Suppose $(R,+,-,*,0,1,\leq)$ is a dense linearly ordered commutative non-trivial ring without infinite or infinitesimal elements. Is $R$ a field?
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"Dense" as in "densely ordered"? Just making sure.... – rschwieb Aug 21 '17 at 16:16
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No. Take, for instance, the ring of dyadic fractions: $$\left\{\frac p{2^n}\mid p\in \Bbb Z, n\in \Bbb N\right\}\subseteq \Bbb Q$$It doesn't contain the multiplicative inverse of $3$, for instance, so it's not a field, but it is easily seen to be non-trivial, linearly ordered and dense, commutative, and having neither infinite nor infinitesimal elements.
Arthur
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