Definition talk:Field Axioms

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If avoiding the commutativity of addition axiom for rings, why include it for fields? --Lord_Farin 18:48, 1 June 2012 (EDT)

Thought it was a good idea at the time. Depends on what source I found the material in. --prime mover 18:53, 1 June 2012 (EDT)
True, true. But then a comment needs to be put that the axiom is derivable from (some of) the others. --Lord_Farin 03:23, 2 June 2012 (EDT)
I'm going to undertake a rewrite. --prime mover 04:28, 2 June 2012 (EDT)

Maybe we should wait until it has crystallised whether or not the commutativity is in fact derivable. --Lord_Farin 04:43, 2 June 2012 (EDT)

Since the Definition:Trivial Ring is already compromised in this respect (i.e. the zero operation on a non-abelian group would be a ring under this definition) it's probably worth doing the conventional thing here. It was probably over-ambitious of me to try and present the axioms as I did. --prime mover 07:21, 2 June 2012 (EDT)
No problem there; not trying is certainly not bringing success either. --Lord_Farin 07:26, 2 June 2012 (EDT)