To make sure that we are talking about the same, I would like to post the relevant definitions I know first.
Definitions:
A pair $(G, +)$ where $G$ is a set and
$+: G \times G \rightarrow G$
is called a commutative group if it has all of the following three characteristics:
- (associativity): $\forall x,y,z \in G: (x+y)+z=x+(y+z)$
- (identity element): $\exists e \in G \forall x \in G: e + x = x = x +e$
- (inverse elements): $\forall x \in G \exists x^{-1} \in G: x^{-1} + x = e = x + x^{-1}$
- (commutativity): $\forall x,y \in G: x+y=y+x$.
A triple $(\mathbb{K}, +, \cdot)$, where $\mathbb{K}$ is a set and
$+: \mathbb{K} \times \mathbb{K} \rightarrow \mathbb{K}$
$\cdot: \mathbb{K} \times \mathbb{K} \rightarrow \mathbb{K}$
is called a field if it has all of the following three characteristics:
- $(\mathbb{K}, +)$ is a commutative group
- $(\mathbb{K} \setminus \{0\}, \cdot)$ is a commutative group
- distributive properties: $\forall x,y,z \in \mathbb{K}: x \cdot (y+z) = (x \cdot y) + (x \cdot z)$ and $\forall x,y,z \in \mathbb{K}: (y+z) \cdot x = (y \cdot x) + (z \cdot x)$
$\mathbb{N^+} := 1, 2, 3, 4, 5 …$ (all positive integers)
$\mathbb{N}_0 := \{0\} \cup \mathbb{N}$ (the natural numbers with zero)
$\mathbb{N}$ can be both: $\mathbb{N^+}$ or $\mathbb{N}_0$.
My question: Does a field exist that has $\mathbb{N}$ as its set?
I know that $(\mathbb{Q}, +, \cdot)$ is a field and I know that $(\mathbb{N}, + )$ is only a commutative semigroup.
But maybe it's possible to define two mappings $\circ, *$ for $\mathbb{N}$ in such a way that $(\mathbb{N}, \circ, *)$ is a field.
Related:
The smallest field containing the integers is the field of rational
numbers.
Source: en-Wikipedia: Integer
I don't know if this is true. If somebody can prove (or at least scribble the proof of) the quote it would be a good answer, I guess.
Best Answer
$\Bbb Q$ is a countably infinite set, so there is a bijection $\varphi:\Bbb N\to\Bbb Q$. For $m,n\in\Bbb N$ define
$$m\circ n=\varphi^{-1}\Big(\varphi(m)+\varphi(n)\Big)$$
and
$$m\ast n=\varphi^{-1}\Big(\varphi(m)\cdot\varphi(n)\Big)\;;$$
then $\langle\Bbb N,\circ,\ast\rangle$ is a field isomorphic to $\langle\Bbb Q,+,\cdot\rangle$.
Added: For your last question, the proof is to note that every field $F$ containing a copy of $\Bbb N$ with the usual operations must also contain a copy of $\Bbb Q$ with the usual operations. In order for $F$ to be an additive group, it must contain an element $-n$ for each $n\in\Bbb N$, and in order for it to be a field, it must contain an element $n^{-1}$ for each $n\in\Bbb Z^+$. Once you have all of the integers and all of the reciprocals of positive integers, it’s easy to see that you must have all of the rationals just to get closure under multiplication.
Note that I’m slurring over a lot of details about isomorphic copies in order to give the fundamental idea.