Quotient by product of prime ideals in polynomial rings In Allen Altman's Commutative Algebra, there is the affirmation that if $p$ is a prime ideal of $R$ and $R$ is a ring, then $(pR[x]+\langle x\rangle)/(pR[x]) \cong \langle x\rangle$. It seems intuitively plausible, but $pR[x]\cap\langle x\rangle \neq \emptyset,$ so I think it is not true.
Should it be $(pR[x]+\langle x\rangle)/(pR[x]) \cong (\langle x\rangle)/(p\langle x\rangle) \cong R/p[x]$?
Thank you in advance for your help
 A: From the isomorphism theorems for rings, we have that
$(S + I)/I \cong S/(S\cap I)$. In this case, we have $S = \langle x\rangle, I = pR[x]$, and hence we have
$(pR[x] + \langle x\rangle)/(pR[x]) \cong \langle x\rangle/ (\langle x\rangle \cap pR[x])$
https://en.wikipedia.org/wiki/Isomorphism_theorems
The question now is--what is $\langle x\rangle \cap pR[x]$?
Elements of $pR[x]$ are of the form $\sum\limits_{i=0}^n a_ip_ix^i$ where $a_i\in R$ and $p_i\in p$. And elements of $\langle x\rangle$ are of the form $\sum\limits_{i=1}^n b_ix^i$, i.e. there is no nonzero constant term in any polynomial in $\langle x\rangle$ except the zero polynomial. Then these two intersect exactly when $\sum\limits_{i=1}^n b_ix^i = \sum\limits_{i=1}^n a_ip_ix^i$ where $a_ip_i = b_i$, $a_i\in R$, and $p_i\in p$. (We should have been using $\mathfrak{p}$ for our notation this whole time...oh well.)
Hence in our quotient, we have that $\sum\limits_{i=1}^n a_ip_ix^i \equiv 0$ (where $a_i\in R, p_i\in p$). We are not quite left with just $(R/p)[x]$, however, because we still don't have polynomials of degree 0. I think it would be more like the ideal $\langle y\rangle\subset (R/p)[y]$. (I'm using another variable here just to make the distinction.)
