TITLE: Vandermonde matrices over a commutative ring.
QUESTION [6 upvotes]: Suppose that $R$ is a commutative ring with identity. I am trying to prove that the two following statements are equivalent.

The ideal generated by all determinants of $n\times n$ Vandermonde matrices with entries in $R$ is the unit ideal.
There exists an invertible $n\times n$ Vandermonde matrix with entries in $R$.

I can see how 2 $\implies 1$. If there is an invertible Vandermonde matrix, then its determinant is a unit, and so the ideal generated by all Vandermonde determinants contains a unit and so must be the unit ideal. I don't see how to prove the converse...
The context for this question is the solution that was posted for the following problem:
https://mathoverflow.net/questions/32217/for-which-rings-does-there-exist-an-invertible-vandermonde-matrix .
In that link, the question of existence of an invertible Vandermonde matrix is equated with "Is the ideal generated by all Vandermonde determinants the unit ideal?". In other words, $1 \iff 2$.

REPLY [4 votes]: $\def\ZZ{\mathbb{Z}}$(1) and (2) are not equivalent. Set $\beta = (1 + \sqrt{-11})/2$ and consider the ring $R = \ZZ[\beta]$. Let $V(x,y,z)$ be the Vandermonde determinant $\det \begin{pmatrix} x^2 & y^2 & z^2 \\ x & y & z \\ 1 & 1 & 1 \end{pmatrix} = (x-y)(x-z)(y-z)$. I claim that $V(x,y,z)$ is never a unit, for $x$, $y$ and $z$ in $R$, but that the set of all $V(x,y,z)$, with $x$, $y$, $z \in R$, generates the unit ideal.
Proof that $V(x,y,z)$ is not a unit Set $u=x-y$, $v=y-z$ and $w = z-x$. If $V(x,y,z) = -uvw$ is a unit, then $u$, $v$ and $w$ are all units.  The only units of $R$ are $\pm 1$ (write a unit of $R$ as $a + b \sqrt{-11}$, with $a$ and $b$ half-integers, and notice that $a^2 + 11 b^2$ must be $1$). But $u+v+w=0$ and we can't have $\pm 1  \pm 1  \pm 1=0$.
Proof that the $V(x,y,z)$ generate the unit ideal  We have $V(2,1,0) = 2$ and $V(\beta,1,0) = -3$. So $V(\beta,1,0) + V(2,1,0) = -1$.
Thought process $2 \times 2$ Vandermondes are trivial so let's try $3 \times 3$. There is a $3 \times 3$ Vandermonde which is a unit if and only if the unit equation is solvable. When we are working with finitely generated rings of algebraic numbers, the unit equation only has finitely many solutions (a hard theorem of Siegel) so it should be easy to construct examples where it has none. 
On the other hand, as explained at the MO link, $3 \times 3$ determinants generate the unit ideal whenever there is no quotient field of size $2$. There are lots of rings of algebraic integers which have no $\mathbb{F}_2$ quotient.
To find an explicit example, I want a ring where it is easy to compute the group of units, and where there is no $\mathbb{F}_2$ quotient. The former suggests the ring of integers in $\mathbb{Q}(\sqrt{D})$, the latter forces $D \equiv 5 \bmod 8$. The first solution I found was $D=13$, but $D=-11$ made for a shorter solution so that's the one I wrote up.