Given a commutative ring $R$ and a monic polynomial $p(x) \in R[x]$ is $R[x]/\langle p(x) \rangle$ always a finite integral extension of $R$?

Question

I suspect this to be true based on the fact that $p(x)$ is monic, so it should be the case that $R[x]/\langle p(x) \rangle$ is a finitely generated module over $R$, but I have no good reference for this.

Answer 1

If $p(x)=x^d+a_{d-1}x^{d-1}+\dots+a_0$, then the quotient is generated (in fact, freely generated) by $S=\{1,x,\dots,x^{d-1}\}$. Indeed, you can prove by induction that $x^n$ is in the submodule generated by $S$ for each $n$. For $n<d$ this is trivial. For $n\geq d$, you have $x^{n-d}p(x)=0$ so $x^n=-a_{d-1}x^{n-1}-\dots-a_0x^{n-d}$, which is generated by lower powers of $x$ and hence by $S$. To see that $S$ freely generates the quotient, note that any nonzero multiple of $p(x)$ has degree at least $d$, so no nontrivial $R$-linear combination of the elements of $S$ can be a multiple of $p(x)$.