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July 21

Valuation ring with idempotent maximal ideal

Does there exist a valuation ring R, not a field, whose maximal ideal m is idempotent (m² = m)? GeoffreyT2000 (talk) 02:27, 21 July 2015 (UTC)[reply]

Per one of the examples in the article "valuation ring", the ring of finite elements of the hyperreals is a valuation ring, and the ideal of infinitesimals is idempotent, as the infinitesimals are closed under square root. (It did make me think a little, though. Thank you.) — Arthur Rubin (talk) 05:51, 21 July 2015 (UTC)[reply]

In fact, the finite elements of any non-Archimedean real-closed field will do, for the same reason. — Arthur Rubin (talk) 06:01, 21 July 2015 (UTC)[reply]