The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.45.5. Let $p$ be a prime number. Let $n, m > 0$ be two integers. There exists an integer $a$ such that $(x + y)^{p^ a}, p^ a(x + y) \in \mathbf{Z}[x^{p^ n}, p^ nx, y^{p^ m}, p^ my]$.

Proof. This is clear for $p^ a(x + y)$ as soon as $a \geq n, m$. In fact, pick $a \gg n, m$. Write

\[ (x + y)^{p^ a} = \sum \nolimits _{i, j \geq 0, i + j = p^ a} {p^ a \choose i, j} x^ iy^ j \]

For every $i, j \geq 0$ with $i + j = p^ a$ write $i = q p^ n + r$ with $r \in \{ 0, \ldots , p^ n - 1\} $ and $j = q' p^ m + r'$ with $r' \in \{ 0, \ldots , p^ m - 1\} $. The condition $(x + y)^{p^ a} \in \mathbf{Z}[x^{p^ n}, p^ nx, y^{p^ m}, p^ my]$ holds if

\[ p^{nr + mr'} \text{ divides } {p^ a \choose i, j} \]

If $r = r' = 0$ then the divisibility holds. If $r \not= 0$, then we write

\[ {p^ a \choose i, j} = \frac{p^ a}{i} {p^ a - 1 \choose i - 1, j} \]

Since $r \not= 0$ the rational number $p^ a/i$ has $p$-adic valuation at least $a - (n - 1)$ (because $i$ is not divisible by $p^ n$). Thus ${p^ a \choose i, j}$ is divisible by $p^{a - n + 1}$ in this case. Similarly, we see that if $r' \not= 0$, then ${p^ a \choose i, j}$ is divisible by $p^{a - m + 1}$. Picking $a = np^ n + mp^ m + n + m$ will work. $\square$


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