We assume all rings considered are commutative. Let $A$ be a ring. Let $B$ be an $A$-algebra of finite presentation. Let $I$ be a small filtered category. Let $C\colon I \rightarrow$ $A$-alg be a functor, where $A$-alg is the category of $A$-algebras. Then colim $Hom_{A-alg}(B, C_i)$ is canonically isomorphic(as a set) to $Hom_{A-alg}(B,$ colim $C_i)$?
On colim $Hom_{A-alg}(B, C_i)$
-
1Related (very) abstract nonsense on the n-lab: [compact object](http://ncatlab.org/nlab/show/compact+object) and [small object](http://ncatlab.org/nlab/show/small+object). For the statement in Zhen Lin's comment they refer to Corollary 3.13 of Adamek and Rosicky's Locally presentable and accessible categories. – 2012-12-27
1 Answers
Let $\mathbb{T}$ be a finitary algebraic theory, such as the theory of commutative $A$-algebras for a commutative ring $A$. Recall, a model $B$ of $\mathbb{T}$ is finitely presented if and only if there exist finite sets $X$ and $Y$ and $\mathbb{T}$-homomorphisms giving a coequaliser diagram $F Y \rightrightarrows F X \rightarrow B$ in the category of $\mathbb{T}$-models, where $F$ is the free $\mathbb{T}$-model functor. More explicitly, if $p_1, p_2 : F Y \to F X$ are the two $\mathbb{T}$-homomorphisms shown above, then $B$ is the $\mathbb{T}$-model generated by $X$ and subject to the equations $p_1(y) = p_2(y)$ for all $y$ in $Y$; conversely, given a finite presentation of $B$, we can construct such a coequaliser diagram.
By the universal property of coequalisers, there is therefore an equaliser diagram $\textrm{Hom}_\mathbb{T}(B, C) \rightarrow \textrm{Hom}_\mathbb{T}(F X, C) \rightrightarrows \textrm{Hom}_\mathbb{T}(F Y, C)$ in the category of sets for every $\mathbb{T}$-model $C$, and by the universal property of free objects, these diagrams are isomorphic to $\textrm{Hom}_\mathbb{T}(B, C) \rightarrow \textbf{Set}(X, U C) \rightrightarrows \textbf{Set}(Y, U C)$ where $U$ is the forgetful functor. Now suppose $C$ is the colimit of some filtered system $C_i$. It is well-known that $U$ preserves filtered colimits, so $U C \cong \varinjlim U C_i$, and $\textbf{Set}(X, -)$ preserves filtered colimits if (and only if!) $X$ is finite, so we get equaliser diagrams $\textrm{Hom}_\mathbb{T}(B, C) \rightarrow \varinjlim \textbf{Set}(X, U C_i) \rightrightarrows \varinjlim \textbf{Set}(Y, U C_i)$ but filtered colimits preserve equalisers, so we also have an equaliser diagram $\varinjlim \textrm{Hom}_\mathbb{T}(B, C_i) \rightarrow \varinjlim \textrm{Hom}_\mathbb{T}(F X, C_i) \rightrightarrows \textrm{Hom}_\mathbb{T}(F Y, C_i)$ and therefore $\textrm{Hom}_\mathbb{T}(B, C) \cong \varinjlim \textrm{Hom}_\mathbb{T} (B, C_i)$, as claimed.
Conversely, suppose $\textrm{Hom}_\mathbb{T}(B, -)$ preserves directed colimits. It is well-known that the lattice of finitely-generated $\mathbb{T}$-submodels $B_i \subseteq B$ form a directed system whose colimit is $B$ itself, so $\textrm{Hom}_\mathbb{T}(B, B) \cong \varinjlim \textrm{Hom}_\mathbb{T}(B, B_i)$ and in particular there exists a finitely-generated $\mathbb{T}$-submodel $B_i \subseteq B$ and a $\mathbb{T}$-homomorphism $s : B \to B_i$ such that $r \circ s = \textrm{id}_B$ where $r : B_i \to B$ is the inclusion; but $r$ is injective, so this implies $B_i = B$ and $r = \textrm{id}_B$. Thus $B$ is finitely-generated, and there is a surjective $\mathbb{T}$-homomorphism $F X \to B$ where $X$ is a finite set.
However, this does not prove that $B$ is finitely-presented. For this, we need to be a bit more clever. Consider all factorisations of $F X \to B$ as $F X \to C_i \to B$ where $C_i$ is finitely-presented and $F X \to C_i$ is surjective. This forms a directed diagram in a fairly obvious way, and it is not hard to see that $B \cong \varinjlim C_i$ as well. The same argument as before gives us $\mathbb{T}$-homomorphisms $r : C_i \to B$ and $s : B \to C_i$ such that $r \circ s = \textrm{id}_B$. Consider the idempotent $\mathbb{T}$-homomorphism $s \circ r : C_i \to C_i$. It is not hard to check that we have a (split!) coequaliser diagram $C_i \rightrightarrows C_i \rightarrow B$ where the two arrows $C_i \to C_i$ are $\textrm{id}_{C_i}$ and $s \circ r$; on the other hand, since $C_i$ is finitely presented, one can easily construct a finitely-presented coequaliser of $\textrm{id}_{C_i}$ and $s \circ r$, so $B$ must itself by finitely presented, since coequalisers are unique up to unique isomorphism.
-
0I didn't know that an algebra of finite presentation could be expressed by a coequalizer diagram(though$I$knew a module of finite presentation could). That's why I said I didn't understand your proof. Yes, it's obvious once you know it. But most of category theory results are like that. – 2012-12-29