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Let $k$ be a field of characteristic 0 (say $\mathbb{C}$).

Consider the ring of polynomials $R = k[X_1,...,X_n]$ and its subring of invariant polynomials $S = R^{S_n}$. It is known that the endomorphism algebra $End_{S}(R)$ is the nil affine Hecke ring (we can give a nice presentation of this ring in terms of Demazure operators). This presentation is as follows: generators:

a) $X_1,...,X_n$ (which act on polynomials by multiplication by the corresponding variable)

b) $T_1,...,T_{n-1}$ which act on polynomials by Demazure operators, i.e.:

$T_i(P) = \frac{P-s_i(P)}{X_{i+1}-X_i}$, where $s_i$ is the usual simple reflection.

The relations are:

a) The $X_i$'s commute

b) $T_i^2=0$ and the $T$'s satisfy the braid relations plus they commute if $|i-j|>1$

c) $[T_i,X_j] = 0$ if $j\neq i,i+1$

d) $T_iX_i-X_{i+1}T_i = -1$ and $T_iX_{i+1}-X_iT_i = 1$.

I would like to know if there is a similar result for the diagonal invariants. Namely: let $A = k[X_1,...,X_n,Y_1,...,Y_n]$ and let $B = A^{S_n}$ where the symmetric group acts by the diagonal embedding $S_n\hookrightarrow S_n\times S_n$. Is there a (nice) presentation of the algebra $End_B(A)$? (does it have a name?) If I understand correctly (from Chevalley-Shephard-Todd theorem) in this case $A$ is not even free over $B$ so the things might get ugly...

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I just wanted to confirm that $A$ is not free over $B$. Take $n=2$. $B$ is generated by $e_1:= x_1+x_2$, $e_2:=x1 x2$, $f_1:= y_1+y_2$, $f_2:= y_1 y_2$ and $g:= x_1 y_2 + x_2 y_1$; these obey the single relation $g^2 - e_1 f_1 g + q$ where $q$ is some fourth degree polynomial in the $e$'s and $f$'s. As a $R:=k[e_1, e_2, f_1, f_2]$-module, we have $B = R \oplus g R$, so $B$ has Hilbert series $h_B:=(1+t^2)/(1-t)^2 (1-t^2)^2$. The Hilbert series of $A$ is $1/(1-t)^4$. As $h_A/h_B = (1+t)^2/(1+t^2)$ is not a polynomial, $A$ is not free over $B$. –  David Speyer Dec 21 '12 at 14:36
    
Thanks David. That's a nice argument! –  Dragos Fratila Dec 21 '12 at 23:04

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