# What is $\Bbb Z[X]/(aX+b)$ isomorphic to?

Let $a,b$ be integers. I would to know what other ring is $R=\Bbb Z[X]/(aX+b)$ isomorphic to?

If $a$ is a unit of $\Bbb Z$, then $R \cong \Bbb Z$. If $a=0$, then $R \cong (\Bbb Z/b\Bbb Z)[X]$. If $a=2,b=0$, then $R \cong \Bbb Z \oplus \Bbb F_2[X]$ as abelian groups at least, but I'm not sure as rings. If $b$ is a multiple of $a$, we could use the Chinese remainder theorem, I think.

But in general, for instance $\Bbb Z[X]/(2X+3)$ or $\Bbb Z[X]/(6X+4)$ I don't know how to manage. It would also be interesting to know what happens if we replace $\Bbb Z$ by any other commutative ring (and $a,b$ elements of that ring)...

Thank you!

• As an abelian group, $R\cong \mathbb{Z}\oplus( \mathbb{Z}/a\mathbb{Z})[X]$ if $a\neq 0$, and $R\cong (\mathbb{Z}/b\mathbb{Z})[X]$ if $a=0$. – Batominovski Aug 16 '16 at 19:49
• If $\gcd(a, b) = 1$ then the ring you get is the subring $\mathbb{Z}[1/a]$ of $\mathbb{Q}$. Otherwise it's something more annoying. For example $\mathbb{Z}[x]/(2x)$ has no simpler description, really. – Qiaochu Yuan Aug 16 '16 at 19:49
• I think that you get $\Bbb Z[-b/a]$ when $-b/a\notin \Bbb Z$. In general, $\Bbb Z[X]/(aX+b)=\{ \sum_{k=0}^n a_k\varepsilon^k : n\in\Bbb N,\,(a_0,\dots,a_n)\in\Bbb Z^n\}$ where $\varepsilon=X\mod (aX+b)$ is an "extra element" s.t. $a\varepsilon+b=0$. That's all. – paf Aug 16 '16 at 20:01
• Note that if $-b/a\in\Bbb Z$, this ring isn't a domain because the equation $aX+b=0$ has 2 distinct solutions. – paf Aug 16 '16 at 20:12
• @paf $\$ More directly,  if $\, b/a\in \Bbb Z\$ then $\, a(X+b/a) = 0\,$ so $\,a\,$ is a nontrivial zero divisor in $\,R.\,$ More generally if nonunit $\,d\mid a,b\,$ then $\,d\,(\frac{a}d X+\frac{b}d) = 0\,$ so $\,d\,$ is a nontrivial zero divisor. When it is not a domain, it cannot be $\,\Bbb Z[-b/a]\,$ (or any other subring of $\,\Bbb Q,\,$ since all are domains). – Bill Dubuque Aug 16 '16 at 20:27