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
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