Timeline for Tensor powers of an algebra all isomorphic
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Jan 26, 2014 at 1:19 | history | edited | Jason Starr | CC BY-SA 3.0 |
New counterexample to modified question. Optimized the generators for the ideal in the third edit.
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| Jan 26, 2014 at 1:13 | history | edited | Jason Starr | CC BY-SA 3.0 |
New counterexample to modified question.
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| Jan 26, 2014 at 0:20 | comment | added | Jesse Elliott | Jason, let me clarify. I want an example where $k \longrightarrow A$ is injective. | |
| Jan 26, 2014 at 0:14 | history | edited | Jason Starr | CC BY-SA 3.0 |
Gave second example where $A$ has no ring homomorphism to $K$ (regardless of $k$-algebra structure)
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| Jan 26, 2014 at 0:10 | comment | added | Jason Starr | @PeterSamuelson: That comment is silly. You can make an example out of any integral domain, say $k=\mathbb{Z}$, and any ideal, say $I=2\mathbb{Z}$, with $A=k/I$. I chose to use $\mathbb{C}[t]$ since I assumed that would be a more familiar ring with which to illustrate the problem. | |
| Jan 25, 2014 at 23:42 | comment | added | Jesse Elliott | Agreed. I don't think it's a counterexample. | |
| Jan 25, 2014 at 23:35 | comment | added | Peter Samuelson | I'm confused, probably about something silly. In your first example, isn't $A \cong \mathbb C$? So then $A$ is isomorphic to a subring of $K$? | |
| Jan 25, 2014 at 23:29 | comment | added | Jesse Elliott | And, yes, I would like an example where the map $k \longrightarrow A$ is injective, which abx in his or her answer claims exists. | |
| Jan 25, 2014 at 22:38 | comment | added | Jesse Elliott | $k$ is a Pr\"ufer domain iff every $k$-torsion-free $k$-module is flat, so that is a more general hypothesis that guarantees it for all $k$-subalgebras of $K$. | |
| Jan 25, 2014 at 18:37 | comment | added | Jason Starr | @user43326: "I guess algebras are meant to be augmented ..." That is certainly possible, but that would also rule out most examples where $A$ is a subalgebra of the fraction field $K$. So I am not certain that is what the OP meant. Perhaps the OP wanted $k$ to be a subalgebra of $A$. | |
| Jan 25, 2014 at 17:54 | comment | added | user43326 | I guess algebras are meant to be augmented, which would exclude this kind of counterexamples. | |
| Jan 25, 2014 at 16:09 | history | edited | Jason Starr | CC BY-SA 3.0 |
Added "of which I am aware".
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| S Jan 25, 2014 at 13:07 | history | answered | Jason Starr | CC BY-SA 3.0 | |
| S Jan 25, 2014 at 13:07 | history | made wiki | Post Made Community Wiki by Jason Starr |