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.


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 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. $\endgroup$ – Mark Spivakovsky Jan 24 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? $\endgroup$ – Rinmyaku Jan 24 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 $R$. $\endgroup$ – Mark Spivakovsky Jan 25 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" $\endgroup$ – Mark Spivakovsky Jan 25 at 18:07

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.