The type-A weight lattice, Laurent ring, and divided differences

By [F3] and [F4], the integers have no zero divisors and $0\mathrm{\ne }1$; thus $\mathbb{Z}$ is a domain by [F5]. Applying [F1] to the Laurent variable $Q$ gives that $R_0=\mathbb{Z}[Q,Q^{-1}]$ is a domain, and applying it again gives that $A$ is a domain with $R_0$-basis $(X^{\lambda }{)}_{\lambda \in {\mathbb{Z}}^n}$ and embeds in its fraction field. By [F2], the adjacent transpositions generate the stated $S_n$ action, which permutes the Laurent monomial basis.

Fix $i$ and a Laurent monomial $X^{\lambda }$. Put $a={\lambda }_i$, $b={\lambda }_{i+1}$, and $z=X_i\mathrm{/}{X}_{i+1}$. Then $X^{\lambda }-s_i(X^{\lambda })=X^{{\lambda }^{\mathrm{\prime }}}(z^a-z^b)$, where ${\lambda }^{\mathrm{\prime }}$ agrees with $\lambda$ off coordinates $i,i+1$ and has exponents $0,a+b$ at coordinates $i,i+1$, respectively. For integers $a,b$, the quotient $(z^a-z^b)\mathrm{/}(1-z)$ is a Laurent polynomial: it is $-{\sum }_{k=0}^{a-b-1}{z}^{b+k}$ if $a>b$, zero if $a=b$, and ${\sum }_{k=0}^{b-a-1}{z}^{a+k}$ if $a<b$. Each formula follows by multiplying the finite geometric sum by $1-z$. Thus $X^{\lambda }-s_i(X^{\lambda })$ is divisible in $A$ by $1-z$, including when either exponent is negative.

Every $f\in A$ is a unique finite $R_0$-linear combination of the monomials by [F1]. Linearity of $f\mapsto f-s_i(f)$ and the monomial calculation in step 1.2 give a quotient in $A$ for every $f$. Since $1-z\mathrm{\ne }0$ in the domain $A$, that quotient is unique. This defines $D_i:A\to A$ as stated.

For $f,g\in A$, the identity $fg-s_i(fg)=(f-s_i(f))g+s_i(f)(g-s_i(g))$ holds because $s_i$ is an algebra automorphism. Dividing by the nonzero element $1-z$ in the fraction field and using the definition of $D_i$ gives $D_i(fg)=D_i(f)g+s_i(f)D_i(g)$.

The construction of $D_i$ and its Leibniz identity use only the monomial basis, finite geometric sums, and cancellation in the integral domain $A$; no choice principle is used.