Timeline for Subgroups of the tensor product $A\otimes A$
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 11, 2016 at 18:05 | answer | added | YCor | timeline score: 6 | |
| Oct 11, 2016 at 16:32 | comment | added | YCor | Actually given a finite abelian group $Q$, an endomorphism of $Q$ is an automorphism if and only if all its $p$-components are automorphisms, and assuming that $Q$ is a finite abelian $p$-group, an endomorphism is an automorphism iff the endomorphism induced on $Q/pQ$ is an automorphism. Applying this to $Q=A\otimes A$, this allows to completely reduce the problem to the field case ($A$ finite-dimensional over $\mathbf{Z}/p\mathbf{Z}$). And in field case the restriction to fields of prime order is somewhat unnecessary as it is often convenient for linear algebra to pass to algebraic closure. | |
| Oct 11, 2016 at 16:27 | comment | added | YCor | In the field case, the question can be studied (after extending scalars) to an algebraically closed field, if it helps. When 2 is invertible, here are two obstructions: (a) $-1$ eigenvalue of multiplicity $\ge 2$. Indeed in this case the projection of $B$ into $\Lambda^2$ is not surjective [image of the map $x\otimes y\mapsto x\wedge (1+\alpha)y$]. (b) there are two (distinct) inverse eigenvalues $t,t^{-1}$. Indeed, given eigenvectors $v,w$, a computation then shows that the images of $v\otimes w$ and $w\otimes v$ are collinear, so injectivity fails. | |
| Oct 11, 2016 at 16:21 | comment | added | YCor | $B$ is the image of the linear endomorphism of $A\otimes A$ given as $(x\otimes y)\mapsto x\otimes y- \alpha(y)\otimes x$. You're asking about its surjectivity, and $B$ is finite, so this is also equivalent to its injectivity. | |
| Oct 11, 2016 at 16:07 | history | edited | YCor |
edited tags
|
|
| Oct 11, 2016 at 15:47 | comment | added | marcos | By $\mathbb{Z}[1/2]$ you mean the extension by 1/2 of the ring of integers right? | |
| Oct 11, 2016 at 15:43 | history | edited | marcos | CC BY-SA 3.0 |
added 44 characters in body
|
| Oct 11, 2016 at 10:15 | comment | added | HeinrichD | The simplest counterexample is $A=\mathbb{Z}[1/2]$. Here, $2$ is invertible and $\forall x,y \in A.\,x \otimes y = y \otimes x$ holds. Thus, $\langle x \otimes y + y \otimes x : x,y \in A \rangle = A \otimes A$. | |
| Oct 11, 2016 at 9:55 | comment | added | marcos | If A is an cyclic abelian group then $\tau=id$, right? But the other way around is wrong, you are right. I am trying to read your link. I think that the argument to apply is that in the first four lines of the first answer (by Will Sawin). I am trying to figure it out. | |
| Oct 11, 2016 at 8:40 | comment | added | HeinrichD | Your claim that $\forall x,y \in A.\,x \otimes y = y \otimes x$ holds in $A \otimes A$ only if $A$ is cyclic is not correct. See mathoverflow.net/questions/119689 | |
| S Oct 11, 2016 at 7:40 | history | suggested | C.F.G | CC BY-SA 3.0 |
improve formatting
|
| Oct 11, 2016 at 7:27 | review | First posts | |||
| Oct 11, 2016 at 7:37 | |||||
| Oct 11, 2016 at 7:25 | review | Suggested edits | |||
| S Oct 11, 2016 at 7:40 | |||||
| Oct 11, 2016 at 7:22 | history | asked | marcos | CC BY-SA 3.0 |