Functoriality of a standard integral domain construction. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T08:05:12Z http://mathoverflow.net/feeds/question/43189 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43189/functoriality-of-a-standard-integral-domain-construction Functoriality of a standard integral domain construction. Stines 2010-10-22T15:13:06Z 2010-10-22T15:13:06Z <p>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. </p> <p>That is to say $G(R) := qf(R)^*/U(R)$. </p> <p>This is a partially ordered abelian group under the ordering $aU(R)\leq bU(R)$ if and only if $\frac{b}{a}\in R^*$. </p> <p>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. </p>