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Question

Let $K$ be a complete nonarchimedean valued field with valuation ring $\mathcal{O}$ and let $L/K$ be a finite extension.

Let $\alpha$ be an element of $L$ and $f\in K[x]$ its minimal polynomial.

Is is true that $f\in\mathcal{O}[x]$?

Comments

First of all, I am asking this, because of a proof I saw, where the author said "...let $f$ be the minimal polynomial of $\alpha$ and let $\overline{f}=f\pmod{\kappa}$, where $\kappa$ is the residue field of $K$."

He is implicitly using the fact that he can send the coefficients of $f$ to the residue field, and we can only do that if the coefficientes are in $\mathcal{O}$.

I strongly suspect that this is not true in general. I know that if $f$ is monic and the independent term is in $\mathcal{O}$ then indeed $f\in\mathcal{O}[x]$, but only with the information that $f$ is the minimal of some element, I cannot deduce that.

I think that the author meant the following:

"Take the minimal polynomial $f$, clean denominators, and now you can send it to $\kappa[x]$, even though $f$ is not monic anymore".

Is this correct?

If so, after sending $f$ to $\kappa[x]$, it feels like we could perfectly make it monic again, right?

Sorry if this question is kind of annoying or obvious, but Neukirch's style is yet kind of hard for me, making a lot of implicit assumptions and I want to understand things correctly.

Thank you in advance for explanations.

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  • $\begingroup$ Let v be the valuation, you can multiply the coefficients of f by an element of O such that the resulting polynomial is in O[x], perhaps it is all what you need in this proof. $\endgroup$ – Tsemo Aristide Nov 4 '15 at 12:54
  • $\begingroup$ Yeah, that is what I meant by "clean denominators". But the answer for the question is then that $f$ not necessarily is in O[x]? $\endgroup$ – Shoutre Nov 4 '15 at 14:44
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Another answer since I didn't understand your original question properly: the result I gave is pretty general. Let me state it clearly:

Let $A$ be an integrally closed integral domain with quotient field $K$, $L$ a finite separable extension of $K$, and $B$ the integral closure of $A$ in $L$. Let $\alpha \in L$ and let $f \in K[X]$ be its minimal polynomial over $K$ (assumed to be monic). Then $\alpha \in B$ if and only if $f \in A[X]$.

I gave the result for the ring of integers of a number field, the proof is the same in general. If $S$ is a multiplicatively closed subset of $A$, then $S^{-1}B$ is the integral closure of $S^{-1}A$ in $L$, and the same result holds with $B$ replaced by $S^{-1}B$ and $A$ replaced by $S^{-1}A$. So e.g. if $\mathfrak p$ is a prime of $\mathcal O_K$, and $S = \mathcal O_K \setminus \mathfrak p$, then $S^{-1} \mathcal O_L$ is the integral closure of $\mathcal O_{K, \mathfrak p}$ in $L$. But $S^{-1} \mathcal O_L$ isn't the valuation ring of valuation of $L$, unless there is only one prime $\mathfrak P$ of $\mathcal O_L$ lying over $\mathfrak p$, in which case $S^{-1} \mathcal O_L = \mathcal O_{L, \mathfrak P}$ (otherwise you just have $\subseteq$).

In general, if $A$ is a Dedekind domain (in particular $A$ is integrally closed so the result holds), and $S = A \setminus \mathfrak p$ for a prime $\mathfrak p$ of $A$, then $S^{-1}B$ is a principal ideal domain with finitely many prime ideals $S^{-1}\mathfrak P_1, ... , S^{-1} \mathfrak P_g$, where $\mathfrak P_1, ... , \mathfrak P_g$ are all the primes of $B$ lying over $\mathfrak p$.

In your case, the fact that $K$ is complete with respect to its valuation means that there is only one prime $w$ lying over $v$, so the integral closure of $\mathcal O$ in $L$ is exactly the valuation ring of $w$.

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Let $\alpha \in L$ with minimal polynomial $f \in K[X]$ (assuming to be monic). Then $f \in \mathcal O_K[X]$ if and only if $\alpha \in \mathcal O_L$.

Pf: $\Leftarrow$: We have $\mathcal O_K = \mathcal O_L \cap K$, and the coefficients of $f$ are symmetric functions of $\alpha$ and its conjugates over $K$. Since $\alpha \in \mathcal O_L$ (by definition $\mathcal O_L$ is the set of things in $L$ which are integral over $\mathcal O_K$), so do all its conjugates. So the coefficients of $f$ are in both $\mathcal O_L$ and $K$.

$\Rightarrow$: $\alpha$ is a root of a monic polynomial with coefficients in $\mathcal O_K$, so $\alpha$ is in $\mathcal O_L$.

This is true more generally when you replace $\mathcal O_K$ by an integrally closed ring.

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  • $\begingroup$ Is $\mathcal{O}_{K}$ the ring of integers of $K$ or its valuation ring? These notions coincide in this context? And just for curiosity: looks like $\mathcal{O}_{L}$ is the integral closure of $\mathcal{O}_{K}$ in $L$. If I extended $v$ uniquely to $L$, would $\mathcal{O}_{L}$ coincide with the valuation ring of $L$? $\endgroup$ – Shoutre Nov 4 '15 at 14:52
  • $\begingroup$ Just realized there is a proposition in Neukirch about that ^ $\endgroup$ – Shoutre Nov 4 '15 at 21:36
  • $\begingroup$ In general, a valuation on $K$ does not extend uniquely to $L$. If $\mathfrak p$ is a prime of $\mathcal O_K$, then the localization of $\mathcal O_K$ at $\mathfrak p$ is the valuation ring. $\endgroup$ – D_S Nov 12 '15 at 3:15
  • $\begingroup$ The extensions of the $\mathfrak p$-adic valuation correspond to the primes $\mathfrak P$ lying over $\mathfrak p$. If (and only if) there is only one prime $\mathfrak P$ lying over $\mathfrak p$, then the valuation ring of $\mathfrak P$ (that is, $\mathcal O_{L, \mathfrak P}$) is the integral closure of $\mathcal O_{K, \mathfrak p}$ in $L$; otherwise, $\mathcal O_{L, \mathfrak P}$ isn't even integral over $\mathcal O_{K, \mathfrak p}$. $\endgroup$ – D_S Nov 12 '15 at 3:15
  • $\begingroup$ Maybe not in general, but I put up a context in the question. Honestly, when I first saw your answer I was not sure if you were using general notions of algebraic number fields applied to local fields that satisfied the hypothesis needed or if you just didn't read the question at all. $\endgroup$ – Shoutre Nov 12 '15 at 3:23

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