant have a^2+b^2=0 otherwise we'd get a^2=-b^2 then (a/b)^2=-1 contradicting the above: that being said, you can check that (a/(a^2+b^2), -b/(a^2+b^2) ) is the inverse of (a,b) for every a,b in K.