# Coefficients of the monomials appearing in a Schubert polynomial

It is known that the coefficients of the monomials appearing in a Schubert polynomial are always positive. My question is: Is it always true that at least one such coefficient must be $1$? If that is not always true, then is it true that the gcd of the coefficients of the monomials appearing in a Schubert polynomial must always be $1$?

• I suspect that for each permutation $w \in S_n$, each monomial appearing in the Schubert polynomial $\mathfrak{S}_w$ is lexicographically smaller-or-equal to the monomial $\mathbf{x}^{L\left(w\right)} := \prod\limits_{i=1}^n x_i^{\ell_i\left(w\right)}$, where $\ell_i\left(w\right)$ denotes the number of all $j > i$ satisfying $w\left(i\right) > w\left(j\right)$. Moreover, I suspect that the coefficient of this monomial $\mathbf{x}^{L\left(w\right)}$ is $1$. – darij grinberg Jun 22 '18 at 15:40
• @darij grinberg : can you please explicitly state what lexicographic order you are talking about here ? – user111492 Jun 22 '18 at 15:42
• Probably it should not be too hard to see this from the "pipe dreams" definition of Schubert polynomials. – Sam Hopkins Jun 22 '18 at 15:53
• Look at Corollary 3.9 of "RC-Graphs and Schubert Polynomials" by Bergeron and Billey: projecteuclid.org/download/pdf_1/euclid.em/1048516036. It describes the (unique) leading term of a Schubert polynomial. – Sam Hopkins Jun 22 '18 at 17:11
• Darij's suspicion is correct: The leading coefficient is $x^{L(w)}$. This is transparent to see from the Kohnert formula for Schubert polynomials (e.g. arxiv.org/abs/1703.00088). (Some people don't believe some proofs of this formula, but I think everyone believes the formula!) – Oliver Jun 23 '18 at 13:10

As mentioned in the comments, there is an ordering such that the leading coefficient in this ordering is $1$ - this property implies that the basis is integral (is it perhaps equivalent to this property)?