# minimal polynomial $\ne$ integral polynomial

Let $$A$$ be an integrally closed domain, $$K=\mathrm{Quot}(A)$$ and $$L|K$$ a finite field extension. I am looking for an element $$x \in L$$, which is integral over $$A$$, such that the (monic) minimal polynomial $$m_{x}$$ of $$x$$ is not equal to the monic polynomial $$f \in A[X]$$ with $$f(x) = 0$$, which there exists by the definition of integral over $$A$$.

Do you know a specific example?

• this is not possible Dec 5, 2017 at 10:23
• If that is the case, then I do not understand the following Neukirch states: Same setting as in the question and $B$ is the integral closure of $A$ in $L$. The fact that $A$ is integrally closed has the effect that an element $\beta \in L$ is integral over $A$ if and only if its minimal polynomial $p(x)$ takes its coefficients in $A$. In fact, let $\beta$ be a zero of the monic polynomial $f(x) \in A[x]$.Then $m(x)$ divides $f(x)$ in $K[x]$, so that all zeroes $\beta_1, ..., \beta_n$ of $m(x)$ are integral over A, hence the same holds for all the coefficients, so $m(x) \in A[x]$. Dec 5, 2017 at 10:37
• When there is no $x$ like in the question, does this not mean, that always $m(x)=f(x)$? Dec 5, 2017 at 10:39
• well $g$ can be the product of $f$ with any monic polynomial in $A[x]$ I guess ? it's not clear why you say "the" monic polynomial in $A[x]$ both times without saying anything strong enough about it to restrict it to only one possible polynomial Dec 5, 2017 at 10:42
• Ah I see. Thank you! If $f$ has minimal degree then $f=m$. Otherwise I just can multiply $f$ by $x$ for example. Dec 5, 2017 at 10:50