The Local-Global Principle

The Local-Global Principle that was discovered in the 1920s by Helmut Hasse (so it is also known as the Hasse Principle) was the first major discovery that pointed to the utility of p-adic numbers.

The set Q of the rational numbers is a (topological) field which is expanded to either R, the field of the reals, or to the various fields Q p, depending on the norm used. The expansion fields are quite different, but each expands Q algebraically, meaning that, for any pair of rational numbers r and s, the results of the operations r ± s, r×s, and r/s, being rational numbers, are preserved as such in all the expansions. So that, for example, r + s = t, if true in Q, is also true in R and all Q p. The terminology is this. Q is said to be a global field while all its expansions, R included, are said to be local. Thus any relation between a number of rational numbers which is true globally (i.e., in Q) is also true locally, i.e., in R and all the Q p's.

The Global-Local Principle asserts a partial converse for the equations involving quadratic forms with integer coefficients: ∑aijxiyj + ∑bixi + c = 0. If such an equation has solutions (even if a priori different) in R and, for every prime p, in Q p, then it has a rational solution in Q. In other words, a quadratic equation with integer coefficients has a global solution (i.e., in Q) if and only if it has solutions in all the local fields (i.e., in R and all the Q p.)

As an example, consider the equation x² - 2 = 0. We'll show that this equation has no solution in Q 5. The Local-Global Principle then will imply that the equation has no solution in Q. As a result, 2 is irrational.

First observe, that any solution of x² - 2 = 0 in Q 5 is a 5-adic integer. Indeed, |2|5 = 1 and, if α is a 5-adic solution, |α²|5 = |α|5|α|5 = 1, so that |α|5 = 1, meaning that α is a 5-adic integer. In particular, this means that α² ≡ 2 (mod 5k), for any integer k > 0. Let a be the zero term (an integer!) in the p-adic expansion of α which, as an expansion of a 5-adic integer, starts with index 0, then a² ≡ 2 (mod 5). But 2 is not a quadratic residue modulo 5: for none of the residues 0, 1, 2, 3, 4 satisfies a² ≡ 2 (mod 5). It follows that the equation x² - 2 = 0 has no solution in Q and the reason is that 2 is not a quadratic residue modulo 5!.

Note

  1. It is known that every real number has a finite p-adic norm for any prime p: R ⊂Q p. However, as the foregoing discussion shows, the embedding is not algebraic. rs = t in R does not imply rs = t in Q p, even though, rs = t in Q does.

  2. So, x² - 2 = 0 has no solution in Q 5. Curious as it may appear, another well known equation, x² + 1 = 0, does have a solution in Q 5. In no way this solution relates to the imaginary i = -1.

References

  1. J. R. Goldman, The Queen of Mathematics, A K Peters, 1998
  2. F. Q. Gouvêa, Local and Global in Number Theory, in The Princeton Companion to Mathematics T. Gowers (ed.), Princeton University Press, 2008

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Copyright © 1996-2017 Alexander Bogomolny

Strange as it may appear, we'll write

q = 1 + p + ... + pm + ...

As we just said, q is well defined as a limit of the sequence {xk} and this is how the identity should be understood. All arithmetic operations are permitted to carry over the limits so that

 q= 1 + p + ... + pm + ...
 pq= p + p² + ... + pm+1 + ...
 pq + 1= 1 + p + ... + pm + ... = q

So that pq + 1 = q, implying q = 1 / (1 - p). Taking p = 2 we obtain:

-1 = 1 + 2 + 4 + ...

One certainly needs time and effort to get used to such weirdness.

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Copyright © 1996-2017 Alexander Bogomolny

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