Thanks a lot for the reply, Frieder! This answers my question completely.

Great, it works! The bilinear form I mentioned above is not non-degenerate. I am accepting your answer.

You are right. I'll just edit my question and add the above as a motivation.

A reference is James's book on representations of general linear group.

This appears naturally 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. 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.

This may not work though as $\chi$ may not vanish on $R b_0$. In fact we have a non-degenerate $\mathbb{F}_p$-valued bilinear form on $R$ given by $(a,b) = f(ab)$ where $f \colon R \to \mathbb{F}_p$ is "sum of coefficient function". Choose a non-trivial character $\chi$ of $\mathbb{F}_p$. Then the map from $R$ to the Pontriyagin dual given by $a \mapsto \psi_a$ is an isomorphism. Here $\psi_a(b) = \chi((a,b))$ . So if all this is right then the result should be true for the ring you mentioned.