Let $R$ be a commutative ring. Let $A$ be an $R$-algebra (i.e., an $R$-module equipped with an $R$-bilinear multiplication map that turns $A$ into a unital ring). We do not require $A$ to be commutative. Assume that $A$ is free as an $R$-module, with a finite basis. Let $\left( e_{1},e_{2},\ldots,e_{n}\right) $ be a basis of the $A$-module $R$.

We define a trace map $\operatorname{Tr}_{A/R}:A\rightarrow R$ as follows: For every $b\in A$, we let $\operatorname{Tr}_{A/R}b$ be the trace of the endomorphism $A\rightarrow A,\ a\mapsto ba$ of the $R$-module $A$. (We are using the fact that $A$ has a finite basis here.) Clearly, the map $\operatorname{Tr}_{A/R}$ is $R$-linear.

Let $\Delta=\det\left( \left( \operatorname{Tr}_{A/R}\left( e_{i}e_{j}\right) \right) _{1\leq i\leq n,\ 1\leq j\leq n}\right) \in R$.

Is it true that $\Delta=u^{2}+4v$ for some elements $u$ and $v$ of $R$ ?

The above is an elaborate generalization of Stickelberger's discriminant theorem. Indeed, if we assume $A$ to be commutative, then $\Delta =u^{2}+4v$ is true; this is proven in Remark 5.4 of Owen Biesel, Alberto Gioia, A new discriminant algebra construction, arXiv:1503.05318v3, Documenta Mathematica 21 (2016), pp. 1051--1088. (More precisely, they prove it in the case when $n\geq2$; but the remaining case is obvious.) If we furthermore assume that $R=\mathbb{Z}$ and $A$ is the integer ring of a number field $K$, then $\Delta$ becomes the discriminant $\Delta_{K}$ of $K$, and the $\Delta=u^{2}+4v$ claim becomes $\left( \Delta_{K}\equiv0 \mod % 4\ \vee\ \Delta_{K}\equiv1 \mod 4\right) $, which is the classical claim of Stickelberger's discriminant theorem.

The claim does not significantly depend on the choice of basis $\left( e_{1},e_{2},\ldots,e_{n}\right) $, since any change of basis causes $\Delta$ to be multiplied by a square in $R$ (namely, by the square of the determinant of the matrix responsible for the change of basis... or of its inverse, depending on how you define that matrix).

I do not know whether to expect the conjecture to be true or not. All examples I have tried (the complexity of computing $\Delta$ for high $n$ limits my abilities here) satisfy $\Delta=u^{2}+4v$ for rather stupid reasons (often, $\Delta$ will either be a square or be divisible by $4$ to begin with); but this says more about the poverty of my examples than about the correctness of the conjecture. For what it's worth, here are my examples:

  • If $A$ is the group ring of a group $G$, and $\left(e_1, e_2, \ldots, e_n\right)$ is the standard basis $G$ of $A$ (abusing notation as usual), then $\Delta = \left(-1\right)^{n\left(n-1\right)/2} n^n$. This is either of the form $4v$ or of the form $1+4v$; thus, the conjecture holds here.

  • If $A$ is the matrix ring $R^{m\times m}$ (so that $n=m^2$), and $\left(e_1, e_2, \ldots, e_n\right)$ is the basis consisting of the matrix units in their usual order, then $\Delta = \left(-1\right)^{m\left(m-1\right)/2} m^n$. This is either of the form $4v$ or of the form $1+4v$; thus, the conjecture holds here.

  • If $A$ is the (standard) quaternion algebra, then $\Delta = -4^4$; thus, the conjecture holds here. Other quaternion algebras may have different $\Delta$, but the factor $4^4$ will still be present.

Notice that the trace map $\operatorname{Tr}_{A/R}$ defined above has a "right analogue": the map $\operatorname{Tr}_{A/R}^{\prime }:A\rightarrow R$ sending each $b\in A$ to the trace of the endomorphism $A\rightarrow A,\ a\mapsto ab$ of the $R$-module $A$. If we do not require commutativity of $A$, these maps will in general not be identical (for a specific example, extend a left-trivial semigroup by a $1$ to obtain a monoid, then take the monoid algebra of this monoid). I do not know whether the $\Delta$ computed via the other map will be different; this is another interesting question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.