Can $\mathbb{Z}$ be endowed with operations that give it the structure of a field?

Question

Does there exist some definition of addition and multiplication for which the set of all integers is a field?

Answer 1

Yes, but not a very natural one.

Since $\mathbb{Z}$ and $\mathbb{Q}$ are both countable, there exists a bijection $\Phi: \mathbb{Z} \to \mathbb{Q}$. With the operations $$m \oplus n := \Phi^{-1}(\Phi(m) + \Phi(n)), \qquad m \otimes n := \Phi^{-1}(\Phi(m)\Phi(n)),$$ $(\mathbb{Z}, \oplus, \otimes)$ is a field, but NB that $\oplus$ and $\otimes$ won't have much to do with the usual addition and multiplication operations on $\mathbb{Z}$.

Answer 2

Yes, but the answer isn't really satisfying. The set of integers is countably infinite. It's therefore in bijection with $\mathbb{Q}$, the field of rational numbers. By transporting the field operations (addition and multiplication) through the bijection, you get a field structure on the set of integers.

Answer 3

There is a countably infinite field of characteristic $2$ whose elements can be identified with the non-negative integers in a natural way.

Addition can be defined by $2^n+2^m=2^n+2^m$ if $m\neq n$ and $= 0$ if $m=n$ and extended by the commutative and associative laws. This is "nim addition"

Multiplication can be defined by $2^{2^m}\cdot 2^{2^n}=2^{2^m+2^n}$ if $m\neq n$ and $3\cdot 2^{2^n-1}$ if $m=n$ again extended by associativity and by the distributive law.

There are other ways of defining the field operations. See Nimbers.

Clearly any countable field can by definition be put into a bijection with the integers. I thought you might be interested in this rather different construction, even though it doesn't quite do everything you want.