Show that $U/M$ is a finite field. Let $U$ be a finitely generated $\mathbb{Z}_p$-algebra and let $M$ be the maximal ideal of $U$, where $\mathbb{Z}_p$ is the ring of $p$-adic integers.

Is $U/M$ a finite field ? If no, then what is the best possible structure on $U/M$ ?

In this post, the analogous question has been answered, the answer by @Daniel Litt, was easy to understand. However unlike $\mathbb{Z}$, the $p$-adic integers $\mathbb{Z}_p$ is not Jacobson ring and hence inverse-image of the maximal ideal $M$ of $U$ under the ring homomorphism $f: \mathbb{Z}_p\to U$ may not (to check ?) be maximal ideal in $\mathbb{Z}_p$. This creates problem.
In otherwords, if the inverse image $f^{-1}(M)$ of the maximal ideal $M$ of $U$ becomes maximal then $\mathbb{Z}_p/f^{-1}(M)$ would be a finite field of characteristic $p$. In that case, since $U$ is finitely generated over $\mathbb{Z}_p$, we get that  $U/M$ is a finite extension of the finite field $\mathbb{Z}_p/f^{-1}(M)$. Therefore $U/M$ is a finite field of prime characteristic.
So the question is-
how to make sure that the inverse image $f^{-1}(M)$ of the maximal ideal $M$ of $U$ becomes maximal ideal in $\mathbb{Z}_p$  ?
Are there other way to answer the first question ?
For example,if $k$ is a field, then by Zariski Lemma, any finitely generated $k$-algebra $R$ and maximal ideal $\mathfrak{m} \subseteq R$, we have that $R/\mathfrak{m}$ is a finite field extension of $k$. Unfortunately in our case it is $\mathbb{Z}_p$  which is an integral domain but not field.
Thanks
Edit: Following answer shows if we assume $U$ to be $\text{finite}$ $\mathbb{Z}_p$-algebra instead of just $\text{finitely generated}$ $\mathbb{Z}_p$-algebra, then we are done.
 A: First of all, the claim you are trying to prove is somewhat ambiguous or misleading:

Let $U$ be a finitely generated $\mathbb{Z}_p$-algebra and let $M$ be the maximal ideal of $U$, where $\mathbb{Z}_p$ is the ring of $p$-adic integers.
Show that $U/M$ is a finite field of characteristic prime, $p$ say.

In general a finitely generated $\Bbb{Z}_p$-algebra can have more than one maximal ideal, so it is unclear what 'the' maximal ideal of $U$ should be.
Next note that $U:=\Bbb{Q}_p$ is a finitely generated $\Bbb{Z}_p$-algebra because $\Bbb{Q}_p\cong\Bbb{Z}_p[X]/(pX-1)$. Its maximal ideal is $M=0$, but clearly $U/M\cong\Bbb{Q}_p$ is not a finite field of prime characteristic.

If you require $U$ to be a finitely generated $\Bbb{Z}_p$-module then indeed $U/M$ is a finite field for every maximal ideal $M\subset U$. In this case $U/M$ is finitely generated as a $\Bbb{Z}_p/(M\cap\Bbb{Z}_p)$-module, where $M\cap\Bbb{Z}_p\subset\Bbb{Z}_p$ is a prime ideal of $\Bbb{Z}_p$. Because $\Bbb{Z}_p$ is local its only prime ideals are $0$ and $p\Bbb{Z}_p$, but if $M\cap\Bbb{Z}_p=0$ then $U/M$ contains $\Bbb{Z}_p$ as a subring, and hence also $\Bbb{Q}_p$, which is not finitely generated as a $\Bbb{Z}_p$-module, a contradiction. Hence $M\cap\Bbb{Z}_p=p\Bbb{Z}_p$, and from here your argument goes through.
