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I think that these two sets cannot be turned into fields by re-defining addition or multiplication (or both) but I am not sure how to prove this only from axioms of the field and (if needed) some elementary properties of fields that follow directly from axioms of the field.

My intuition is that $\mathbb N_0$ and $\mathbb Z$ are not "dense enough", as is, for example, $\mathbb Q$, to be able to become fields.

How would you prove that we can(or cannot) turn them into fields?


marked as duplicate by Sil, Theoretical Economist, Lord Shark the Unknown, José Carlos Santos, Namaste Aug 12 '18 at 15:34

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  • 1
    $\begingroup$ If you take any bijection $\mathbb{Z}\to\mathbb{Q}$, then pulling back the addition and multiplication from $\mathbb{Q}$, you get a field. However, you cannot expect to preserve other properties, like being an ordered field, with the original order in $\mathbb{Z}$. It will be an ordered field if you also pull back the order from $\mathbb{Q}$. $\endgroup$ – user583012 Aug 11 '18 at 13:58
  • $\begingroup$ en.wikipedia.org/wiki/Transport_of_structure $\endgroup$ – Kenny Lau Aug 11 '18 at 14:08
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    $\begingroup$ A field need not be dense at all. Finite fields are examples. The smallest field $F_{2}$ has cardinality 2! $\endgroup$ – NivPai Aug 11 '18 at 14:09
  • $\begingroup$ $\Bbb{N}_0$ becomes a field when the addition is bitwise XOR and the multiplication is defined by a method due to Conway. See this answer and this question for links to more (you can safely ignore my answer to that question, it is not relevant here). $\endgroup$ – Jyrki Lahtonen Aug 11 '18 at 18:16

Let $X$ be any countably infinite set. Then we have a bijection $\alpha : X \to \mathbb{Q}$. Define $$x\oplus y = \alpha^{-1}(\alpha(x) + \alpha(y)) \quad \quad x\otimes y = \alpha^{-1}(\alpha(x)\alpha(y))$$ where the operations on the right are the natural ones in $\mathbb{Q}$, then you have $(X, \oplus, \otimes)$ a field (isomorphic to the rationals).

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    $\begingroup$ so if $\alpha: \ldots,-3,-2,-1,0,1,2,3,4,5,6,\ldots \to \ldots,-\frac12,-2,-1,0,1,2,\frac12,3,\frac13,4,\ldots$ you get $1\oplus2 = 4$ and $2 \otimes 2 = 6$ $\endgroup$ – Henry Aug 11 '18 at 14:08
  • $\begingroup$ This answer is exactly the same as the one posted here $\endgroup$ – D. G. Aug 12 '18 at 2:10

For prime $p$, $\mathbb{Z}_{p}$ is a field, under addition and multiplication modulo $p$.


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