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Let

$R \colon= {\Bbb Z}_p[[S_1,...,S_n,X_1,...,X_d]]$

be the $(n+d)$-variable formal power series ring over ${\Bbb Z}_p$. Choose an arbitrary ideal ${\frak b}$ of $R$ such that

${\mathrm{ht}}({\frak b}) > d$.

Conjecture. There exits some specialisation as $S_i = \alpha_i$ with $\alpha_i \in {\Bbb Z}_p$ and $|{\alpha_i}| < 1$ for $i = 1,\ldots,n$ such that

${\frak b} \ni p^m$

for some $m >0$ after specialisation (i.e., after inserting those values $\alpha_i$ in $S_i$, respectively).

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  • $\begingroup$ Are you working with the ring of arbitrary power series, or are you working with a subring of power series that are $p$-adically convergent on the unit disk? $\endgroup$ – Jason Starr Jan 17 '17 at 8:42
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    $\begingroup$ @JasonStarr: the coefficients are in $\mathbb{Z}_p$ so an arbitrary power series converges on the open unit disc. $\endgroup$ – Kevin Buzzard Jan 17 '17 at 9:07
  • $\begingroup$ @KevinBuzzard. You are right. For some reason I was thinking of a disk centered at $1$. $\endgroup$ – Jason Starr Jan 17 '17 at 9:14
  • $\begingroup$ Yes, I am working with an arbitrary power series. Pierre $\endgroup$ – Pierre MATSUMI Jan 17 '17 at 9:37
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    $\begingroup$ Dear Jason Starr, thank you very much. Could you please explain it to me more plainly. I know that Hilbert-Samuel function HS(m) is the one which for a local ring (R,m) such that R/m = k gives HS(m) := dim_k R/m^n. I see that the degree of HS for $R/{\frak b}$ is smaller than that of ${\Bbb Z}_p[[S_1,...,S_n]]$. Why can ensure the existence of aforementioned f ? $\endgroup$ – Pierre MATSUMI Jan 17 '17 at 11:56
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I apologize for my comment above. The proof is a bit more involved than I originally anticipated. Here is a stronger assertion: for every $\underline{e}=(e_1,\dots,e_n)$ with $e_1 \gg \dots \gg e_n \gg 0,$ there exists a positive integer $m=m(\underline{e})$ such that for every $(\alpha_1,\dots,\alpha_n)$ with $\alpha_i\in p^{e_i}\mathbb{Z}_p\setminus p^{e_i+1}\mathbb{Z}_p,$ $p^m$ is in $\mathfrak{b}+\langle S_1-\alpha_1,\dots,S_n-\alpha_n\rangle.$ The result is proved by induction on $n.$ When $n$ equals $0,$ then $\text{ht}(\mathfrak{b}) \geq d+1 = \text{dim}\mathbb{Z}_p[[X_1,\dots,X_d]].$ Thus $\mathfrak{b}$ is primary for the maximal ideal. In particular, since $p$ is in the maximal ideal, there exists $m>0$ such that $p^m$ is in $\mathfrak{b}.$

Now, by way of induction, assume that $n>0$ and assume that the result has been proved for smaller values of $n$. Since $R$ is a Noetherian ring, there exists an integer $\ell>0$ such that $(\text{rad}(\mathfrak{b}))^\ell \subset \mathfrak{b}$. Assume that there exists $m>0$ such that $$p^m \in \text{rad}(\mathfrak{b}) + \langle S_1-\alpha_1,\dots,S_n-\alpha_n\rangle.$$ Then also $$(p^m)^\ell \in \mathfrak{b}+\langle S_1-\alpha_1,\dots,S_n-\alpha_n \rangle.$$ Thus, it suffices to consider the case that $\mathfrak{b}$ is radical.

