Can we make $\mathbb{Z}$ into a field?

Question

This is probably an elementary question about fields, but I think it is a little tricky.

Can we make the integers $\mathbb{Z}$ into a field?

Let me be more precise. Is it possible to make $\mathbb{Z}$ into a field so that the underlying additive group structure is the usual addition in $\mathbb{Z}$? In other words, we just need to define "multiplication" on $\mathbb{Z}$ that makes it into a field. But the question is: How to define such a multiplication?

The ordinary multiplication (that we learn in elementary school) doesn't work as $2$ does not have a multiplicative inverse.

Answer 1

The multiplicative structure in $\mathbb{Z}$ is determined by the additive structure. We start with the axiom of the multiplicative identity: $1 \times n = n$ for all $n \in \mathbb{Z}$. The axiom of distributivity implies $$2 \times n = (1 + 1) \times n = (1 \times n) + (1 \times n) = n + n = 2n$$

This determines multiplication by positive integers. I'll leave it to you to show that it is also determined for the negative integers and zero.

Morally: The distributive property tells us how to multiply in terms of how to add, assuming we know how to multiply certain "basic elements". In the case of $\mathbb{Z}$, the entire ring is generated by the basic element $1$, and multiplication by $1$ is governed by the axioms.

Answer 2

Theorem: If $F$ is a field and $x \in F$ with $x \neq 0$, then either:

• there exists $y \in F$ such that $y + y = x$, or
• $x + x = 0$.