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Suppose ${\mathrm{ch}}(K) = p > 0$ and we consider the formal power series ring $K[[X_1,\ldots,X_{np}]]$ over $K$ in $np$ variables $X_1,\ldots, X_{np}$. Let $\Lambda$ be the set defined as follows$\colon$ \begin{align*} & \Lambda \colon\!= {\mathrm{all~ sets~}} \{(i_1,\ldots,i_p), (i_{p+1},\ldots,i_{2p}), \ldots, (i_{(n-1)p + 1},\ldots,i_{np}) \}, \\ & {\mathrm{where}}, ~ \{1, 2, 3, 4, \ldots \} = \{i_1, i_2, i_3, i_4, \ldots \} \phantom{I} {\mathrm{s.t.}} \phantom{I} i_k \not= i_l \phantom{I} {\mathrm{for}} \phantom{i} k \not= l. \end{align*}

Namely, $\Lambda$ is the set of the divisions of $(1,\ldots, np)$ into $n$ $``p$-tuples''.

For $\lambda = \{(i_1,\ldots,i_p), (i_{p+1},\ldots,i_{2p}), \ldots, (i_{(n-1)p + 1},\ldots,i_{np}) \} \in \Lambda$, we shall associate the following ideal $I_{\lambda}$ of $A_{\infty}$$\colon$ \begin{equation*} I_{\lambda} \colon\!= (X_{i_1} + \ldots + X_{i_p}, X_{i_{p+1}} + \ldots + X_{i_{2p}}, \ldots, X_{(n-1)p + 1} + \ldots + X_{np}). \end{equation*}

We shall define the ideal $S_n$ of the ring $K[[X_1,\ldots,X_{np}]]$ by the following$\colon$ \begin{equation*} S_n \colon= \underset{\lambda \in \Lambda}{\bigcap} I_{\lambda}. \end{equation*} Further, we shall specify the generators of $S_n$ as follows$\colon$ \begin{equation*} S_n = (\theta, s_2, \ldots, s_{m(n)}), \end{equation*} where $\theta \colon= X_1 + \ldots + X_{np}$.

Conjecture. The degrees ${\mathrm{deg}}(s_2), \ldots, {\mathrm{deg}}(s_{m(n)})$ diverge when $n \to \infty$.

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  • $\begingroup$ A naive guess: isn't your ideal generated by $\theta$ and all products of $np-n+1$ distinct variables? $\endgroup$ – Ilya Bogdanov Oct 12 '19 at 8:40
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    $\begingroup$ "diverge" is meant for "tends to infinity" or "does not converge"? $\endgroup$ – YCor Oct 12 '19 at 9:18
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I still hope that my guess is correct, but here is a proof of weaker statement.

Let $q$ be a generator different from $\theta$.. Reduce it by $(\theta)$ substituting $X_1=-X_2-X_3-\dots$; it will not vanish. So assume it to be reduced (this does not increase $\deg q$).

Order the variables as $X_1\succ X_3\succ\dots$. The set of generators of $I_\Lambda$ is a Groebner base wrt this order.

Now $q$ should be reducible wrt any of those bases. This means that its leading term should contain one of $X_2, \dots, X_n$ (say, $X_i$), also one of $X_2,\dots, X_{i-1},X_{i+1},\dots, X_{n+1}$, and so on. Hence $\deg q\geq n$, which yields the result.

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  • $\begingroup$ Great thanks. Pierre Matsumi $\endgroup$ – Rinmyaku Oct 13 '19 at 13:27

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