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I'm looking for a definition like this：

In integral domains, if $ b\neq 0 $ and $b \mid a $, we define $\frac{a}{b}$ (or $a/b$ ) to be the only element $x$ such that $bx=a$.

I see many people on MSE use this division in their proofs (see 1 and 2) without declaration, but I can't find any information about it in formal writings such as algebra books from Artin, Lang, or Hungerford.

My question is that: Is the definition suitable?

If this division is well-defined, what useful properties does it have?

Any direct answers or links to other places are much appreciated.

Sky subO
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  • Have you tried proving the definition is well-defined? – Somos Aug 24 '21 at 02:20
  • Yes. Since integral domains satisfy the cancellation law, I think we can easily prove that x is unique. – Sky subO Aug 24 '21 at 07:58
  • Furthermore, if $b$ is a unit, this new division is consistent with our usual definition of division, which defines $\frac{a}{b}$ to be $ab^{-1}$. – Sky subO Aug 24 '21 at 16:49

1 Answers1

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For any ring $R$ and $a\in R$, the map $\mu_a:R\to R,\: x\mapsto ax$ is a morphism from the additive group $(R,+)$ to itself (indeed, it is even a morphism of (right) $R$-modules $R_R\to R_R$). An integral domain can be characterised as a commutative ring with identity element such that $\mu_a:R\to R$ is injective for all $a\ne0$. For $b\in \mathrm{im}(\mu_a)$, it certainly does make sense to define $b/a := \mu_a^{-1}(b)$.

Most of the usual properties can be proved by working with the maps $\mu_a$. For example, since $\mu_{ab}(x) = (ab)x = a(bx) = \mu_a(\mu_b(x))$ for all $x$, we have $\mu_{ab} = \mu_a\circ\mu_b$, and hence $(c/b)/a = \mu_a^{-1}(\mu_b^{-1}(c)) = (\mu_{b}\mu_a)^{-1}(c) = \mu_{ba}^{-1}(c) = c/(ba)\;$ (whenever defined).

Erik D
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