Power series ring and monomials

Question

Let $A_n \colon= K[[X_1,\ldots,X_n]]$ be a formal power series ring over a field $K$ of characterisc $p > 0$ in $n$ variables.

For a given positive number $\epsilon > 0$ we call a monomial $X_{i_1}^{e_{i_1}} \cdots X_{i_n}^{e_{i_n}}$ an $\epsilon$-monomial if it satisfies the following conditions simultaneously$\colon$ \begin{align*} & e_{i_1}/p^{i_1} < \epsilon \\ & e_{i_2}/p^{i_2} < \epsilon \\ & \cdots \\ & e_{i_n}/p^{i_n} < \epsilon. \end{align*} Let $\alpha$, $\beta$ be two $\epsilon$-monomials. Then for two elements $x, y \in (X_1,\ldots,X_n)$, i.e., the unique maximal ideal of $A_n$, we consider the product defined by \begin{equation*} P_{\alpha,\beta}(x,y) \colon= (\alpha + x)(\beta + y). \end{equation*}

Q. Suppose that $x$ (resp. $y$) comprises monomials different from $\alpha$ (resp. $\beta$). That is, $x$ (resp. $y$) is composed of monomials neither of which is equal to $\alpha$ (resp. $\beta$).

Then does the product $P_{\alpha, \beta}(x, y)$ always contain a non-zero $2\epsilon$-monomial?

Answer 1

No. For instance, take $n=\epsilon=1$, $p=2$. Then the only $\epsilon$-monomials are 1 and $X_1$ and the $2\epsilon$-monomials are 1, $X_1$, $X_1^2$ and $X_1^3$. Take $\alpha=\beta=X_1$, $x=y=X_1^2-X_1$. We have $(\alpha+x)(\beta+y)=X_1^2X_1^2=X_1^4$, which involves no $2\epsilon$-monomials.

Answer 2

This answer replacees my earlier, incorrect one. The answer is still "no", even if $x$ does not involve $\alpha$ and $y$ does not involve $\beta$. Example: take $n=2$, $p=3$, $\alpha=\beta=X_1^2X_2^8$, $\epsilon=1$, $x=X_1^4$, $y=-X_2^{16}$. We have $(\alpha+x)(\beta+y)=X_1^6X_2^8-X_1^2X_2^{24}$ which contains no $2\epsilon$-monomials.