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?

  • 1
    $\begingroup$ this is not possible $\endgroup$
    – mercio
    Dec 5, 2017 at 10:23
  • $\begingroup$ 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 $m(x) \in A[x]$. $\endgroup$
    – Sqyuli
    Dec 5, 2017 at 10:37
  • $\begingroup$ When there is no $x$ like in the question, does this not mean, that always $m(x)=f(x)$? $\endgroup$
    – Sqyuli
    Dec 5, 2017 at 10:39
  • $\begingroup$ 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 $\endgroup$
    – mercio
    Dec 5, 2017 at 10:42
  • $\begingroup$ Ah I see. Thank you! If $f$ has minimal degree then $f=m$. Otherwise I just can multiply $f$ by $x$ for example. $\endgroup$
    – Sqyuli
    Dec 5, 2017 at 10:50


You must log in to answer this question.

Browse other questions tagged .