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

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.

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.