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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 $C_i)$?

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Yes. This is actually an "if and only if" and is true for all kinds of finitary algebraic structures. – Zhen Lin Dec 27 '12 at 13:23
@ZhenLin Proof? – Makoto Kato Dec 27 '12 at 13:33
Related (very) abstract nonsense on the n-lab: compact object and 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. – Martin Dec 27 '12 at 13:59
up vote 2 down vote accepted

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.

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$B$ is an $A$-algebra of finite presentation if and only if it is isomorphic to $A[X_1,\dots,X_n]/I$, where $A[X_1,\dots,X_n]$ is a polynomial ring over $A$ and $I$ is a finitely generated ideal. I don't see why $B$ is a finitely presented model in $\mathbb{T}$. Could you explain it? – Makoto Kato Dec 28 '12 at 0:02
That's a somewhat high-tech definition if you think about it. Fundamentally, an algebra of finite presentation is something that is generated by finitely many elements and finitely many equations. This is what the coequaliser diagram expresses. – Zhen Lin Dec 28 '12 at 0:10
Sorry I don't understand. If you or someone else posts an answer which proves the statement of the title question directly, I will accept it. – Makoto Kato Dec 28 '12 at 0:20
A direct proof is ugly. You of all people should know that. Anyway, an $R$-module $M$ is of finite presentation if and only if there is an exact sequence of the form $R^n \to R^m \to M \to 0$. My coequaliser diagram is the non-additive version of this fact. – Zhen Lin Dec 28 '12 at 0:28
I suppose,in our case, $F X$ is a polynomial ring over $A$ in the coequalizer diagram $F Y \rightrightarrows F X \rightarrow B$. Could you explain what $F Y$ is in our case? – Makoto Kato Dec 28 '12 at 0:44

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