TITLE: Why is $\mathbb{Z}[x]/(1-x,p)$ isomorphic to $\mathbb{Z}_{p}$, where $p$ is a prime integer.
QUESTION [12 upvotes]: I want to know why $\mathbb{Z}[x]/(1-x,p)$ is isomorphic to $\mathbb{Z}_{p}$, where $p$ is a prime integer?

Here's what I have so far, but I am unsure if I am correct.   Every $f\in \mathbb{Z}[x]$ can be written as $(1-x)q+ r$ where $q\in \mathbb{Z}[x]$ and $r$ is in $\mathbb{Z}$.   Does It follows that there are $p$ cosets of $(1-x,p)$  ( namely 0+(1-x,p), 1+(1-x,p),2+(1-x,p)  etc ..) 
That would imply $\mathbb{Z}[x]/(1-x,p)$ is isomorphic to $\mathbb{Z_{p}}$

REPLY [2 votes]: LEMMA $\rm\quad S\ :=\ R[x]/(x-a,b)\ \cong\ R/b\ \ $ for $\rm\ a,b\in R\ $ any ring. 
Proof $\ \ $ In $\rm\:S\ $ we have $\rm\ x = a\ $ hence $\rm\ f(x) = f(a)\:.\ $ Therefore the natural  map of $\rm R $ into $\rm S$ is onto, with kernel $\rm K \supset b\:R\:.\ $  If $\rm\ c\in K\ $ then $\rm\ c = (x-a)\ f(x) + b\ g(x)\ \Rightarrow\ c\in b\:R\ \ $  via $\ $ eval $\: $ at $\rm\ x = a\:.$ Therefore $\rm\ \  \ K = b\:R\ \ $ so $\rm\: \ S \cong R/K =  R/b\:.$