Timeline for Filtered Colimit of associative $k$-algebras that are domains
Current License: CC BY-SA 3.0
11 events
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| Oct 21, 2019 at 19:59 | comment | added | Dave | Sorry for the delay, I was on vacation. I have posted a question here: math.stackexchange.com/questions/3403322/…. Any help you could provide would be greatly appreciated, and thanks for your help thus far. | |
| Oct 12, 2019 at 15:50 | comment | added | Jeremy Rickard | @Dave Could you post a question (maybe on math.stackexchange) if you want more details? I'd be happy to answer, but writing explanations in comments is a bit of a pain! Thanks. | |
| Oct 11, 2019 at 22:14 | comment | added | Dave | I think I understand most of this now, but I am still having a hard time seeing why if $x_j\in A_j$ and $y_k\in A_k$ represent the same element of $A$ then they must map to the same element in some $A_l$? | |
| Oct 10, 2019 at 14:59 | comment | added | Dave | 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. | |
| Oct 9, 2019 at 8:32 | comment | added | Jeremy Rickard | @Dave If $x_j\in A_j$ and $y_k\in A_k$ represent the same element of $A$, then they map to the same element in some $A_l$. The key case is when $x_j=a(w)$ and $y_k=b(w)$ for some maps $a,b$ in the filtered system. Then there are maps $f:A_j\to A_l$ and $g:A_k\to A_l$, but it perhaps $fa\neq gb$, so $f(x_j)$ and $g(y_k)$ may be different. However, the second part of the definition of "filtered" says that there is a map $h:A_l\to A_m$ in the filtered system with $hfa=hgb$, and so $hf(x_j)=hg(y_k)$. So any equation that holds in $A$ holds in some single $A_m$. | |
| Oct 9, 2019 at 8:24 | comment | added | Jeremy Rickard | @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 $A_l$. | |
| Oct 8, 2019 at 19:55 | comment | added | Dave | 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 $f(x_j)g(y_k)=0$? | |
| Oct 8, 2019 at 8:07 | comment | added | Jeremy Rickard | @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. | |
| Oct 7, 2019 at 19:08 | comment | added | Dave | 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 $A_l$? | |
| Mar 5, 2016 at 23:38 | vote | accept | CommunityBot | ||
| Mar 2, 2016 at 9:38 | history | answered | Jeremy Rickard | CC BY-SA 3.0 |