Consider the language L = { + , ⋅ , 0 , 1 } of rings. It is easy to show using compactness that if σ is a sentence that holds in all fields of characteristic 0, there is some N ∈ N such that σ holds for all fields of characteristic p ≥ N. A sort of converse would be, if σ is a sentence that holds in all fields of positive characteristic, σ is true in all fields of characteristic 0.

Question

Consider the language   L  =  {  +  ,  ⋅  ,  0  ,  1  } of rings. It is easy to show using compactness that if   σ is a sentence that holds in all fields of characteristic   0, there is some   N  ∈      N   such that   σ holds for all fields of characteristic   p  ≥  N. A sort of converse would be, if   σ is a sentence that holds in all fields of positive characteristic,   σ is true in all fields of characteristic   0.

Carson Mueller · Accepted Answer

The four square theorem yields the following theorem in positive characteristic:  ∃  a  ,  b  ,  c  ,  d  :      a    2    +      b    2    +      c    2    +      d    2    +  1  =  0

Tatiana Cook · Accepted Answer

The canonical solution is to use the fact that       x    2    +      y    2    =  −  1 has a solution in every finite field, which can be proved by a simple counting argument. That equation is solvable in any field of positive characteristic, because it is solvable in the prime subfield. But not in       R  .This is the same idea as the other answer, but the proof of solvability is easier than the   4-squares theorem.

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