Suppose the field axioms include 0-1. Prove that, in this case, every element is equal to 0. Thus the existence of 0-1 would contradict the field axiom that 1≠0.
The Attempt at a Solution
My question regarding the proof is, why bother to show that every element in the field is 0 in order to show that 1≠0. In other words, isn't it easier to say:
Previously proven lemma: For all x in F, 0*x = 0.
Suppose there exists 0-1 in F such that 0*0-1=1. Then by the lemma above, the left side of the previous equation simply reduces to 0, and hence we are left with 1 = 0, a contradiction.