Vandermonde matrices over a commutative ring. 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, $1 \iff 2$.
 A: $\def\ZZ{\mathbb{Z}}$(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.
A: The answer has ideas written as I think them. Latter I will try to rewrite it nicely.
The answer you link sort of contains the answer to your question.
A Vandermonde is the product of all differences (without repetitions) of the entries in the degree $1$ row. So what you need is to show that there are $n$ elements whose differences are all units or assume there are not such elements (for a counterpositive). 
"not (2)" is equivalent (as it is said there) to having a maximal ideal $m$ with $R/m$ having less than $n$ elements. 
Assume there is such $m$, then all Vandermondes are going to be in $m$, because some of the differences you put in the degree $1$ row are going to be in $m$. So, "not (1)".

We are missing to prove that "not (2)" implies that there is a maximal ideal $m$ such that $R/m$ has less than $n$ elements. We prove the counterpossitive. 
Assume that for every maximal ideal $m$, $R/m$ has at least $n$ elements. 
Take one maximal ideal $m_1$ and $x_1,\ldots,x_n$ $n$ representatives of $n$ different classes mod $m_1$. Then The Vandermonde of those elements $V(x_1,\ldots,x_n)\notin m_1$. 
Take another maximal $m_2$ and $b_1\in m_1\setminus m_2$. If $x_1,\ldots,x_n$ are in different classes mod $m_2$ we keep them as $V(x_1,\ldots,x_n)\notin m_2$ as well. If some of the $x_i$ belong to the same class mod $m_2$ we can add to some of them an elements of the form $b_1a$, for $a\in R$ such that $x_1+b_1a_1,\ldots,x_n+b_1a_n$ belong to different classes mod $m_2$ (we take the $a_1,\ldots,a_n$ belonging to different classes mod $m_2$ for this to happen, but we can do it because there are more than $n$ different classes mod $m_2$ by assumption). Notice they still belong to the same (different) classes mod $m_1$ as before. Then take this set of elements are the new $x_1,\ldots,x_n$ and then $V(x_1,\ldots,x_n)\notin m_2$ as well.
Take the next maximal ideal $m_3$ and $c_{1}\in m_1\setminus m_3$ and $c_2\in m_2\setminus m_3$. If $V(x_1,\ldots,x_n)\notin m_3$ we are happy. If it belongs to $m_3$ we can add to the $x_i$ elements of the form $c_1c_2$, with $a\in R$, to make them belong to different classes mod $m_3$, while keeping the classes mod $m_1$ and mod $m_2$ to which they belong. 
Continuing this way for all maximal ideals of $R$ we get elements $x_1,\ldots,x_n$ such that for every maximal ideal $m$ they belong to different classes mod $m$. Therefore the Vandermonde $V(x_1,\ldots,x_n)\notin m$ for every maximal $m$. An element that is not in any maximal ideal is a unit.
Let me thing what happens if we have infinitely many maximal ideals.

