If $f(x)\in\mathbb{Z}[X]$ is irreducible,then $\mathbb{Z}[X]/⟨f(x)⟩\cong\mathbb{Z}[a]$,where a is a root of f in some field extension

Question

We know that :

Let $\mathbb{F}$ be a field, $f(x)$ is a irreducible polynomial in $\mathbb{F}[x]$ and has a root $\alpha$ in some extension then $\mathbb{F}[x]/⟨f(x)⟩\cong\mathbb{F}(\alpha)$

But I saw here that OP is using same arguement for ring of integers $\mathbb{Z}$ even though $\mathbb{Z}$ is not a field.So I want to know whether this statement is true:

If $f(x)\in\mathbb{Z}[X]$ is irreducible,then $\mathbb{Z}[X]/⟨f(x)⟩\cong\mathbb{Z}[a]$,where $a$ is a root of $f$ in some field extension

Is this statement correct?If it is so then what is the most general result?

Thanks in advance!

Answer 1

The statement doesn't quite make sense; what do you mean by a field extension? There is no field in your statement...

If $f\in\Bbb{Z}[X]$ has a root $a$ in some field extension of $\Bbb{Q}$, then it has a root in the field extension $\Bbb{Q}\subset\Bbb{Q}(a)$, and we have $\Bbb{Z}[a]\subset\Bbb{Q}(a)$. The kernel of the surjective map $$\varphi:\ \Bbb{Z}[X]\ \longrightarrow\ \Bbb{Z}[a]:\ X\ \longmapsto\ a,$$ clearly contains $f$, and so $(f)\subset\ker\varphi$. Because $f$ is irreducible it follows that $\ker\varphi=(f)$. (Why is $\ker\varphi=(1)$ impossible?) Now by the first isomorphism theorem $\Bbb{Z}[a]\cong\Bbb{Z}[X]/(f)$.

More generally, if $R\subset S$ is an extension of integral domains and $f\in R[X]$ is irreducible and $s\in S$ is such that $f(s)=0$, then $R[s]\cong R[X]/(f)$.