Minimality of field of fractions expressed by functor

435 Views Asked by At

Express the construction of the field of fractions as a functor.

Mac Lane's book Categories for the Working Mathematician

I'll call $\mathbf{Int}$ the category of integral domains, and $\mathbf{Field}$ the category of fields, each with monomorphisms as arrows. It's easy to realize that, once we define the field of fractions $\mathcal K(R)$ of the integral domain $R$, this defines a functor $$\mathrm{K}:\mathbf{Int}\to\mathbf{Field}$$ objects: $$\mathrm{K}R = \mathcal K(R)$$

arrows: $$\mathrm Kf:\mathrm K R\to\mathrm KR' \atop \qquad\qquad \,\,a/b\mapsto f(a)/f(b)$$ It is indeed well defined, respecting identity arrows and compositions.

Conversely though, I don't think that any functor $\bf Int\to Field$ defines the field of fractions of an integral domain, seeing as there are many other fields I can associate to an integral domain. For example, the mapping $\mathrm{X}:\mathbf{Int}\to\mathbf{Field}$ defined on objects by $$\mathrm XR = \mathcal K(R[X]) := R(X)$$ and on arrows by $$\mathrm Xf:\mathrm XR\to\mathrm XR' \atop \qquad\quad p(x)\mapsto ((\mathrm Kf)p)(x)$$ with $(\mathrm{K}f)p$ defined as "do $\mathrm Kf$ to the coefficients of $p$", is also a functor. My question is, what condition do I have to assume of such a functor, if the resulting object is to be the field of fractions of $R$, i.e. the smallest field containing $R$? I have slipped a bit in using elements, of course my next goal will be to use the "arrows definition" of a ring. For now though I'd accept an answer in these terms.

1

There are 1 best solutions below

1
On BEST ANSWER

You want your functor $K$ to be left adjoint to the forgetfulish$^*$ functor $U: \mathbf{Field} \to \mathbf{Int}$, defined on objects by $U(F) = F$. This means that for any integral domain $R$ and any field $F$, \begin{align} \DeclareMathOperator{\Hom}{Hom} \Hom_{\mathbf{Field}}(KR, F) &\cong \Hom_{\mathbf{Int}}(R, UF) \\ &= \Hom_{\mathbf{Int}}(R, F) \end{align}

Perhaps you will enjoy looking at this list of examples.


$^*$ Usually, a forgetful functor maps into $\mathbf{Set}$ by forgetting all algebraic structure, but here we are not losing so much information.