Suppose $R$ is a commutative ring, and $S \subset R^{n\times n}$ is an $R$-module. We are given $H\_0,\dots,H\_n \in R^{n\times n}$, and we know that for all $r \in R$,
$$H\_0 + r H\_1 + \dots r^n H\_n \in S$$
The question is: when can we conclude that $H\_i \in S$ for all $i=0,1,\dots,n$ ?


Clearly this is true when $R = \mathbb{R}$, because we can choose real numbers $r\_0,r\_1,\dots,r\_n$ and write:
$$ H\_0 + r\_i H\_1 + \dots r\_i^n H\_n \in S \qquad \text{for $i=0,1,\dots,n$}$$
The corresponding Vandermonde matrix $V$ is invertible provided the $r\_i$ are distinct. Now take a linear combination of the left-hand-sides of the above relation, using coefficients provided by the ith row of $V^{-1}$. Since $S$ is closed under linear combinations, we conclude that $H\_i \in S$, as required.


The same argument won't work for a general commutative ring, but is the result still true? If not, what constraints must be imposed on $R$ to make it so?



For which rings does there exist an invertible Vandermonde matrix?
The fact that the $H\_i$ are matrices is irrelevant; the question is whether a module generated by elements $H\_0\dots,H\_n$ is necessarily generated by all the elements of the form $\Sigma\_{0\le i\le n}r^iH\_i$. If this is true in modules of matrices then it's true in all free modules, and if so then it's true when the $H\_i$ form a basis for a free module, and if it's true in that case then it's true in all modules, in particular modules of matrices.


When the $H\_i$ are a basis for $R^{n+1}$ then the question becomes "Is the ideal generated by all $(n+1)\times (n+1)$ Vandermonde determinants the unit ideal?" That fails if and only if there is a maximal ideal $m$ containing all such determinants, so if and only if there exists $m$ such that over the field $R/m$ there is no invertible Vandermonde matrix, so it fails if and only if some residue field of $R$ has at most $n$ elements.


