Intuition behind factor ring, ideals with multiple generators and isomorphism.

Question

Given the following statement:

$\mathbb{Z}[x]/(1-x,n) \cong \mathbb{Z}_{n}$ for $n \in \mathbb{N}$.

How would one go around wrapping their head around it?

For instance, when dealing with factor rings like $\mathbb{R}[X]/(X^2 + 1)$ everything is of the form $a+bX$ and then you can apply the "think-of-generator-as-zero" trick and obtain $X^2=-1$ and see how it's isomorphic to $\mathbb{C}$.

But what happens when you factor a ring by an ideal generated by two elements? What's the form of the elements inside $\mathbb{Z}[x]/(1-x,n)$?

Answer 1

The isomorphism works for any $n$, and is true because of the 3rd isomorphism theorem: $$\mathbf Z[X]/(np,1-X)\simeq (\mathbf Z[X]/n\mathbf Z[X])/\bigl((n,1-X)/p\mathbf Z[X]\bigr)$$ and of this isomorphism: $\;\mathbf Z[X]/n\mathbf Z[X]\simeq (\mathbf Z/n\mathbf Z)[X]$.

In practice, a polynomial $f(X)$ modulo $(n, 1-X)$ is the same as$\bar f(\bar1)$ where $\bar f$ is the polynomial with coefficients reduced modulo $n$.

Answer 2

Well, the intuition is pretty much the same, you want morphism $\varphi:\Bbb Z[x]\to \Bbb Z_p$ such that $1-x$ and $p$ map to $0$, i.e. $\varphi(x) = \varphi(1)$, $\varphi(p) = p\varphi(1) = 0$. Reasoning is that ideal is sent to $0$ if and only if its generators are. Now, if you send $1$ to $1$, you get map $$\varphi(a_0+a_1x+\ldots+a_nx^n) = a_0+a_1+\ldots+a_n.$$ This is obviously surjective map, all it now needs to be shown is $\ker \varphi = (1-x,p)$ to establish the requested isomorphism. Now, if $f\in\ker\varphi$, then the sum of it's coefficients is divisible by $p$. Let $\{a_{i_1},\ldots a_{i_k}\}$ be all coefficients of $f$ divisible by $p$ and $\{b_{j_1}\,\ldots, b_{j_l}\}$ those that are not. Now, $$f=\sum_r a_{i_r}x^{i_r} + \sum_r b_{j_r}x^{j_r}$$ where the left sum is divisible by $p$, and the right has $1$ as its root, hence is divisible by $1-x$. We conclude that $f = g(x)(1-x)+h(x)p$ and thus $f\in (1-x,p)$ completing the proof.

As you can see, $p$ being prime had absolutely no role, except that for $p$ prime you can conclude that $(1-x,p)$ is maximal ideal. If $p$ weren't prime, $\Bbb Z_p$ would not be integral domain and $(1-x,p)$ wouldn't be prime ideal.