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        riddles >> putnam exam (pure math) >> Only the trivial ring adds this way
        
(Message started by: ecoist on Dec 13^th, 2006, 2:28pm)

Title: Only the trivial ring adds this way
Post by ecoist on Dec 13^th, 2006, 2:28pm

Let R be a ring whose additive structure is the rationals under addition modulo 1. Show that all products in R equal 0.

Title: Re: Only the trivial ring adds this way
Post by Eigenray on Dec 14^th, 2006, 2:26pm

(a/b)*(c/d)
= (a/b)*(bc/bd)
= (a/b)*(c/bd + ... + c/bd) [b times]
= (a/b)*(c/bd) + ... + (a/b)*(c/bd)
= (a/b + ... + a/b)*(c/bd)
= (a)*(c/bd)
= 0*(c/bd)
[ = (0+0)*(c/bd)
= 0*(c/bd) + 0*(c/bd)
= (a/b)*(c/d) + (a/b)*(c/d)]

so (a/b)*(c/d) = 0.

Or: A multiplication on a ring with additive group G is an element of Hom_Z( G (x)_Z G, G). But if G is divisible and torsion, then G (x)_Z G = 0.