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If $A$ is a finite dimensional commutative, associative, unital algebra over a field $\mathbb{K}$ then does there exist a non-zero vector $v\in A\otimes_{\mathbb{K}}A$ such that $(a\otimes_{\mathbb{K}}1)v=(1\otimes_{\mathbb{K}}a)v$ for all $a\in A$?

If $A$ happens to be a Frobenius algebra then we can take $v$ to be the dual of the associated non-degenerate bilinear form. Moreover, I think I have shown that for a $\mathbb{K}$-basis of $A$ given by $e_{i}$ if there exists $v=\sum_{ij}\beta^{ij}e_{i}\otimes_{\mathbb{K}}e_{j}$ and $\beta^{ij}$ is non-degenerate then $\beta^{ij}$ gives a Frobenius form.

My main interest is the case where $\mathbb{K}=\mathbb{R}$ however I can't see this making much of a difference in general.

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  • $\begingroup$ What about $v = 0$, or, more general, $v \in \mathbb{K}$? $\endgroup$ – Dirk Jun 4 '18 at 10:55
  • $\begingroup$ 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. $\endgroup$ – Campbell Jun 4 '18 at 11:14
  • $\begingroup$ @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 $0$. $\endgroup$ – Jason Starr Jun 4 '18 at 11:20
  • $\begingroup$ Ah, yes, my error, sorry. $\endgroup$ – Dirk Jun 4 '18 at 11:54
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Let $I$ be the ideal in $A\otimes A$ generated by elements of the form $a\otimes 1-1\otimes a$. (Equivalently, this is the kernel of the multiplication map $A\otimes A\to A$.) We will assume that $0\neq A\neq \mathbb{K}$, so that $0<I<A\otimes A$. As everything is finite-dimensional, the sequence of powers $I^n$ must eventually stabilise. Let $n$ be least such that $I^n=I^{n+1}=I.I^n$. By the standard determinant trick, this gives an element $u$ with $u=1\pmod{I}$ and $u.I^n=0$. Let $m$ be least such that $u.I^m=0$. As $u=1\pmod{I}$ we have $u\neq 0$, so $m>0$. Now $u.I^{m-1}\neq 0$ and $u.I^{m-1}\leq\text{ann}(I)$ so $\text{ann}(I)\neq 0$.

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    $\begingroup$ "the standard determinant trick" also known as Nakayama's lemma :) $\endgroup$ – darij grinberg Jun 4 '18 at 11:31
  • $\begingroup$ @darijgrinberg Matsumura calls that result Nakayama's Lemma, but many other people use that name just for the statement $\mathfrak{m}M=M\Rightarrow M=0$. I am no historian and do not really have an opinion. $\endgroup$ – Neil Strickland Jun 4 '18 at 11:50
  • $\begingroup$ Suggestion: Nakayama's determinant trick :) Not a big deal, but I've gotten used to seeing at least one determinantal trick per proof, and those aren't all the same. $\endgroup$ – darij grinberg Jun 4 '18 at 11:57

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