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 $n \to \infty$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ "diverge" is meant for "tends to infinity" or "does not converge"? $\endgroup$ – YCor Oct 12 '19 at 9:18

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.