The commutative ring $R$ is defined where every element $a^n=a$ for some $ n\in Z^{+}$ .
Let $cl(b) \in R/P$. As $b \in R \implies b^{n-1}=1$.
Now $cl(b)=b+P$.
$ (cl(b))^n =(b+P)^{n} $
= $b^{n} +.....+(P)^n$
= $b+P = cl(b)$
As $R/P$ is an integral domain there are no divisors of zero . Now $(cl(b))^{n-1}=1$. Hence $(cl(b))^{n-2}$ is the inverse of $cl(b)$. Is this attempt correct?
I'm going to write $\bar{b}$ instead of $cl(b)$ to make it easier to look at.
There is also a fatal problem with the original statement that "$a^n=a$ for some $n\in \mathbb Z^+$." This is trivially true in any ring for $n=1\in\mathbb Z^+$. What was intended is that the integer $n > 1$.
As noted by the other solution, it is unwarranted to say that $b^{n-1}=1$ in $R$. But also, you don't even need that.
The remainder is correct except for a technicality handling the case when $\bar{b}=\bar{0}$.
You noted that $\bar{b}^n=\bar{b}$ in $R/P$, which does allow cancellation, since it is a domain. Therefore you can say that either $\bar{b}=\bar 0$ or $\bar{b}$ cancels, and in the latter case $\bar{b}$ is invertible in $R/I$, with inverse $\bar{b}^{n-2}$. That means every element of $R/P$ is invertible or zero, so $R/P$ is a field, and therefore $P$ is maximal.
Actually this holds for a larger class of rings called von Neumann regular rings. (Ones that are commutative, that is.) A ring is called von Neumann regular if for every $a$ there exists an $x$ such that $axa=a$. In our case, here, $x=a^{n-2}$. (This exponent is always available since we have assumed $n > 1$.) there is a famous theorem by Jacobson that also says your ring must be commutative, too.
I would put the proof like this: let $P$ be a prime ideal of a von Neumann regular ring. Note that $(ax)^2=ax$ so $ax$ is an idempotent. Now, in the domain $R/P$, an idempotent must be $\bar 0$ or $\bar 1$, so $\overline{ax}=\bar 0$ or $\overline{ax}=\bar 1$. In the first case, $\bar a=\overline{axa}=\bar 0$, and in the second case $\bar x=\bar{a}^{-1}$. Thus each element is zero or invertible, $R/P$ is a field, and $P$ is maximal.
NO, you cannot simplify by $b$ in $R$. However, you can do it in $R/P$ since its an integral domain, PROVIDED THAT $cl(b)\neq 0$.
If you want to write things properly, you should say first that you take an element $cl(b)\neq 0$ in the quotient, and show that it is invertible.
