Suppose that $R$ is a commutative ring with identity. I am trying to prove that the two following statements are equivalent.

  1. The ideal generated by all determinants of $n\times n$ Vandermonde matrices with entries in $R$ is the unit ideal.
  2. 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, $1 \iff 2$.

  • $\begingroup$ While it's true that the ideal generated by a unit is the unit ideal, you seem to be implying that the converse is also true. i.e. non-units cannot generate the unit ideal. But that is false. For example, in the ring of integers, the ideal generated by {2,3} is the unit ideal, yet 2 and 3 are non-units. $\endgroup$ – Laurent Lessard Sep 3 '13 at 5:07
  • $\begingroup$ Your question is interesting, although I think in the linked answer they don't claim (1) and (2) to be equivalent. Notice that for answering that question they just need (1) since $S$ is an $R$-module and then they can take not only Vandermondes but also combinations of Vandermondes (i.e. elements in the ideal generated by Vandermondes) multiplied by the vector of $H_i$'s. That is why they only need (1). $\endgroup$ – OR. Sep 4 '13 at 15:09
  • $\begingroup$ I see your point. Having an invertible Vandermonde matrix is sufficient but not necessary. What we need is for there to exist matrices $A_1,\dots,A_n$ and Vandermonde matrices $V_1,\dots,V_n$ such that $\sum_{i=1}^n A_i V_i = I$. How do we make the leap from combinations of Vandermonde matrices to combinations of determinants of Vandermonde matrices (i.e. (1)). ? $\endgroup$ – Laurent Lessard Sep 4 '13 at 19:19
  • $\begingroup$ I think I may have answered my own question --- Cauchy-Binet formula. If $V$ is an $m\times p$ matrix with $m\ge p$, then $V$ has a left-inverse iff some linear combination of its $p\times p$ minors is equal to 1. So if $V$ is a (nonsquare) Vandermonde matrix, it is left-invertible iff 1 belongs to the ideal generated by a determinants of square matrices found by selecting rows of $V$. But each of these is itself a square Vandermonde, so iff 1 belongs to the ideal generated by all Vandermonde determinants. $\endgroup$ – Laurent Lessard Sep 4 '13 at 20:02
  • $\begingroup$ Yes, that is why the answer to the linked question is what it is. Now, the question on (1) being equivalent to (2), i.e. this question, is still interesting on its own right. I think. $\endgroup$ – OR. Sep 4 '13 at 20:37

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.

  • $\begingroup$ I understand that the existence of a maximal ideal $m$ with $R/m$ having fewer than $n$ elements is equivalent to "not (1)". What I'm struggling with is to understand why the existence of such a maximal ideal is equivalent to "not (2)". $\endgroup$ – Laurent Lessard Sep 3 '13 at 6:37
  • $\begingroup$ Take any $n$ elements and form a Vandermonde with them. Since they belong to less than $n$ different classes mod $m$, then at least one of the difference between two of those elements belong to $m$. Now use that the Vandermonde is equal to the product of all those differences. That product most then belong to $m$. So, every Vandermonde will belong to $m$ which then contains the ideal generated by all Vandermondes. $\endgroup$ – OR. Sep 3 '13 at 13:17
  • $\begingroup$ Yes -- that shows half of the equivalence. Namely, if there is an $m$ such that $R/m$ has fewer than $n$ elements, then every Vandermonde determinant belongs to $m$, and thus there are no invertible Vandermonde matrices. What about the other direction? $\endgroup$ – Laurent Lessard Sep 3 '13 at 16:29
  • $\begingroup$ You are right. This needs an explanation. I will add it in the answer because it is rather long. $\endgroup$ – OR. Sep 3 '13 at 22:02
  • $\begingroup$ Hmm... I am only being able to probe it for rings with finitely many maximal ideals (semilocal rings). $\endgroup$ – OR. Sep 3 '13 at 22:30

$\def\ZZ{\mathbb{Z}}$(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.

  • 1
    $\begingroup$ @LaurentLessard This should be the answer to your question now. $\endgroup$ – OR. Sep 8 '13 at 20:54

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