Every radical ideal in the Noetherian ring $R$ is a finite intersection of minimal primes over that ideal, $\mathfrak{q}_1\cap \dots \cap\mathfrak{q}_r.$ Since $\text{ht}(\mathfrak{b})$ equals $\min(\text{ht}(\mathfrak{q}_1),\dots,\text{ht}(\mathfrak{q}_r)),$ every prime ideal $\mathfrak{q}_j$ satisfies the same hypothesis as $\mathfrak{b}$. For every $j=1,\dots,r,$ assume that there exists $m_j=m_j(\underline{e})>0$ such that $p^{m_j} \in \mathfrak{q}_j + \langle S_1 -\alpha_1,\dots,S_n-\alpha_n\rangle$. Set $m= m_1+\dots+m_r$. Then also $$p^m=p^{m_1}\cdots p^{m_r}\in(\mathfrak{q}_1\cap \dots \cap \mathfrak{q}_r) + \langle S_1-\alpha_1,\dots,S_n-\alpha_n \rangle \subset \mathfrak{b} + \langle S_1-\alpha_1,\dots,S_n-\alpha_n \rangle.$$ Thus, it suffices to prove the result in the special case when $\mathfrak{b}$ is a prime ideal.

If the prime $\mathfrak{b}$ contains $S_1$, then for $m=e_1$, for every $\alpha_1 = p^{e_1}u$ for a unit $u$, $$p^{e_1} = u^{-1}\cdot p^{e_i}u = u^{-1}S_1 - u^{-1}(S_1-\alpha_1)\in \mathfrak{b} + \langle S_1-\alpha_1,\dots,S_n-\alpha_n \rangle.$$ So the result is proved when $S_1\in \mathfrak{b}$.

Finally, assume that $\mathfrak{b}$ is a prime and $S_1\not\in \mathfrak{b}$. Then the image of $S_1$ in the integral domain $R/\mathfrak{b}$ is nonzero, hence it is a regular element. By Krull's Hauptidealsatz, the quotient ring $R/(\mathfrak{b} + \langle S_1 \rangle)$ has dimension one less. Thus for the ring $R'=R/\langle S_1 \rangle = \mathbb{Z}_p[[\overline{S}_2,\dots,\overline{S}_n,\overline{X}_1,\dots,\overline{X}_d]]$ with $n-1$ variables $\overline{S}_i$, $i=2,\dots,n$, the ideal $\mathfrak{b}' = (\mathfrak{b}+ \langle S_1 \rangle)/\langle S_1 \rangle$ again satisfies the hypothesis. By induction on $n$, the result is true for $\mathfrak{b}'$. Thus, for every $e_2\gg \dots \gg e_n \gg 0$, there exists $m>0$ such that for every $(\alpha_2,\dots,\alpha_n)$ with $\alpha_i\in p^{e_i}\mathbb{Z}_p\setminus p^{e_i+1}\mathbb{Z}_p$, $$p^{m}\in \mathfrak{b}' + \langle \overline{S}_2-\alpha_2, \dots,\overline{S}_n-\alpha_n\rangle.$$ Thus, there exists $f_1,\dots,f_n\in R$ such that $$ p^m-(f_1S_1 + f_2(S_2-\alpha_2) + \dots +f_n(S_n-\alpha_n)) \in \mathfrak{b}.$$ For every $e_1>m$, for $\alpha_1=p^{e_1}u$ for $u$ a unit, then $p^m-p^{e_1}uf_1$ equals $p^m(1-p^{e_1-m}uf_1)$. Since the constant coefficient of $1-p^{e_1-m}uf_1$ is a unit, $1-p^{e_1-m}uf_1$ is an invertible element $v$ of $R$. Thus, $$p^m v -(f_1(S_1-\alpha_1) + f_2(S_2-\alpha_2) + \dots + f_n(S_n-\alpha_n))\in \mathfrak{b}.$$ Multiplying by $v^{-1}$, $p^m$ is in $\mathfrak{b}+ \langle S_1-\alpha_1,\dots,S_n-\alpha_n \rangle$. Therefore the result is proved by induction on $n$.

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  • $\begingroup$ Dear Professor Jason Starr, great thanks! Pierre Matsumi $\endgroup$ – Pierre MATSUMI Jan 18 '17 at 8:54

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