# Simple field proof

## Homework Statement

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.

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Fredrik
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## Homework Statement

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.
I agree, your way is simpler. Not sure why they want you to do this. Maybe because 1≠0 isn't always included in the field axioms. If you're working with a definition that doesn't include that, it looks like you can only conclude that the field is trivial (i.e. that it has only one element).