Why is $\mathbb{Z}/(1-x,p)$ isomorphic to $\mathbb{Z}_{p}$, where $p$ is a prime integer.

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 $\mathbb{Z_{p}}$

Solutions Collecting From Web of "Why is $\mathbb{Z}/(1-x,p)$ isomorphic to $\mathbb{Z}_{p}$, where $p$ is a prime integer."

Basically, yes.

It is a bit easier if you consider it in steps, though. First, consider $\mathbb{Z}[x]/(p)$, which is isomorphic to $\mathbb{Z}_p[x]$; then consider $\mathbb{Z}_p[x]/(1-x)$, which is isomorphic to $\mathbb{Z}_p$ under the map “evaluate at 1” (or using the division algorithm, like you do above).

Or you can do it the other way. First consider the quotient $\mathbb{Z}[x]/(x-1)$, which is isomorphic to $\mathbb{Z}$ under evaluation at $x=1$; then mod out by $(p)$.

(What you are doing if you go this route is using the homomorphism theorems; the ideal $(p + (x-1))$ of $\mathbb{Z}[x]/(x-1)$ corresponds to the ideal $(p,x-1)$ of $\mathbb{Z}[x]$, and $(R/I)/(J/I)\cong R/J$, where $I$ and $J$ are ideals of $R$ with $I\subseteq J$).

Define a map $\mathbb{Z}[x]\to \mathbb{Z}_p$ that sends $f(x)$ to the equivalence class of $f(1)$ modulo $p$. This is a homomorphism of rings. Since, every $f(x)$ can be written as $(x-1)q+r$ as you did, this map is surjective (just look at the images of the constant functions $0,1,…,p-1$), and the kernel is precisely $(x-1,p)$. This gives the required isomorphism by the first isomorphism theorem.

Counting cosets only gives you an abstract isomorphism, while this gives you a canonical one.

Just to add something to the nice answers you already have. In fact this is a special case of a more general proposition. It can be shown that all the maximal ideals $\mathfrak{m}$ of the polynomial ring $\mathbb{Z}[x]$ have the form $\mathfrak{m} = (g(x), p)$ where $p \in \mathbb{Z}$ is a prime number and $g(x) \in \mathbb{Z}[x]$ satisfies that its reduction modulo $p$ is an irreducible element $\bar{g}(x)$ in the polynomial ring $\mathbb{Z}_{p}[x]$. As a consequence of this you get that the quotient ring $ \mathbb{Z}[x] / \mathfrak{m} $, which in fact is a field since $\mathfrak{m}$ is maximal, is a finite algebraic extension field of $\mathbb{Z}_p$.

You can find this theorem for instance as proposition 1.5 in page 22 of Miles Reid’s Undergraduate Commutative Algebra. Another place where you may want to look is Section II.4.3 in page 84 of Eisenbud and Harris’ The Geometry of Schemes, where this is discussed more generally in the context of the prime spectrum $\text{Spec} \, \mathbb{Z}[x]$ and you can see a nice picture of it and how its “points” correspond to the points in $\text{Spec} \, \mathbb{Z}$.

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 $\rm\: \ S \cong R/K = R/b\:.$

Remember that finite fields are isomorphic if and only if they have the same number of elements.