For which rings does there exist an invertible Vandermonde matrix? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T08:09:23Z http://mathoverflow.net/feeds/question/32217 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32217/for-which-rings-does-there-exist-an-invertible-vandermonde-matrix For which rings does there exist an invertible Vandermonde matrix? Laurent Lessard 2010-07-16T19:40:38Z 2010-07-17T00:29:12Z <p>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 $i=0,1,\dots,n$ ?</p> <p>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.</p> <p>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?</p> http://mathoverflow.net/questions/32217/for-which-rings-does-there-exist-an-invertible-vandermonde-matrix/32227#32227 Answer by Tom Goodwillie for For which rings does there exist an invertible Vandermonde matrix? Tom Goodwillie 2010-07-16T21:19:36Z 2010-07-17T00:29:12Z <p>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.</p> <p>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.</p>