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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.

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    $\begingroup$ Define $a*b=a+b+1$ $\endgroup$ – happymath Feb 23 '14 at 7:36
  • $\begingroup$ One potential solution is to consider $\mathbb{Z}_p$, i.e. the integers modulo $p$, where $p$ is a prime, which is a field. However, I acknowledge that this involves a redefinition of $\mathbb{Z}$ rather than multiplication; so it may not be quite what you're looking for. $\endgroup$ – Newb Feb 23 '14 at 7:36
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    $\begingroup$ @user129901 - that won't work. $0*b=0+b+1\neq 0$. $\endgroup$ – nbubis Feb 23 '14 at 7:41
  • $\begingroup$ @nbubis here the multiplicative identity is $-1$. And we need $a*e=e*a=a$. Putting $e=-1$ satisfies it. $\endgroup$ – happymath Feb 23 '14 at 7:48
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    $\begingroup$ As the answers (+1) indicate, you cannot do this with normal addition. You need to redefined both addition and multiplication. A simple alternative is to use a bijection $\Bbb{Z}\to\Bbb{Q}$, and pull back the operations on the rationals. For my money a more interesting variation is to restrict to non-negative integers, and use the NIM-addition (i.e. the usual base-2 addition, but WITHOUT CARRY). Surprisingly then there is a matching multiplication that turns $\Bbb{N}$ into a field. See here. $\endgroup$ – Jyrki Lahtonen Feb 23 '14 at 7:59
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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.

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    $\begingroup$ You aren't determining the multiplicative structure from the additive: you're determining the multiplicative structure from the additive and a choice of integer to be the identity. An example of another multiplicative structure is $a \times b = -ab$. Those might be the only two: I haven't worked through the details. $\endgroup$ – Hurkyl Feb 23 '14 at 7:41
  • $\begingroup$ Thanks for your answer! The multiplicative identity needs not be the usual $1$, as Hurkyl points out. But it is a very useful answer! +1 $\endgroup$ – Prism Feb 23 '14 at 7:42
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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
  • $x + x = 0$.
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    $\begingroup$ Thanks! (+1) I don't immediately see where this takes me, but give me a little bit of time to work it out :) $\endgroup$ – Prism Feb 23 '14 at 8:22

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