Minimal polynomial with coefficients in an integral domain

Question

Let $R$ be an integral domain, $K$ its field of fractions and $L/K$ a finite field extension. Let $b \in L$ such that its minimal polynomial $f$ is in $R[X]$. I want to show that $R[X]/(f)R[X]$ isomorphic to $R[b]$ as rings.

Note that $(f)R[X]$ denotes the ideal generated by $f$ in $R[X]$.

My ideas: Let be $i: R[X] \to K[X], p \mapsto p$ the canonical inclusion and the K-algebra homomorphism $\phi: K[X]\to K[b]$ defined by setting $X=b$. Obviously we have $\def\Im{\operatorname{Im}}\Im\phi \circ i = R[b]$, as well as $(f)R[X] \subseteq (f)K[X]$ and $K[b] \cong K[X]/(f)K[X]$. Moreover $(f)R[X] \subseteq\ker$ $\phi \circ i$.

It might happen that $fg \in R[X]$, where $g \in K[X]$, but $g \notin R[X]$ (i.e., $g$ has fractions as coefficients). This would imply that $fg \notin (f)R[X]$ and so the conclusion cannot be deduced. Can this happen?

Answer 1

You should get your notation straight. I assume $f=P$ and $a=b$.

The key is that $f$ is monic. Thus division with remainder works in $R[X]$. If $g(b)=0$, we can write $g = hf+r$ with $r \in R[X]$ being of lower degree than $f$. On the other hand, we obtain $r(b)=0$, which is a contradiction to the minimality of $f$, hence shows $r=0$ and thus $g \in (f)$. So $(f)$ is indeed the kernel.

Answer 2

The key here is that minimal polynomials are only defined over fields, so although you did not say so explicitly $f$ is the minimal polynomial of $b$ over$~K$, and that minimal polynomials are by definition monic. This implies (together with the given $f\in R[X]$) that $\langle f\rangle_{K[X]}\cap R[X]=\langle f\rangle_{R[X]}$: every $K[X]$-multiple $fg$ of $f$ that lies in $R[X]$ in fact has $g\in R[X]$. If this were not the case, take a minimal degree counterexample$~g$; since the leading coefficient of $g$ is also that of $fg$, hence in $R$, one could remove the leading term of $g$ so as to obtain a smaller degree example, which is absurd.

Now the obvious way to define an isomorphism works: consider the evaluation morphism $e_b:K[X]\to L$ that is identity on$~K$ and sends $X\mapsto b$, which has kernel $\langle f\rangle_{K[X]}$. Restrict it to$~R[X]$ to obtain a morphism with image $R[b]$ (by definition) and kernel $\langle f\rangle_{K[X]}\cap R[X]=\langle f\rangle_{R[X]}$, and conclude by the isomorphism theorem that $R[b]\cong R[X]/\langle f\rangle_{R[X]} $.

Note however that if rather than defining minimal polynomials to be monic, one had defined them to be classes of associate polynomials (just like least common divisors are defined only up to associates) then finding in the class of $f$ an element of $R[X]$ does not suffice for the conclusion sought here.