Action of certain endomorphisms on Pontriyagin dual Let $R$ be a finite ring and $F$ be an algebraically closed field in which $|R|$ is invertible. Does there exists an $F$-valued character $\chi$ of $(R, +)$ such that every character $\psi$ is of the form $\psi(a) = \chi(ab)$ for some $b \in R$? If not, does the statement hold when $R$ is commutative (or under any other reasonably nice assumption)?
This is true when $R$ is a finite field. I believe that the statement in this case follows from the non-degeneracy of the trace form over the prime field contained in $R$.
Any references would be really helpful.
Motivation: This appears when we study representation theory of unipotent groups. Suppose $U$ be the subgroup of unipotent lower triangular matrices in $GL_n(R)$. Then elements of the form $E_{\chi} = \sum_{M \in U} \chi(M_{2,1})M \in F[U]$ appear naturally where $\chi$ is an $F$-valued character of $(R, +)$. There is nothing special about $M_{2,1}$, one can work with any closed "root subgroup". Now conjugating $E_{\chi}$ by an element of the cartan subgroup yields $E_{\chi(.b)}$ for some $b$ and so it becomes useful to know whether all characters can be obtained in this way. 
 A: Call a character $\chi$ (left) generating if every character is of the form $\psi(a)=\chi(ab)$ for some $b\in R$. It turns out that, when $R$ is finite, a character is left generating if and only if it is right generating (every character has the form $\psi(a) = \chi(ba) $ for some $b\in R$).  
The answer to your question is as follows:  

A finite ring $R$ has a generating character if and only if $R$ is a finite Frobenius ring.  

For a proof, see, for example:  
Thomas Honold, MR 1831096 Characterization of finite Frobenius rings, Arch. Math. (Basel) 76 (2001), no. 6, 406--415.  
When the ring is finite and commutative, this is equivalent to $R$ having a multiplicity-free socle, or $R$ being a direct product of finite local rings with simple socle. (see T. Y. Lam, MR 1653294 Lectures on modules and rings , Thm. 15.27, and more generally §§15, 16 for more characterizations and examples and non-examples.) 
Your question is quite natural and appears in other contexts. For example, it appears naturally when studying Gauss sums over arbitrary finite rings, as in a series of papers by Lamprecht (cited in Honold's paper), and in:
Fernando Szechtman, MR 1916077 Quadratic Gauss sums over finite commutative rings, J. Number Theory 95 (2002), no. 1, 1--13.  
This property of finite Frobenius rings is also relevant to coding theory, see
Jay A. Wood, MR 1738408 Duality for modules over finite rings and applications to coding theory, Amer. J. Math. 121 (1999), no. 3, 555--575.
A: The following is a commutative counter example.
Fix a prime $p$, let $k=\mathbb{F}_p$ and let $R=k[x,y]/(x^2,y^2,xy)$.
Let $m=(x,y)<R$. Then $R$ is a 3-dimensional $k$-vector space, $m$ a 2-dimensional subspace and every additive character on $R$ could be identified with a $k$-functional.
Fix a character $\chi$ and view it as a functional. Fix an element $0\neq b_0\in m\cap \ker(\chi)$. Note that $\chi$ vanishes on $Rb_0=m\cap\ker(\chi)$.
It follows that $|\{\chi(\cdot b)\mid b\in R\}|<|R|$. Since the dual group has $|R|$ elements, there exists some character $\phi$ such that there is no $b$ satisfying $\phi(\cdot)=\chi(\cdot b)$.
