9
$\begingroup$

Consider the ring of polynomials $\mathbb{k} [x_1, x_2, \ldots , x_n]$ as a module over the ring of symmetric polynomials $\Lambda_{\mathbb{k}}$. Is $\mathbb{k} [x_1, x_2, \ldots , x_n]$ a free $\Lambda_{\mathbb{k}}$-module? Can you write down "good" generators explicitly? (I think that it has to be something very classical in representation theory).

Comment: My initial question was whether this module flat. But since all flat Noetherian modules over polynomial ring are free (correct me if it is wrong), it is the same question.

There is much more general question, which seems unlikely to have good answer. Let $G$ be a finite group and $V$ finite dimensional representation of G. Consider projection $p: V \rightarrow V/G$. When is $p$ flat?

$\endgroup$
10
$\begingroup$

Let $s_i= \sum x_1 \ldots x_i$ be the fundamental symmetric polynomials. We have a sequence of free extensions $$k[s_1, \ldots, s_n] \subset k[s_1, \ldots, s_n][x_1]\subset k[s_1, \ldots, s_n][x_1][x_2] \subset \cdots \\ \subset k[s_1,\ldots ,s_n] [x_1] \ldots [x_n] = k[x_1, \ldots ,x_n]$$ of degrees $n$, $n-1$, $\ldots$ ,$1$. At step $i$ the generators are $1, x_i, \ldots, x_i^{n-i}$. Therefore $$k[s_1, \ldots, s_n] \subset k[x_1, \ldots, x_n]$$ is free of degree $n!$ with generators $x_1^{a_1} x_2^{a_2} \cdots x_n^{a_n}$ with $0 \le a_i \le n-i$.

More generally for a finite reflection group of transformations $G$ acting on a vector space $V$ over a field $k$ of characteristic $0$ (to be safe) the algebra of invariants $k[V]^G$ is a polynomial algebra and $k[V]$ is a free $k[V]^G$ module of rank $|G|$ --see the answer of @stephen: .

$\endgroup$
  • 1
    $\begingroup$ I think some more details are in order: why $1,x_1,\dots,x_1^{n-1}$ is a basis of the first ring extension? $\endgroup$ – user26857 Nov 3 '14 at 19:41
  • 2
    $\begingroup$ And why don't satisfy any equation of degree $<n$? $\endgroup$ – user26857 Nov 3 '14 at 19:50
  • 2
    $\begingroup$ @user26857: Another useful observation using just basic Galois theory is : $k(s_1, \ldots, s_n) \subset k(x_1, \ldots, x_n)$ is Galois with group $S_n$. $\endgroup$ – orangeskid Nov 3 '14 at 19:57
  • 1
    $\begingroup$ Great answer! Just a remark: The basis orangeskid describes is different from the Schubert basis - geometrically, the images of the $x_i$ in the integral cohomology of the flag variety are the Chern classes of the $n$ tautological line bundles (and the iterated computation above has a geometric counteprart, by viewing the flat variety as an iterated projective bundle and using Leray-Hirsch at each step), while as I said above the Schubert polynomials correspond to Bruhat cells. $\endgroup$ – Hanno Nov 3 '14 at 20:02
  • 1
    $\begingroup$ Regarding your last paragraph, it can certainly happen that in characteristic $p$ the ring of invariants is not polynomial even when the group is generated by reflection; on the other hand polynomials invariants is enough to imply that the group is generated by reflections in any characteristic. $\endgroup$ – Stephen Nov 3 '14 at 20:34
12
$\begingroup$

Yes, that's indeed a classical result of representation theory: ${\mathbb Z}[x_1,...,x_n]$ is graded free over ${\Lambda}_n$ of rank $n!$ (the graded rank is the quantum factorial $[n]_q!$), and a basis is given by Schubert polynomials defined in terms of divided difference operators.

See for example the original article of Demazure, in particular Theorem 6.2.

Passing to the quotient, one obtains the graded ring ${\mathbb Z}[x_1,...,x_n]/\langle\Lambda_n^+\rangle$ which is isomorphic to the integral cohomology ring of the flag variety of ${\mathbb C}^n$, and the ${\mathbb Z}$-basis of Schubert polynomials coincides with the basis of fundamental classes of Bruhat cells in the flag variety. This is explained in Fulton's book 'Young Tableaux', Section 10.4.

