Let $\pi \colon R \twoheadrightarrow {\Bbb T}$ be a surjective ring homomorphism between finite algebras over ${\Bbb Z}_p$. Further, we suppose the following three conditions$\colon$

  1. $R$ is a complete intersection, i.e. $R = {\Bbb Z}_p[[X_1,\ldots,X_d]]/(f_1,\ldots,f_d)$.

  2. $\pi$ induces an isomorphism $\pi^{*} \colon {\mathrm{Hom}}_{{\Bbb Z}_p}({\Bbb T}, \overline{{\Bbb Z}_p}) = {\mathrm{Hom}}_{{\Bbb Z}_p}(R, {\overline{\Bbb Z_p}})$, where we denote by $\overline{{\Bbb Z}_p}$ the integral closure of ${\Bbb Z}_p$ in the algebraic closure $\overline{{\Bbb Q}_p}$.

  3. ${\Bbb T}$ is reduced.

We shall denote by $R^{\mathrm{red}}$ the reducification of $R$. Then, I would like to ask

Q. Do the above three conditions imply the isomorphism $\pi \colon R^{\mathrm{red}} \cong {\Bbb T}$?

In the case where $\overline{{\Bbb Z}_p}$ is replaced with ${\Bbb Z}_p$, the question was answered in the negative by Professor Spivakovsky.


2 Answers 2


The answer is ``no''. Let $R=\frac{\mathbb Z_p[X]}{(X^2-p)(X^3-p)}$ and $\mathbb T=\frac{\mathbb Z_p[X]}{(X^2-p)}$. The ring $R$ is a hypersurface, hence a complete interserction; in addition, it is reduced, so $R=R^{red}$. The ring $\mathbb T$ is also reduced. The map $\pi^*$ is the zero map between two modules, each of which is equal to $(0)$, so it is an isomorphism. Yet, $R=R^{red}$ is not isomorphic to $\mathbb T$.

  • 1
    $\begingroup$ Why is T a quotient of R? $\endgroup$
    – GTA
    Jan 24, 2020 at 11:18
  • $\begingroup$ Because the ideal $(X^2-p)(X^3-p)$ is contained in the ideal $(X^2-p)$. There is a natural surjective homomorphism from $R$ to $\mathbb T$ that sends every element of $(X^2-p)$ to 0. $\endgroup$ Jan 24, 2020 at 15:20

The answer to the modified question (that is, with $\overline{\mathbb Z_p}$ instead of $\mathbb Z_p$) is "yes".

The prime $p$ is not a zero divisor in $R$. Replacing $R$ by $R^{red}$ does not change the problem (we no longer claim that $R$ is a complete intersection, but $p$ is still not a zero divisor). We will prove that $\pi$ is an isomorphism. Assume the contrary, aiming for contradiction. Let $I=Ker(\pi)$. Then there exists a minimal prime $P$ of $R$ such that $I\not\subset P$, so $\pi$ induces a map $\frac RP\rightarrow\frac R{I+P}$ with $\frac R{I+P}$ zero-dimensional.

The ring $\frac RP$ contains $\mathbb Z_p$ and is integral over it, hence admits a non-zero homomorphism to $\overline{\mathbb Z_p}$. This homomorphism does not factor through $\mathbb T$, contradicting the hypothesis 2.

  • $\begingroup$ Dear Mark, I would like you teach me how you deduce the map from minimal primes of T to those of R? Is it obvious? As far as I know, the intersection of all minimal primes coincides with nilpotent elements. However, if R is reduced from the beginning, it must be $0$. I don't see where you used the condition 1. (or 3.) Could you please explain the body of the proof more closely? $\endgroup$
    – Pierre
    Jan 24, 2020 at 15:46
  • $\begingroup$ Proof that $p$ is not a zero divisor in $R$. The ring $R$ is a one-dimensional Cohen-Macaulay ring. Hence all of its associated primes are minimal and have coheight 1. The ring $\frac R{(p)}$ is finite over $\mathbb F_p$, hence zero-dimensional. Thus $p$ is not contained in any minimal prime of $R$, hence in no associated prime of $R$. This proves that $p$ is not a zero divisor in $R$. $\endgroup$ Jan 25, 2020 at 16:19
  • $\begingroup$ I use hypothesis 1 to prove that $p$ is not a zero divisor in $R$. The intersection of ALL minimal primes of $R$ is $(0)$; the intersection of SOME of them need not be $(0)$. If $\pi$ is not an iso, its kernel is such an intersection. I use hypothesis 1 once again to conclude that $\frac RP$ admits a map to $\mathbb Z_p$ (since $p\notin P$). I use hypothesis 3 in the sentence "Replacing ... does not change the problem" $\endgroup$ Jan 25, 2020 at 18:07

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