Taylor expansion on an allowed next digit

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This proves L154. Write $J\preceq_pN$ for the digitwise domination in the statement, and put $K=N-J$. Then $0\le J\le N$, so $0\le j=s+p^eJ\le n$. Since $0\le s\le t<p$, the lower $e$ digits of $j$ are dominated by those of $n$, and the remaining digits are dominated by $J\preceq_pN$. Thus $p\nmid\binom nj$. For completeness, the Lucas criterion follows by expanding

$$(1+X)^n\equiv\prod_{r\ge0}(1+X^{p^r})^{n_r}\pmod p,$$

where $n_r$ are the base-$p$ digits. The coefficient of $X^j$ is $\prod_r\binom{n_r}{j_r}$, which is nonzero exactly under domination.

Taylor expansion at the integer point $(s,t-s)$ has integer coefficients. By the definition of $\mu$ and $H$,

$$F(j,n-j)=p^{e\mu}H(J,K)+p^{e(\mu+1)}Z, \qquad Z\in\mathbf Z.$$

Set $x=J\bmod p$. Domination gives $0\le x\le d$, and $K\equiv d-x\pmod p$. The valuation assertion and its equality criterion follow from this display, including the convention for zero.

Because $p\nmid N$, $d\ne0$ in $\mathbf F_p$. The polynomial $\phi(X)=H(X,d-X)\in\mathbf F_p[X]$ has degree at most $\mu$ and is nonzero. Indeed, homogeneity gives the identity in $\mathbf F_p(X,Y)$

$$H(X,Y)=\left(\frac{X+Y}{d}\right)^\mu \phi\!\left(\frac{dX}{X+Y}\right).$$

If $\phi$ vanished identically, every coefficient of $H$ would vanish modulo $p$, contrary to the hypothesis. It has at most $\mu$ roots. When $d\ge\mu$, the $d+1$ distinct residues $0,\ldots,d$ therefore include a nonroot $x$. Take $J=x$, with all higher digits zero. Then $J\preceq_pN$, and the equality criterion gives $v_pF(j,n-j)=e\mu$ as required. $\square$