# If $A= \mathbb{Z}[X]/(X^n+1)$, is it true that $A/mA \cong \mathbb{Z}_m[X]/(X^n+1)$?

Let $R= \mathbb{Z}[X]/(X^n+1)$ for some sufficiently large $n$. For $q \geq 2$, I want to show that $R/qR \cong \mathbb{Z}_q[X]/(X^n+1)$.

I've tried to prove it, but I dont know the construction of $qR$. In fact,

$qR=q \left( \mathbb{Z}[X]/(X^n+1)\right)$; is it equal to $(q\mathbb{Z})[X]/(X^n+1)$?? because $X^n+1 \notin q\mathbb{Z}[X]$. Please if someoene can understand this things, and prove the isomorphism (ring isomorphism).

Consider the following exact sequence of abelian group: $$\{0\}\to(X^n+1)\Bbb Z[X]\to\Bbb Z[X]\to R\to\{0\}$$ By tensoring with $\Bbb Z/q\Bbb Z$ we get the following exact sequence: $$(\Bbb Z/q\Bbb Z)\otimes(X^n+1)\Bbb Z[X]\to(\Bbb Z/q\Bbb Z)\otimes\Bbb Z[X]\to(\Bbb Z/q\Bbb Z)\otimes R\to\{0\}$$ Since tensoring commutes with polynomial ring extension (see here for a proof) and with quotients (see here) we get the exact sequence $$(\Bbb Z/q\Bbb Z)\otimes(X^n+1)\Bbb Z[X]\to(\Bbb Z/q\Bbb Z)[X]\to R/qR\to\{0\}$$ Since the image of $(\Bbb Z/q\Bbb Z)\otimes(X^n+1)\Bbb Z[X]$ into $(\Bbb Z/q\Bbb Z)[X]$ is the ideal $(X^2+1)(\Bbb Z/q\Bbb Z)[X]$, we get the required isomorphism $$R/qR\cong(\Bbb Z/q\Bbb Z)[X]/(X^2+1)(\Bbb Z/q\Bbb Z)[X]$$