by Andi Bauer
Last Updated May 15, 2019 16:20 PM - source

I'm wondering whether the following addition and multiplication over the set $(\mathbb{R}/{0}\times \mathbb{Z}) \cup \{0\}$ define a field:

$$ (a,a')+(b,b')= \begin{cases} (a,a') \text{ if } a'>b'\\ (b,b') \text{ if } b'>a'\\ (a+b,a') \text{ if } b'=a' \text{ and } a\neq -b\\ 0 \text{ if } b'=a' \text{ and } a= -b\\ \end{cases} $$ $$ (a,a')(b,b')=(ab,a'+b') $$ $$ -(a,a')=(-a,a') $$ $$ (a,a')^{-1}=(a^{-1},-a') $$

[$0$ is the additive unit, which fixes addition and multiplication with $0$. $(1,0)$ the multiplicative unit.]

If yes, does this field have a name? If no, which of the axioms fail?

I'm a bit confused because I thought there are only "relatively few" different fields, such as the rational, real, complex numbers, or finite fields.

No, addition is not associative. For instance, $$((1,0)+(-1,0))+(1,-1)=0+(1,-1)=(1,-1)$$ but $$(1,0)+((-1,0)+(1,-1))=(1,0)+(-1,0)=0.$$

Note that you can tell something must be wrong with just the additive axioms, since your operation $+$ does not allow cancellation and so cannot be a group operation. Since there clearly is an identity and inverses, associativity must fail.

May 15, 2019 16:19 PM

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