8
$\begingroup$

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.

$\endgroup$

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.