# Isomorphism between finite algebras over ${\Bbb Z}_p$

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

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 $$\mathbb T$$.

• Why is T a quotient of R? – GTA Jan 24 '20 at 11:18
• 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. – Mark Spivakovsky Jan 24 '20 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.

• 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? – Rinmyaku Jan 24 '20 at 15:46
• 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 $R$. – Mark Spivakovsky Jan 25 '20 at 16:19
• 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" – Mark Spivakovsky Jan 25 '20 at 18:07