$\endgroup$
7
$\begingroup$

The answer by @orangeskid (+1) is the most classical and direct, and the answer by @Hanno connects your question to Schubert calculus and the geometry of flag varieties (+1). Hoping this isn't too self-promoting, you might also have a look at my paper Jack polynomials and the coinvariant ring of $G(r,p,n)$ (I worked in somewhat more generality)

http://tinyurl.com/l8u9wh5

where I showed that certain non-symmetric Jack polynomials give a basis as well (I was working with more general reflection groups and over the complex numbers, but a version should work over any field of characteristic $0$; I have not thought much about this in characteristic $p$). The point of my paper was really to connect the descent bases (yet another basis!) studied much earlier by Adriano Garsia and Dennis Stanton in the paper Group actions of Stanley-Reisner rings and invariants of permutation groups

http://tinyurl.com/lhhtney

to the representation theoretic structure of the coinvariant algebra as an irreducible module for the rational Cherednik algebra. Of course this structure becomes much more complicated in characteristic smaller than $n$; in particular the coinvariant algebra will in general not be irreducible as a module for the Cherednik algebra.

Towards your second question: by definition $p$ is flat if and only if $k[V]$ is a flat $k[V]^G$-module. This certainly holds if $k[V]$ is free over $k[V]^G$; a sufficient condition for this is given in Bourbaki, Theorem 1 of section 2 of Chapter 5 of Lie Groups and Lie algebras (page 110): in case the characteristic of $k$ does not divide the order of $G$, it suffices that $G$ be generated by reflections. This is false (in general) in characteristic dividing the order of the group. But see the paper Extending the coinvariant theorems of Chevalley, Shephard-Todd, Mitchell, and Springer by Broer, Reiner, Smith, and Webb available for instance on Peter Webb's homepage here

http://www.math.umn.edu/~webb/Publications/

for references and what can be said in this generality (this is an active area of research so you shouldn't expect to find a clean answer to your question).

Conversely, assuming $p$ is flat and examining the proof of the above theorem in Bourbaki, it follows that $k[V]$ is a free $k[V]^G$-module. Now Remark 2 to Theorem 4 (page 120) shows that $G$ is generated by reflections. So to sum up: if $p$ is flat then $G$ is generated by reflections; in characteristic not dividing the order of $G$ the converse holds.

$\endgroup$
  • $\begingroup$ MO level, neat. +1! $\endgroup$ – orangeskid Nov 3 '14 at 20:00
  • $\begingroup$ I have a problem about your links. Article by Webb and others is just about connected topics. They do not answer my question (or I just do not see it, then please explain). About Bourbaki book I have a deeper trouble. It sounds silly but I even can not find what is called theorem 1 in this section. Could you post here what it is about (proof is not necessary). $\endgroup$ – quinque Nov 3 '14 at 23:32
  • $\begingroup$ @user167762, The statement is Theorem 1 on page 110 of my English edition---are you able to locate it now? I referenced the paper of of BRSW not because it answers your question, but because it is the most recent thing I can think of that deals with the subject in detail and I thought the references it contained might prove useful for you or other future readers. I will edit my answer a bit to give an indication of what I believe to be the state of the art. As far as I know, in positive characteristic there is no clean characterization of groups such that the projection is flat. $\endgroup$ – Stephen Nov 4 '14 at 12:26
2
$\begingroup$

Another perspective:

The ring extension $S=k[s_1,\dots,s_n]\subset k[x_1,\dots,x_n]=R$ is finite, and since $R$ is Cohen-Macaulay then it is free of rank say $m$. We thus have a Hironaka decomposition $R=\oplus_{i=1}^m S\eta_i$ for some homogeneous $\eta_i\in R$. Using this the Hilbert series of $R$ is $$H_R(t)=\sum_{i=1}^mH_S(t)t^{\deg\eta_i}=\sum_{i=1}^mt^{\deg\eta_i}/\prod_{i=1}^n(1-t^i).$$ On the other side, $H_R(t)=1/(1-t)^n$ and therefore $\sum_{i=1}^mt^{\deg\eta_i}=\prod_{i=1}^{n-1}(1+t+\cdots+t^i)$ which for $t=1$ gives $m=n!$.

$\endgroup$
  • $\begingroup$ You used a theorem about Cohen-Macaulay. I do not know it. Could you please write it? $\endgroup$ – quinque Nov 3 '14 at 22:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.