If k is a field in which 1 + 1 &#x2260;<!-- ≠ --> 0 , prove that <msqrt>

vasorasy8

vasorasy8

Answered question

2022-07-12

If k is a field in which 1 + 1 0, prove that 1 x 2 is not a rational function. Hint. Mimic the classical proof that 2 is irrational
It seems rather odd to discuss the square root function in the context of an arbitrary field, but here goes nothing!
Proof:
Suppose that 1 x 2 is a rational function in k ( x ). Then there exist polynomials p ( x ) and q ( x ) 0 that are relatively prime with 1 x 2 = p ( x ) q ( x ) ; clearly we may take p ( x ) 0. Then 1 x 2 = p ( x ) 2 q ( x ) 2 or q ( x ) 2 ( 1 x 2 ) = p ( x ) 2 , which says q | p 2 , and since q | q 2 , we can infer q | p 2 . Moreover, since p | p 2 and ( p , q ) = 1, then p q | p 2 or q | p, which contradicts the fact that the polynomials are relatively prime. Hence 1 x 2 cannot be a rational function.
How does this sound? Aside from the problem of discussing –     in the context of an arbitrary field, I am worried about not using the fact that 1 + 1 0, at least not explicitly. Where exactly is this assumed used, if at all?
EDIT:
Suppose that q ( x ) = 1, and let f ( x ) = a n x n + . . . a 1 x + a 0 . Then 1 x 2 = f ( x ) 2 . If x = 0, 1 = a 0 2 and therefore a 0 must be a unit. Letting x = 1 we get 0 = ( a n + . . . + a 1 + a 0 ) 2 ; and letting x = 1 we get 0 = ( a n . . . a 1 + a 0 ) 2 . Since we are working in a field, there can be no nonzero nilpotent which means that a n + . . . + a 1 + a 0 and a n . . . a 1 + a 0 are both zero. Adding the two equations together yields 2 a 0 = 0, and since 1 + 1 0, a 0 = 0 which contradicts the fact that a 0 is a unit.
How does this sound?
Another attemtpt:
If f ( x ) 2 = 1 x 2 , then 2 = deg ( f 2 ) = 2 deg ( f ) implies deg ( f ) = 1. Yet if x = 1, we get f ( 1 ) 2 = 1 1 = 0 and therefore f ( 1 ) = 0 since there are no nonzero nilpotent elements in a field; similarly, f ( 1 ) = 0. Since 1 + 1 0, f has two distinct roots in k yet is only a 1-st degree polynomial, a contradiction.

Answer & Explanation

Elijah Benjamin

Elijah Benjamin

Beginner2022-07-13Added 10 answers

Your proof is not quite complete, because q | p is not a complete contradiction to ( p , q ) = 1: we could have q = 1. We must rule out that case, i.e., we must show that 1 x 2 is not the square of a polynomial. That is where we will use the assumption that 1 + 1 0.

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