Lemma 10.46.5. 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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).