Timeline for The existence of $v\in A\otimes_{\mathbb{K}}A$ such that $(a\otimes_{\mathbb{K}}1)v=(1\otimes_{\mathbb{K}}a)v$

9 events

when toggle format	what		by	license	comment
Jun 4, 2018 at 11:54	comment	added	Dirk		Ah, yes, my error, sorry.
Jun 4, 2018 at 11:28	answer	added	Neil Strickland		timeline score: 5
Jun 4, 2018 at 11:20	comment	added	Jason Starr		@DirkLiebhold. For $v\in \mathbb{K}\setminus\{0\}$, typically $a\otimes_{\mathbb{K}}1\cdot v$ does not equal $1\otimes_{\mathbb{K}} a\cdot v$. Denote by $I$ the kernel of $A\otimes_K A\to A,$ $a\otimes b \mapsto a\cdot b$. Then $I$ is a nonzero ideal. By Nakayama, $I^\ell = I^{\ell+1}$ only if $I^\ell=0$. Since $A\otimes_K A$ is Artinian, some $I^\ell$ is zero, i.e., $I$ is nilpotent. Let $\ell$ be the largest integer such that $I^\ell$ is nonzero. Then for every nonzero $v\in I^\ell$, for every $a\otimes 1 - 1\otimes a$ in $I$, the product with $v$ equals $0$.
Jun 4, 2018 at 11:15	history	edited	Campbell	CC BY-SA 4.0	added 9 characters in body
Jun 4, 2018 at 11:14	comment	added	Campbell		For $v\in\mathbb{K}-\{0\}$ we don't have $(a\otimes_{\mathbb{K}}1)v=(1\otimes_{\mathbb{K}}a)v$ for all $a\in A$ but I should have mentioned non-zero.
Jun 4, 2018 at 10:55	comment	added	Dirk		What about $v = 0$, or, more general, $v \in \mathbb{K}$?
Jun 4, 2018 at 10:55	history	edited	Martin Sleziak	CC BY-SA 4.0	typo in the title
Jun 4, 2018 at 10:51	review	First posts
Jun 4, 2018 at 11:00
Jun 4, 2018 at 10:51	history	asked	Campbell	CC BY-SA 4.0