Why can we use the minimal polynomial to generate the kernel in this case too? This might be a really easy question, but I am confused. Further more, I am not satisfied with the title of my question. Please feel free to improve it.

This is kind of a standard setting when looking at extensions of Dedekind rings (in algebraic number theory):


  Let $A$ be a Dedekind ring and $K$ its field of fractions. Let further $L/K$ be a finite separable field extension and $B$ the integral closure of $A$ in $L$.


In this situation we always have a primitive element $\beta \in B$ such that $L = K(\beta)$. And of course $\beta $ has a minimal polynomial — let's call it $m_{\beta,K} \in K[X]$. Since $\beta$ is integral over $A$, we even know $m_{\beta,K} \in A[X]$.

If we look at the homomorphism

$$ \begin{align}\Psi \colon K[X] &amp;\longrightarrow K[\beta] \\
 f &amp;\longmapsto f(\beta) \end{align}$$
which substitutes $\beta$ for $X$, we have $\ker(\Psi)=(m_{\beta,K})$, i.e. the principal ideal $K[X]m_{\beta,K}$. So far so good.

But now I often read that for the homomorphism

$$ \begin{align}\varphi \colon A[X] &amp;\longrightarrow A[\beta] \\
 f &amp;\longmapsto f(\beta) \end{align}$$

we also have $\ker(\varphi)=(m_{\beta,K})$, where $(m_{\beta,K})$ now of course means the principal ideal $A[X]m_{\beta,K}$.

It is clear to me that $A[X]m_{\beta,K} \subseteq \ker(\varphi)$. But why exactly do we have $A[X]m_{\beta,K} \supseteq \ker(\varphi)$? I am also confused because $A[X]$ is not necessarily a PID (right?), which would explain $\ker(\varphi)=A[X]m_{\beta,K}$.
In general, when you have a field extension $L/K$ and $\beta \in L$ algebraic over $K$, the minimal polynomial $m_\beta \in K[X]$ of $\beta$ over $K$ divides any polynomial that has $\beta$ as a root i.e. is in the kernel of the evaluation map. To see this remember that since $K$ is a field, $K[X]$ is a PID. Now in your case with the evaluation map $\phi: A[X] \rightarrow A[\beta]$, $\ker(\phi)$ may not be principal but we have $(m_\beta) \subset \ker(\phi)$ because $\beta$ is integral over $A$. Now since $m_\beta$ divides any polynomial in $K[X]$ (including $A[X]$) that vanishes at $\beta$, it divides everything in $\ker(\phi)$. Here I made a mistake and assumed it was obvious that after division by $m_\beta$, the remaining polynomial would lie in $A[X]$. One way to see that this is true because $A$ is integrally closed and $K$ is its fraction field (Gauss's Lemma). Knowing this, we have $\ker(\phi) \subset (m_\beta)$. 

Another way to see that division by $m_\beta$ is valid in $A[X]$ is to note that in any polynomial ring the division algorithm works when dividing by a monic polynomial. This only works because we already know $m_\beta \in A[X]$.


  Monic Polynomial Division (Algorithm):
  Let $R$ be a commutative ring with $1$ and $f,g \in R[X]$ with $g \neq 0$ and monic. Then there exists $h,r \in R[X]$ such that $f=gh +r$ with $\deg(r)&lt;\deg(g)$ or $r=0$. 


Proof: The cases when $f=0$ and $\deg(g)&gt;\deg(f)$ are obvious so assume otherwise. Inductively assume that for polynomials of degree less than $f$, this is true. Then simply get rid of the highest order term. If $f=a_n x^n+a_{n-1} x^{n-1}+\cdots$ and $g$ has degree $m \leq n$, then $f-a_n x^{n-m}g$ has lower degree so we can find $h,r$ with $f-a_n x^{n-m} g=gh+r$. Then $f=(a_n x^{n-m}+h)g+r$, as required.
I wanted to write a short answer to my own question for future reference.

Let $f \in A[X]$ such that $f(\beta)=0$ and let $a \neq 0$ be the leading coefficient of $f$ (w.l.o.g. $f \not\equiv 0$).

Proof I We have $\left(\frac{1}{a}f\right)(\beta) = 0 $ and therefore $f \in \ker(\Psi) = K[X] m_{\beta,K}$. Hence there is a $g \in K[X]$ such that
  $$ \frac{1}{a}f = g \cdot m_{\beta,K}. $$
Since both $\frac{1}{a}f$ and $m_{\beta,K}$ are monic, $g$ has to be monic as well. Now Gauss's Lemma grants $g \in A[X]$ and we get
  $$f = \underbrace{ag}_{\in A[X]} \cdot m_{\beta,K}.  $$
This shows $f \in A[x]m_{\beta,K}$ and further $\ker(\varphi) \subseteq A[x]m_{\beta,K}$.

Proof II By polynomial division in $A[X]$ (note $m_{\beta,K} \in A[X]$ is monic) we find $g,r \in a[X]$ with
  $$ f = g \cdot m_{\beta,K} + r$$
and $\deg r \lt \deg m_{\beta,K}$. Since $f(\beta)=0$ we must have $r(\beta)=0$, which means $r \in \ker(\Psi)=K[x]m_{\beta,K}$. Because of $\deg r \lt \deg m_{\beta,K}$ it follows that $r \equiv 0$ and thus
  $$ f = g \cdot m_{\beta,K} $$
This shows $f \in A[x]m_{\beta,K}$ and further $\ker(\varphi) \subseteq A[x]m_{\beta,K}$.