Existence of a ring with some properties All our rings are nonzero, commutative and unital. Given a ring $R$ and an element $p\in R^*$, let
$$\begin{align*}
A_p=&\,\{(p-1)r+1\colon\ r\in R\};\\
B_p=&\,\{p^j\colon\ j\in\mathbb{Z}\}.
\end{align*}$$
For $r\in R$ and $n\in\mathbb{N}$, let
$$w_n(r)=1+r+\cdots+r^{n-1},$$
so that
$$\begin{equation}\begin{aligned}
(p-1)\cdot w_n(p)+1=&\,p^n;\\
(p-1)\cdot [-p^{-n}w_n(p)]+1=&\,p^{-n}.
\end{aligned}\end{equation}\tag{$\ast$}\label{wn}$$
This shows that the inclusion $A_p\supseteq B_p$ always holds.

QUESTION: Does there exist a  ring $R$ containing an invertible element $p$ such that $A_p=B_p?$

If we allow $p-1$ to be a zerodivisor, the answer is affirmative: take $p=1$, for example. For a nontrivial example, consider $R=\mathbb{Z}[t]/I$, being $I=\langle 2(t-1),t^{2}-1\rangle$, and take $p=\overline t$: it can be easily verified that $A_p=B_p=\{1,p\}$ in this case. I think that similar examples can be constructed with the element $p$ having any prescribed finite multiplicative order, but I am not interested on these.

MINOR QUESTION: Can $p$ be taken with infinite multiplicative order (and such that $p-1$ is a zerodivisor)?


The real challenge rises when we impose the additional condition "$p-1$ is regular".
My thoughts so far:

Let $R$ be a ring containing an element $p\in R^*$ with $p-1$ regular such that $A_p=B_p$.

1. $p-1$ cannot be invertible: otherwise we would have $(p-1)\cdot[-(p-1)^{-1}]+1=0\in A_p$, and clearly $0\notin B_p$.
2. $R$ must be infinite (regular elements in finite rings are always invertible).
3. $p$ has infinite multiplicative order: in fact, the ring $R$ is equipotent with the set $A_p$, via the map $r\mapsto (p-1)r+1$ (which is injective by regularity of $p-1$). Therefore $B_p=A_p$ is infinite by 2.
4. $R=\{w_n(p)\colon\ n\geq 0\}\cup\{-p^{-n}w_n(p)\colon\ n\geq 0\}.$
In fact, if $r\in R$, then $(p-1)r+1=p^j$ for some $j\in\mathbb{Z}$. Using the regularity of $p-1$ together with \eqref{wn} we obtain the desired result.

From now on, let $q=p^{-1}$. Then for all $n\geq1$ we have
$$-p^{-n}w_n(p)=-qp^{1-n}(p^{n-1}+p^{n-2}+\cdots+p+1)=-qw_n(q),$$
and therefore
5. $R=\{w_n(p)\colon\ n\geq0\}\cup\{-qw_n(q)\colon\ n\geq0\}.$
6. We also have
$$R=\{w_n(q)\colon\ n\geq0\}\cup\{-pw_n(p)\colon\ n\geq0\},$$
which shows the (expected) symmetry between $p$ and $q$ in the problem. This follows immediately from 5, together with the equality $R=-pR$.

We have $p\in R$, hence for some $j\geq 0$ we have $p=w_j(p)$ or $p=-qw_j(q)$.
7. If $p=w_j(p)$, then $j\geq 4$. In fact:


*

*$j=0\implies p=w_0(p)=0$, which is absurd;

*$j=1\implies p=w_1(p)=1$, which is absurd by 3 (or, more elementarily, because $p-1$ is regular, hence nonzero);

*$j=2\implies p=w_2(p)=1+p\implies 0=1$, which is absurd;

*$j=3\implies p=w_3(p)=1+p+p^2\implies p^2=-1\implies p^4=1$, which is absurd by 3.


