Q:

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$\colon$

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

A:

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 $\mathbb T$. 

