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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 $\mathbb{Z_{p}}$

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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.

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