8. If $p=-qw_j(q)$, then $q=w_{j+2}(q)$. In fact, we have $w_{j+2}(q)=1+q+q^2w_j(q)$, so $p=-qw_j(q)$ implies
$$w_{j+2}(q)=1+q-q[-qw_j(q)]=1+q-qp=q.$$

As a consequence, by changing $p$ by $q$ if necessary (item 6), we may assume that
$$p=w_j(p),\ \style{font-family:inherit;}{\text{for some}}\ j\geq 4$$
(by item 7).
9. $R$ is finitely generated, as $\mathbb{Z}$-module, by $\{1,p,\ldots,p^{j-2}\}$. In fact, by 5 we have $R=\mathbb{Z}[p,q]$, and from the equality $p=1+p+p^2+\cdots+p^{j-1}$ we get $1=-p^2w_{j-2}(p)$, which shows that $q=p^{-1}=-pw_{j-2}(p)\in\mathbb{Z}[p]$. Therefore $R=\mathbb{Z}[p]$, and the basics of the theory of integral dependence show that $R=\sum_{i=0}^{j-2}\mathbb{Z}p^i$ in this case.
10. $\operatorname{char}(R)=0$: in fact, if $\operatorname{char}(R)=m>0$, then 9 would imply that $R$ is a finitely generated $\dfrac{\mathbb{Z}}{m\mathbb{Z}}$-module, hence a finite ring, contradicting 2.

