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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.

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  • $\begingroup$ I put an answer below, but I must say that now I regret it. The question seems entirely random and unmotivated. $\endgroup$
    – Uri Bader
    Commented Oct 9, 2016 at 13:01
  • $\begingroup$ 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. $\endgroup$
    – rohitna
    Commented Oct 9, 2016 at 18:52
  • $\begingroup$ A reference is James's book on representations of general linear group. $\endgroup$
    – rohitna
    Commented Oct 9, 2016 at 18:53
  • $\begingroup$ My previous comment was harsh, but too many questions on this site are not well motivated. There are $\aleph_0$ many arbitrary questions, and a great deal of those appear here. I see now that yours is indeed motivated. It will be useful (for you, first of all) if next time you put a question you will try to explain its importance in the first place. $\endgroup$
    – Uri Bader
    Commented Oct 9, 2016 at 19:23
  • $\begingroup$ You are right. I'll just edit my question and add the above as a motivation. $\endgroup$
    – rohitna
    Commented Oct 9, 2016 at 19:34

2 Answers 2

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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.

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  • $\begingroup$ Thanks a lot for the reply, Frieder! This answers my question completely. $\endgroup$
    – rohitna
    Commented Oct 11, 2016 at 2:19
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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 $\phi(\cdot)=\chi(\cdot b)$.

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  • $\begingroup$ 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. $\endgroup$
    – rohitna
    Commented Oct 9, 2016 at 18:23
  • $\begingroup$ @rohitna for your last comment, with your notation, I believe $\psi_0=\psi_{x-y}$. Thus $a\to \psi_a$ is not an isomorphism. Note that indeed $x-y$ is in the kernel of your bilinear form. $\endgroup$
    – Uri Bader
    Commented Oct 9, 2016 at 19:34
  • $\begingroup$ @rohitna note that for every $b\in m$, $Rb=kb$. It follows that if $b\in\ker(\chi)$ then $Rb\subset\ker(\chi)$. In particular, $Rb_0\subset\ker(\chi)$, that is $\chi$ does vanish on $Rb_0$. $\endgroup$
    – Uri Bader
    Commented Oct 9, 2016 at 19:46
  • $\begingroup$ Great, it works! The bilinear form I mentioned above is not non-degenerate. I am accepting your answer. $\endgroup$
    – rohitna
    Commented Oct 9, 2016 at 19:58
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    $\begingroup$ Maybe I should mention: a character $\chi$ on a ring $R$ will have the property you seek iff its kernel does not contain any non-trivial ideal. On a field for example, any non-trivial character will do. $\endgroup$
    – Uri Bader
    Commented Oct 9, 2016 at 20:03

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