Is that the whole question? Given that F is a field, there are two operations defined on the set F. But "FxF" is a set, there is NO operation automatically defined on it. I might suspect that you mean to define them "coordinate" wise: (a, b)+ (c, d)= (a+c, b+d) and (a, b)(c, d)= (ac, bd) but you just gave the complex numbers as (F,F) where that is not true. But perhaps that is what you mean (in which case C is NOT RxR as you wrote) since if RxR is NOT a field with those definitions.Then, on the second part of it, they say:
Let F be a field. Let's assume there's an a(that belongs to)F, so that a^2+1=0 . Prove that FxF isn't a field.
I can understand that it's like asking me to prove that a,b on Z=a+bi has to be that (a,b(belong to)R) on a field, because in this question it's asked to prove that it's impossible that there would be a field with (a,b) so that (a,b) can be imaginary.
Now, in order to prove that a group is NOT a field, it's enough to show that it has a,b(belong to)F so that a,b(are not equal to zero), and a*b=0.
I just couldn't find any way to prove that ^...
Would any of you mind to help me?
Thank you very much
One important property of a field is that every non-zero member has a multiplicative inverse. (a, 0) for a non-zero is not 0 for this field because (a, 0)+ (b, c)= (a+b, c) which is not equal to (b, c). Does (a, 0) have a multiplicative inverse?
Or, since you say "it's enough to show that it has a,b(belong to)F so that a,b(are not equal to zero), and a*b=0", what is (a,0)*(0,b)?
(Notice, again, that stating the operation is important. Using the definition of "multiplication" that makes RxR into C, (a,0)*(0,b)= a*0- b*0+ a*b+ 0*0= ab which is NOT 0.)