Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In this question it is shown that being able to compute reciprocals (together with sums and differences) is enough to do do multiplication in a field of characteristic $\ne 2$. That made me wonder:

Can we formulate a set of field axioms that are based on the reciprocal function rather than a multiplication operation? Something like

A field is an abelian group $G$ (written additively) together with a designated element $1\ne 0$ and an involution $x\mapsto\frac{1}{x}$ of $G\setminus\{0\}$ such that

  • $\frac 11=1$
  • ???

Obviously we could derive a sufficient set of axioms by translating the usual field axioms using the algorithm from the earlier question, but that would end up being very ugly (not least because there are several cases that one must handle specially in order not to divide by zero). Is there a nicer way to characterize involutions that arise as the reciprocal of some field?

Partial answers (for example, ones that work only for characteristic 0) would also be interesting.

share|improve this question
1  
I'm not sure that's what's shown in that question. Both answers (J.D.'s and N. S.'s in a comment) make assumptions about the operands; in particular, in both answers $A+B\ne1$ is assumed. –  joriki Mar 18 '12 at 1:15
    
@joriki: Sure, but it still looks like it would amount to an algorithm if you're allowed to compare intermediate results to known constants and change your strategy when you hit a special case. For example, check whether $A+B=1$ and if so compute $(A+1)B-B$ instead. –  Henning Makholm Mar 18 '12 at 1:29
2  
Related: The corresponding characterization of fields using the map $x \mapsto 1-x$ can be found here: math.uga.edu/~pete/Dicker1966.pdf : Dicker, A set of independent axioms for a field and a condition for a group to be the multiplicative group of a field. –  Martin Brandenburg Mar 18 '12 at 1:29
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.