# $\mathbb Z [X] / (X)$ isomorphic to $\mathbb Z[X] / (X+1)$ isomorphic to $\mathbb Z [X] / (X+2015)$

I have to prove that $\mathbb Z [X] / (X) \cong \mathbb Z[X] / (X+1) \cong \mathbb Z [X] / (X+2015)$.

I think that one answer could be that $\mathbb Z[X]/(X) \cong \mathbb Z(0)$, $\mathbb Z[X]/(X+1)\cong Z(-1)$ and $\mathbb Z [X] /(X+2015)\cong \mathbb Z(-2015)$, as $0, -1, -2015$ are roots of these polynomials. Also, as $0, -1, -2015\in \mathbb Z$, then all of them are isomorphic to $\mathbb Z$, hence isomorphic between them.

But I don't know very well how to justify this rigorously and my teacher has said to me that there is a simpler way to proof this, without using the roots of these polynomials.

$$\phi:\Bbb Z[x]\to\Bbb Z[x]/\langle x+1\rangle\;,\;\;\phi(p(x)):=p(x+1)+\langle x+1\rangle$$
• ,is the approach by OP correct?I doubt it,since I have studied that $\mathbb{F}[X]/⟨p(X)⟩\cong\mathbb{F(a)}$ where $\mathbb{F}$ is field and p is an irreducible polynomial in $\mathbb{F}[X]$,since in OP's case $\mathbb{Z}$ is not a field,can we still apply this theorem? Oct 9, 2020 at 10:06