Let $C$ be a filtered subcategory of the category of commutative algebras over a fixed field $k$ whose objects are all integral domains. Then the colimit of the obvious diagram is an integral domain. Does this statement also hold in the case where we drop the commutativity condition?
1 Answer
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I don't see how commutativity matters.
Suppose $A$ is the filtered colimit of algebras $A_i$ and $x,y\in A$ with $xy=0$. Then $x$ is represented by $x_j\in A_j$ and $y$ by $y_k\in A_k$ for some $j$ and $k$. By "filtered", these map to elements $x_l,y_l\in A_l$ for some $l$ such that $x_ly_l=0$. Since $A_l$ is a domain, either $x_l=0$ (and therefore $x=0$) or $y_l=0$ (and therefore $y=0$).
answered Mar 2, 2016 at 9:38
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$\begingroup$ I understand I'm a little late to the party here, but I am a bit confused by the answer here (I am new to category theory). What is meant by "$x$ is represented by $x_j\in A_j$" (similarly for $y$)? How do we represent an element of the colimit by an element of an object in $C$? And I understand how you've used the filtered condition, but how do we know that $x_ly_l=0$ in $A_l$? $\endgroup$– DaveCommented Oct 7, 2019 at 19:08
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1$\begingroup$ @Dave The colimit $A$ of the $A_j$ comes with maps $A_j\to A$ for every $j$. By "$x$ is represented by $x_j$", I mean that $x\in A$ is the image of $x_j\in A_j$ under the map $A_j\to A$. For general colimits (of algebras, groups, ...), such elements generate $A$, but for filtered colimits, every element of $A$ is of this form for some $A_j$. And two elements $x_i\in A_i$ and $x_j\in A_j$ represent the same element of $A$ iff they map to the same element in some $A_l$ in the filtered system. This is essentially why filtered colimits have nicer properties than general colimits. $\endgroup$ Commented Oct 8, 2019 at 8:07
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$\begingroup$ Thanks for the response! I guess I figured that "representing" was using the morphisms $\psi_j:A_j\to A$, but I suppose my confusion is how we know that there exists some $j$ such that $x$ is in the image of $\psi_j$? Also, if this is the case then we get $0=xy=\psi_j(x_j)\psi_k(y_k)$, and then there are maps $f:A_j\to A_l$ and $g:A_k\to A_l$ by the filtered condition, but how do we know that $f(x_j)g(y_k)=0$? $\endgroup$– DaveCommented Oct 8, 2019 at 19:55
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$\begingroup$ @Dave In any (not necessarily filtered) colimit $A$ of algebras (or groups, ...) is generated by (the images under $\psi_j$ of) the elements of the $A_j$, subject to relations stating that for any $f:A_j\to A_i$ in the filtered system, and any $x_j\in A_j$, $x_j=f(x_j)$. So any element of $A$ can be expressed in terms of elements of finitely many $A_j$. What the first part of the definition of "filtered" simplifies is that all of these $A_j$ map into a single $A_l$, and so your element of $A$ is represented by a single element of $A_l$. $\endgroup$ Commented Oct 9, 2019 at 8:24
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1$\begingroup$ Okay, thanks very much for the clarification! It's all still new to me, so I'll keep thinking about this. I appreciate the help. $\endgroup$– DaveCommented Oct 10, 2019 at 14:59