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The evident forgetful functor from fields to integral domains has a left adjoint, namely the construction of the quotient field for a given integral domain. Another standard construction is taking the group of divisibility of an integral domain, which involves taking the quotient of the group of units of the quotient field by the group of units of the original ring.

That is to say $G(R) := qf(R)^*/U(R)$.

This is a partially ordered abelian group under the ordering $aU(R)\leq bU(R)$ if and only if $\frac{b}{a}\in R^*$.

Does anyone know a functor from integral domains to partially ordered abelian groups that adequately describes this operation. I do not think that you can get away with the full category of integral domains however. I have described a functor when restricting the morphisms to the monomorphisms of integral domains.

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    $\begingroup$ I'm not sure I understand the question. Even the formation of the fraction field is not functorial with respect to arbitrary homomorphisms of domains (consider e.g. the map from $\mathbb{Z} \rightarrow \mathbb{Z}/2\mathbb{Z}$). Conversely, an injection of domains inducesa morphism of fraction fields and a morphism on unit groups, hence induces a morphism on divisibility groups, which is seen to be order-preserving. Does this answer your question, or is there more to it? $\endgroup$
    – Pete L. Clark
    Commented Oct 22, 2010 at 15:22
  • $\begingroup$ Right, that is what I am saying, is that there is no way to take the entire category of integral domains and do this. I was wondering if the category with only the monomorphisms was the best way to do it. In that case I would also be interested in hearing if anyone knew any of the properties of that particular category. $\endgroup$
    – Stines
    Commented Oct 22, 2010 at 15:44
  • $\begingroup$ I suspect you ought to consider more general localizations than the field-of-fractions (which is a less confusing name than "quotient field"). In other words, you want to consider pairs $(R,S)$ where $R$ is an integral domain and $U(R)\subset S\subset R$ and $S$ is closed under multiplication. The field of fractions corresponds to $S = R\backslash\{0\}$; when $\mathfrak{J}$ is prime, localizing at $\mathfrak{J}$ means taking $S=R\backslash \mathfrak{J}$. Inverting an element $2$ means taking $S=\bigcup_n 2^n U(R)$... $\endgroup$
    – some guy on the street
    Commented Oct 22, 2010 at 16:05
  • $\begingroup$ some guy-There are many standard terms for "localizing a domain at the nonzero elements", I learned it as quotient field. I am not interested in other localizations, just this one. Furthermore I am not interested in the localization itself, merely the functor and the category for which $R \mapsto G(R)$ can be adequately descibed. $\endgroup$
    – Stines
    Commented Oct 22, 2010 at 17:29
  • $\begingroup$ @Stines: we are somewhat out in space here, as you have not talked about about any particular motivation or application. In this vacuum, I suppose I would say that yes, the category of monomorphisms of domains is a natural one to take as the domain of this functor. But, again: I can't say whether it is "the best": what do you want to do with it? Also, asking if anyone knows any properties of some particular category seems like a rather unfocused question. All in all I think you'll need to put in more to get more of an answer than you already have. $\endgroup$
    – Pete L. Clark
    Commented Oct 22, 2010 at 23:42

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