It turns out that there is a nice explicit basis for $\mathbb{C}[x_1,\ldots,x_n]/(p_1,\ldots,p_n)$: the *Vandermonde determinant* $\Delta$ and all its partial derivatives. The $n\times n$ Vandermonde determinant is the determinant of the following matrix:

$$\left(\begin{array}{cccc}

1 & 1 & \cdots & 1 \\

x_1 & x_2 & \cdots & x_n \\

\vdots & \vdots & \ddots & \vdots \\

x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1}

\end{array}\right)$$

Explicitly, it equals $\Delta=\prod_{i < j}(x_j-x_i)$. So in the case of two variables, $\Delta=x-y$, and in the case of three variables, $\Delta=(x-y)(y-z)(x-z)$. Now, consider all partial derivatives of $\Delta$, that is, all polynomials of the form $$\frac{\partial}{\partial x_{i_1}}\frac{\partial}{\partial x_{i_2}}\cdots\frac{\partial}{\partial x_{i_k}} \Delta$$ for some $k$. The claim is that a subset of these form a $\mathbb{C}$-basis for the ring of coinvariants; to do so, we need to show that there are enough of them that are mutually linearly independent. We begin by computing the dimension of the $d$th graded piece of the coinvariant ring for all $d$. Consider again the decomposition $$\mathbb{C}[x_1,\ldots,x_n]=\mathbb{C}[p_1,\ldots,p_n]\otimes_{\mathbb{C}[S_n]}\mathbb{C}[x_1,\ldots,x_n]/(p_1,\ldots,p_n),$$ and consider the Hilbert series of both sides. (Recall that the Hilbert series of a graded ring $R=\oplus_d R_d$ is the generating function $H_R(q)=\sum_d \dim_{\mathbb{C}}(R_d) q^d$.) On the left hand side, it is a simple combinatorial exercise to see that the Hilbert series of $\mathbb{C}[x_1,\ldots,x_n]$ is $$\frac{1}{(1-q)^n}.$$ On the right hand side, recall that the bases for the ring of symmetric functions in $n$ variables are indexed by partitions with at most $n$ parts. So if $p_n(d)$ denotes the number of partitions of $d$ having at most $n$ parts, then the Hilbert series of $\mathbb{C}[p_1,\ldots,p_n]$ is the generating function $$\sum_d p(d)q^d=\frac{1}{(1-q)(1-q^2)\cdots(1-q^n)}.$$ Finally, since Hilbert series are multiplicative across tensor products, we have the equation $$\frac{1}{(1-q)^n}=\frac{1}{(1-q)(1-q^2)\cdots(1-q^n)}\operatorname{Hilb}(\mathbb{C}[x_1,\ldots,x_n]/(p_1,\ldots,p_n)).$$ Solving, we find $$\operatorname{Hilb}(\mathbb{C}[x_1,\ldots,x_n]/(p_1,\ldots,p_n))=\frac{(1-q)(1-q^2)\cdots(1-q^n)}{(1-q)^n}=(n)_q!,$$ which is the $q$-factorial number. This means that, in particular, the highest degree elements in the coinvariant ring have degree $\binom{n}{2}$, and this degree component has dimension $1$. Thus the Vandermonde determinant $\Delta$ spans the top degree component, and as an $S_n$-module it is a single copy of the sign representation. (The Hilbert series formula also immediately implies that the dimension of the entire coinvariant ring is $n!$.)

Finally, to show that the partial derivatives of $\Delta$ span the ring of coinvariants, we consider its dual correspondence with the ring of *harmonics*. Consider the inner product on $\mathbb{C}[x_1,\ldots,x_n]$ given by $$\langle f,g \rangle=(\partial_f(g))_0,$$ where $\partial_f$ denotes the differential operator formed by replacing any $x_i$ in the polynomial $f$ by $\frac{\partial}{\partial x_i}$, and the subscript $0$ indicates that we take the constant term. This is a well-defined inner product having an orthogonal basis consisting of the monomials. Under this inner product, we can consider the orthogonal subspace to the ideal $I=(p_1,\ldots,p_n)$.

The orthogonal subspace can be described explicitly as the space of all functions $f$ which are killed by the “power sum operators” $\frac{\partial^k}{\partial x_1^k}+\cdots+\frac{\partial^k}{\partial x_n^k}$ for all $k$. This space is called the ring of *harmonics*, and is isomorphic to the ring of coinvariants since it is the orthogonal space to the ideal $I$ that we quotient by.

Finally, since the Vandermonde determinant $\Delta$ is in the ring of harmonics, all its partial derivatives are also elements of it since the partial derivative operators commute. Moreover, if we consider only the partial derivatives of the form $$\frac{\partial^{a_1}}{\partial x_1^{a_1}}\cdots \frac{\partial^{a_n}}{\partial x_n^{a_n}}\Delta$$ in which $a_i\le i-1$ for all $i$, then there are exactly $n!$ of them, and their leading terms in lexicographic order are all distinct. Thus they are independent and generate the ring of harmonics, which has dimension $n!$.

(Combinatorial Bonus: Notice that this choice of partial derivatives corresponds with the Carlitz Codes described in this post, which are $q$-counted by the $q$-factorial.)

To sum up, it is possible to generalize the nice decomposition of polynomials in two variables into a symmetric and antisymmetric component, by using $n!$ components for the $n$-variable case. These components correspond to the $n!$ partial derivatives of $\Delta=\prod_{i < j} (x_i-x_j)$ that come from the Carliz codes $a_1,\ldots,a_n$ with $a_i\le i-1$ for each $i$. Mystery solved!

I just realized I forgot to reference Francois Bergeron’s book for the proof on the last page of this post: see http://www.amazon.com/Algebraic-Combinatorics-Coinvariant-Mathematics-Mathematiques/dp/1568813244, in the chapter on coinvariants.

I’d much rather use the basis of Schubert polynomials!

Thanks for pointing that out Allen – I had forgotten about the connection with the flag variety. Perhaps I’ll post about Schubert polynomials next week or the week after.

Just discovered this blog– very nice!

Expanding on what Allen K. said about Schubert polynomials, I like to think of at least three different well-known bases for the polynomial ring in n variables over the symmetric polynomials which model the q-factorial in 3 ways:

(1) The Schubert polynomials model the q-factorial as the generating function for permutations w by inv(w), the inversion number or Coxeter group length.

(2) Garsia’s “descent monomial basis” model the q-factorial as the generating function for permutations w by maj(w), the major index.

(3) Then there is what Garsia calls “the Artin basis”, that is the set of all monomials in x_1,…,x_n for which the degree of x_i is at most i-1. They appear in E. Artin’s old slim paperback book “Galois theory”, when he proves the fundamental theorem of symmetric functions. And these monomials model the q-factorial as a product of q-numbers, similarly to your partial derivatives of the Vandermonde product.

Thanks for the detailed reply! I’m sorry it took me so long to see that you commented and approve it. You should be able to comment freely from now on. ðŸ™‚

Maria

I think I’m missing something – why are we tensoring with the base ring $\mathbb{C}[S_n]$? For instance, wouldn’t that make $(x+y)(x+2y) = (-1 \cdot (x+y))\otimes (x+2y) = (x+y)\otimes (-1\cdot (x+2y)) = (x+y)(2x+y)$? (where $-1$ is the nonidentity element of $S_2$ considered inside $\mathbb{C}[S_2]$.

By the way, your blog is awesome!