Timeline for Global dimension of a subalgebra with all units
Current License: CC BY-SA 3.0
27 events
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| S Jun 18, 2014 at 16:27 | history | bounty ended | CommunityBot | ||
| S Jun 18, 2014 at 16:27 | history | notice removed | CommunityBot | ||
| Jun 12, 2014 at 7:45 | comment | added | Jeremy Rickard | So does my example above with $A=k[x^2,x^3]$ and $B=k[x,s_1,s_2,t_1,t_2]$ satisfy what you want? | |
| Jun 12, 2014 at 2:45 | history | edited | ABIM | CC BY-SA 3.0 |
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| Jun 12, 2014 at 2:37 | comment | added | ABIM | I'm happy your stressing the details, its important. Essentially what I want clause $2$ to state, is that: by "more relations" I mean that if $\mathfrak{F}$ is the smallest free algebra projecting (canonically) onto A, then there cannot be a smaller free algebra projecting (canonically) onto B (which is not the case of $\mathbb{C}[x^2,x^3]$ and $\mathbb{C}[x]$ since the smallest free algebra projecting onto the former is $\mathbb{C}<x,y>$ and the smallest projecting onto the later is $\mathbb{C}<x>$ (namely itself, up to isomorphism)). | |
| Jun 11, 2014 at 22:21 | comment | added | Jeremy Rickard | By the way, I'm not just nit-picking for the sake of it. I suspect the example in my last comment might not settle what you want to know (which is why I didn't post it as an answer), although it's a counterexample to one interpretation of what you ask. But if not, then I'm not clear exactly what you do mean by a subalgebra of an algebra having "more relations". | |
| Jun 11, 2014 at 21:09 | comment | added | Jeremy Rickard | There's still the problem that the hypotheses allow non-trivial choice in the selection of $\pi_A$, $\pi_B$ and $\iota'$ which can affect whether or not $\iota'$ and $j$ are injective. For example, for a field $k$, take $A=k[x^2,x^3]$, $\mathfrak{F}_A=k\langle u,v\rangle$ with $\pi_A(u)=x^2$, $\pi_A(v)=x^3$, $B=k[x,s_1,s_2,t_1,t_2]$, $\mathfrak{F}_B=k\langle x,s_1,s_2,t_1,t_2\rangle$ with the obvious $\pi_B$. Then the obvious choice of $\iota'$ isn't injective, but there are choices that are: e.g., $\iota'(u)=x^2+s_1s_2-s_2s_1$ and $\iota'(v)=x^3+t_1t_2-t_2t_1$. | |
| Jun 11, 2014 at 18:51 | comment | added | ABIM | oops, ok I put those crucial assumptions back in | |
| Jun 11, 2014 at 18:50 | history | edited | ABIM | CC BY-SA 3.0 |
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| Jun 11, 2014 at 8:59 | comment | added | Jeremy Rickard | Did you mean to drop the "smallest possible" condition on $\mathfrak{F}_A$ and $\mathfrak{F}_B$ in your last edit? If so, then you can make $\mathbb{C}[x^2,x^3]\subseteq\mathbb{C}[x]$ into a counterexample by choosing a reduntantly large set of generators for $\mathbb{C}[x]$. | |
| Jun 10, 2014 at 17:58 | history | edited | ABIM | CC BY-SA 3.0 |
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| Jun 10, 2014 at 17:56 | comment | added | ABIM | Yes the injectiveness of "$j$" is "part of the hypothesis"; in that, it is an immediate consequence of the assumptions on $\iota$ and $\iota'$ and the snake lemma). Thanks for spotting that typo (I was just attempting to motivate assumption $1$ and $2$ in that last paragraph). | |
| Jun 10, 2014 at 17:38 | history | edited | ABIM | CC BY-SA 3.0 |
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| Jun 10, 2014 at 15:22 | comment | added | Jeremy Rickard | You call $\iota'$ an inclusion, but it's not automatically injective (if it is, then so is $j$, obviously). And $\pi_A$ and $\pi_B$ are not uniquely determined, so it's conceivable that the injectivity depends on the choice. Is the hypothesis that there is some choice of $\pi_A$ and $\pi_B$ such that $\iota'$ is injective? Also, I think some of the maps in your diagram are mislabelled, and is "$A$ is some unit of $B$" in the last paragraph a typo? | |
| S Jun 10, 2014 at 14:49 | history | bounty started | ABIM | ||
| S Jun 10, 2014 at 14:49 | history | notice added | ABIM | Authoritative reference needed | |
| Jun 8, 2014 at 18:57 | comment | added | ABIM | Sorry about that I typed quicker than I thought, ok I fixed up the details and the question should all be in order now :) | |
| Jun 8, 2014 at 18:56 | history | edited | ABIM | CC BY-SA 3.0 |
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| Jun 8, 2014 at 17:16 | comment | added | Jeremy Rickard | I think your edit has several typos, but if $\pi$ is surjective, how can $\pi'$ be surjective if $A\neq B$? | |
| Jun 8, 2014 at 16:58 | comment | added | ABIM | Nope... I rephrased my question, to make it clearer that no new relations can be introduced... so in this sense $\mathbb{C}[x^2,x^3]$ would not be a counterexample since it is not trivially related and yet $\mathbb{C}[x]$ is. | |
| Jun 8, 2014 at 16:56 | history | edited | ABIM | CC BY-SA 3.0 |
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| Jun 8, 2014 at 16:37 | history | edited | ABIM | CC BY-SA 3.0 |
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| Jun 8, 2014 at 16:32 | history | edited | ABIM | CC BY-SA 3.0 |
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| Jun 8, 2014 at 16:24 | history | edited | ABIM | CC BY-SA 3.0 |
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| Jun 6, 2014 at 18:57 | comment | added | Jeremy Rickard | I may be misunderstanding your question, but isn't $\mathbb{C}[x^2,x^3]\subseteq\mathbb{C}[x]$ a counterexample? | |
| Jun 6, 2014 at 17:25 | history | edited | ABIM | CC BY-SA 3.0 |
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| Jun 5, 2014 at 21:18 | history | asked | ABIM | CC BY-SA 3.0 |