Well... I am stuck at this point. I believe that no such ring $R$ exists. Any help will be appreciated.
 A: We will first prove that
1) Some power of $p$ is an integer or the inverse of an integer and therefore
2) some integer $K+1$ is a unit
Then we will possibly modify this integer to prove that some integer is $0$ in $R$, which will end the argument, after a short reflection.
As $q\in \mathbb{Z}[p]$, as you observed, $R$ is a quotient of $\mathbb{Z}[x]$ by an ideal $I$, with $p$ corresponding to $x$. As $x$ is invertible in this quotient, some element of the form $xf-1$ lies in $I$ and therefore $R$ is a finite module over $\mathbb{Z}$, as you observed. Now, if you take its total quotient (tensor with $\mathbb{Q}$), you get a finite-dimensional vector space $R_0$ over $\mathbb{Q}$. It is a quotient of $\mathbb{Q}[x]$ by the ideal corresponding with $I$.
If $(p-1)$ is regular, multiplication by it is injective. This can be seen as a $\mathbb{Q}$-linear endomorphism of $R_0$. Since it is injective (and $R_0$ is a finite-dimensional vector space), it is surjective too and $1$ belongs to its image. Therefore, there exists a polynomial $h\in \mathbb{Q}[x]$ such that $(x-1)h=1$ in $R_0$. By multiplying by the common denominator K, you get an integer K and $g\in Z[x]$ such that $(x-1)g=K$ in $R$. Observe that the number $K$ can be arbitrarily multiplied by any integer, and still grant the existence of an(other) integer polynomial $g$ satistying the last condition.
Observe the equation $(x-1)g=K$ in two cases
Now the case $g=w_n(x)$ implies that $x^n=K+1$ is an integer (clearly we can't have $K=-1$) and if you want $p$ to have distinct powers we also can't have $K=-2$ or $0$
The case $g=-(1/x^n)w_n(x)$ yields $x^n=1/(K+1)$, that is, $x^n$ is the inverse of an integer.
In both cases, $K+1$ is invertible since $x$ is.
Finally, to kill your argument once and for all, we got two polynomials in $I$, that is, $xf-1$ and either $x^n-j$ (case 1) or $jx^n-1$ (case 2), where $j=K+1$ from the above argument. Surely $j$ cannot be $1$ or $-1$, so there is a prime $r$ dividing $j$ (I avoid using letters $p$ and $q$). 
Observe now that, in case $r^2$ divides $j$, $K$ and $j$ can be modified to ensure that $r$ divides $j$ but $r^2$ does not (possibly modifying $n$). As observed in the argument above $K$, being the denominator of $h$, can be multiplied by any integer, still granting the existence of the required $g$. Replace $K$ by $(r+1)K$. Observe that $K$ is coprime with $r$, for $r$ divides $K+1$. Now the new $j$ will be $(r+1)K+1=rK+(K+1)$, a multiple of $r$, which is not multiple of $r^2$, for $K+1$ is and $rK$ is not.
Taking possibly this new $j$, in case 1, we can therefore apply Eisenstein's criterion to $r$ and conclude that $x^n-j$ is irredicible in $\mathbb{Q}[x]$. Moreover, still in $\mathbb{Q}[x]$, it must be comprime to $xf-1$ simply because it does not divide it (just look at their degree $0$ terms).
Then, since $\mathbb{Q}[x]$ is an Euclidian domain, $1$ is a linear combination in $\mathbb{Q}[x]$ of elements of $I$ and therefore there is an integer $N$ in $I$. Then $R$ is a finite $\mathbb{Z}_N$ module, hence a finite set, a contradiction. For case 2 please check that you can just replace $xf-1$ by the polynomial expressing the inverse of $q$ in $\mathbb{Z}[x]$ modulo $I$, that is, seeing $R$ as a quotient of $\mathbb{Z}[x^{-1}]$ through the isomorphism $f(x)\leftrightarrow f(x^{-1})$.
A: I will answer my question, based on the answer provided by Marcus Barão Camarão (the accepted answer). I will use the notations as in the original question (and of course my own "writing style").
First, since $R/\mathbb{Z}$ is an integral extension of rings (by 9), the element $p-1$ satisfies a (monic) nonconstant polynomial in $\mathbb{Z}[X]$, and since $p-1$ is regular, we may assume that such polynomial has nonzero constant term. Isolating such term yields $(p-1)a=K$, for some $a\in R$ and some nonzero integer $K$. (I think that the original argument using tensorization is correct, but even so the justification would be somewhat complicated.)
On the other hand, $(p-1)a=p^n-1$ for some integer $n$ (this is the very initial hypothesis $A_p=B_p$), and $n\neq 0$ because $(p-1)a=K\neq 0$ (by 10). Let $j=K+1$, so $p^n=j$. Obviously $j\neq 0$, and by 3 we have $j\neq\pm 1$. Let $r$ be a prime factor (in $\mathbb{Z}$) of $j$. If $r^2\mid j$, then $K\equiv -1\pmod{r,r^2}$. Defining $j'=(r+1)K+1$ we get
$$j'\equiv(r+1)\cdot(-1)+1=-r\pmod{r,r^2},$$
which shows that $r$ divides $j'$ but $r^2$ does not, and moreover $(p-1)[(r+1)a]=(r+1)K=j'-1$. As mentioned before, the left side is equal to $p^m-1$ for some nonzero integer $m$ (possibly distinct from $n$), hence $p^m=j'$. Changing $j$ by $j'$ and $n$ by $m$ if necessary, we may assume that $p^n=j$, with $r\mid j$ but $r^2\nmid j$. Moreover, changing $p$ by $q$ if necessary, we may assume that $n\geq 1$ (by 6).
Finally, as $p\in R^*$ and $R=\mathbb{Z}[p]$, we have $pf(p)=1$, for some $f(X)\in\mathbb{Z}[X]$; as $j\nmid -1$ in $\mathbb{Z}$, it follows that $X^n-j\nmid Xf(X)-1$ in $\mathbb{Z}[X]$. By Eisenstein's criterion (using the prime $r$), the polynomial $X^n-j$ is irreducible in $\mathbb{Z}[X]$, hence irreducible in $\mathbb{Q}[X]$. We cannot have $X^n-j\mid Xf(X)-1$ in $\mathbb{Q}[X]$ (otherwise, if $Xf(X)-1=h(X)(X^n-j)$, with $h(X)\in\mathbb{Q}[X]$, then after clearing denominators we would obtain that $X^n-j$ divides $b\cdot(Xf(X)-1)$ in $\mathbb{Z}[X]$, for some nonzero integer $b$, and this is impossible because $X^n-j$ is prime and it has degree greater than zero, so it cannot divide $b$). Thus, $A(X)(X^n-j)+B(X)(Xf(X)-1)=1$ for some $A(X),B(X)\in\mathbb{Q}[X]$. if $N$ is a positive integer such that $NA(X),NB(X)\in\mathbb{Z}[X]$, then after multiplying the last equality by $N$ and evaluating at $p$, we get $0=N$ in $R$, hence $\operatorname{char}(R)>0$, contradicting 10.
