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The rank-one cyclotomic quotient has a monic basis

Statement

Let $F$ be a field, let $q, u_{1}, \dots, u_{r} \in F^{\times}$ , and let $A_{r, 1} (F; q, u_{1}, \dots, u_{r})$ be the rank-one specialization of the multiplicative cyclotomic presentation The multiplicative cyclotomic Hecke algebra, with $Q \mapsto q$ and $X_{1} \mapsto X$ . Then

$A_{r, 1} (F; q, u_{1}, \dots, u_{r}) ≅ F [X] / (\prod_{a = 1}^{r} (X - u_{a})),$

and $1, X, \dots, X^{r - 1}$ is an $F$ -basis. If $r \geq 2$ and $u_{1} = \dots = u_{r} = c \neq 0$ , the change of variable $ε = X - c$ identifies this algebra with $F [ε] / (ε^{r})$ . In that case $ε$ is a nonzero nilpotent.

Facts & Assumptions

Given: A field $F$ , units $q, u_{1}, \dots, u_{r} \in F$ , and the rank-one multiplicative cyclotomic presentation.

[F1]

The multiplicative cyclotomic Hecke algebra: the rank-one affine presentation includes the invertible generator $X_{1}$ .

[F2]

The multiplicative cyclotomic Hecke algebra: the cyclotomic quotient imposes $p (X_{1}) = \prod_{a = 1}^{r} (X_{1} - u_{a}) = 0$ .

Proof

Given: $F$ , $q, u_{1}, \dots, u_{r}$ , and $A_{r, 1} (F; q, u_{1}, \dots, u_{r})$ as in the statement.

1.1F1F2algebra

Write $e_{k}$ for the $k$ th elementary symmetric polynomial in $u_{1}, \dots, u_{r}$ and $e_{r} = u_{1} \dots u_{r} \neq 0$ . By [F1] and [F2], the rank-one presentation is $F [X^{\pm 1}] / (p (X))$ . Expanding $p (X) = 0$ gives $X^{r} - e_{1} X^{r - 1} + \dots + (- 1)^{r} e_{r} = 0$ , so in $F [X] / (p)$ the element $X$ has inverse $(- 1)^{r + 1} e_{r}^{- 1} (X^{r - 1} - e_{1} X^{r - 2} + \dots + (- 1)^{r - 1} e_{r - 1})$ . Thus the natural map $F [X] / (p) \to F [X^{\pm 1}] / (p)$ is onto, and the assignment of $X^{- 1}$ to this polynomial defines its inverse.

2.1step 1.1algebra

Since $p$ is monic of degree $r$ , division by $p$ writes every polynomial uniquely as a multiple of $p$ plus a polynomial of degree below $r$ . Uniqueness follows because a nonzero multiple $g p$ over the field $F$ has degree $\deg (g) + r \geq r$ . Therefore the residue classes $1, X, \dots, X^{r - 1}$ are a basis of $F [X] / (p)$ and hence of the Laurent quotient in step 1.1.

3.1step 2.1algebra∎

If all $u_{a} = c$ , then $p (X) = (X - c)^{r}$ . Substitution $ε = X - c$ is an invertible change of polynomial variable, so it induces an isomorphism $F [X] / ((X - c)^{r}) ≅ F [ε] / (ε^{r})$ . Because $r \geq 2$ , step 2.1 gives the nonzero basis vector $ε$ ; its $r$ th power is zero in this quotient.

Depends on

The multiplicative cyclotomic Hecke algebra

Used by

A repeated root collides the generic one-box contents Example
The rank-one level-two cyclotomic quotient Example

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Georges Neaime, Geodesic normal forms and Hecke algebras for the complex reflection groups G(de,e,n) (standard reference, not scraped)
Andrew Mathas, Cyclotomic quiver Hecke algebras of type A (lecture notes) (standard reference, not scraped)