# Intersection of a certain linear ideals of $K[[X_1,\ldots,X_{np}]]$ for ${\mathrm{ch}}(K) = p > 0$

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 $$n \to \infty$$.

• A naive guess: isn't your ideal generated by $\theta$ and all products of $np-n+1$ distinct variables? Oct 12, 2019 at 8:40
• "diverge" is meant for "tends to infinity" or "does not converge"?
– YCor
Oct 12, 2019 at 9:18

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.