
> 
> 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](https://math.stackexchange.com/questions/394785/proof-of-stickelberger-s-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](http://arxiv.org/abs/1503.05318v3) (published in [Documenta Mathematica 21 (2016), pp. 1051--1088](https://www.emis.de/journals/DMJDMV/vol-21/28.html)).
(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.



Is the discriminant of a free (as a module) $R$-algebra always congruent to a square modulo 4?
I see this question is still being viewed. To avoid misleading anyone, let me put a placeholder answer here: The question has been **answered positively** (with a beautiful proof) by John Voight, Asher Auel and Owen Biesel:


[Asher Auel, Owen Biesel, John Voight, *Stickelberger's discriminant theorem for algebras*, submitted to AMM, arXiv:2208.06138v1](https://arxiv.org/abs/2208.06138v1).


(Theorem 3.28 proves my conjecture.)


