AnswerBun.com

A problem about an unramified prime in a Galois extension

MathOverflow Asked by NeoTheComputer on December 9, 2020

I asked this question in MathStackExchange, but I didn’t receive any answer.

Let $K/mathbb{Q}$ be a Galois extension of degree $n$, and denote its ring of integers by $mathcal{O}_K$. Let $mathfrak{p}$ be an arbitrary prime ideal of $mathcal{O}_K$, which is unramified over $mathbb{Z}$, and prime to $n!$. We will denote the residue field of $mathfrak{p}$ by $kappa(mathfrak{p})$, its characteristic by $p$, and its residue degree by $f$. Let $x in mathcal{O}_K$, and let $bar{x}$ be its image in $kappa(mathfrak{p})$,
and assume that $P in mathbb{Z}[X]$ is a monic minimal polynomial of $bar{x}$, such that $P(x) in mathfrak{p} backslash mathfrak{p}^2$, and $deg(P)=f$.

(Q): Show that $mathcal{O}_K/mathfrak{p}^2$ is generated by the image of $x$ over $mathbb{Z}/p^2mathbb{Z}$.


My attempts: Since $P$ has minimal degree among the polynomials which are vanishing $x$ module $mathfrak{p}$, it should be irreducible over the field $mathbb{Z}/pmathbb{Z}$.
Therefore $1, x, cdots, x^{f-1}$ are linearly independent over $mathbb{Z}/p$.
Also, notice that $$dfrac{dfrac{mathbb{Z}}{pmathbb{Z}}}{P(X)} equiv dfrac{mathbb{Z}}{pmathbb{Z}} oplus x dfrac{mathbb{Z}}{pmathbb{Z}} oplus cdots oplus x^{f-1}dfrac{mathbb{Z}}{pmathbb{Z}}$$
is a field between $dfrac{mathbb{Z}}{pmathbb{Z}}$ and $dfrac{mathcal{O}_K}{mathfrak{p}}$, with $dfrac{mathbb{Z}}{pmathbb{Z}}$-degree equal to $f=[dfrac{mathcal{O}_K}{mathfrak{p}}:dfrac{mathbb{Z}}{pmathbb{Z}}]$, so it should equal to $dfrac{mathcal{O}_K}{mathfrak{p}}$. So we can conclude that $dfrac{mathcal{O}_K}{mathfrak{p}}$ is generated by the image of $x$ over $dfrac{mathbb{Z}}{pmathbb{Z}}$. (My proof of this fact may contain extra details; if so, please let me know). But I don’t have any idea why $mathcal{O}_K/mathfrak{p}^2$ is generated by the image of $x$ over $mathbb{Z}/p^2mathbb{Z}$?

I’m looking to figure out how, in this case, "the assumption $P(x) in mathfrak{p} backslash mathfrak{p}^2$" helps me solve the problem. I have this issue with similar problems; for instance, I had trouble dealing with exercises 19-22 from chapter 4 of Marcus’s Number Fields. (In these exercises I had to deal with "the assumption $pi in Q backslash Q^2$", finally I solved them after a long hard try and search). Also, I tried to look for some versions of Nakayama’s lemma, but I have not succeeded.

One Answer

A set of representatives for $mathcal{O}$ modulo $mathfrak{p}$ is given by $$S:={a_0+a_1x+dotsb+a_{f-1}x^{f-1} : a_0,a_1,dotsc,a_{f-1}in{0,1,dotsc,p-1}}.$$ As $P(x)$ lies in $mathfrak{p}setminusmathfrak{p}^2$, a set of representatives for $mathfrak{p}$ modulo $mathfrak{p}^2$ is given by $$Scdot P(x)={b_0P(x)+b_1xP(x)+dotsb+b_{f-1}x^{f-1}P(x) :\ b_0,b_1,dotsc,b_{f-1}in{0,1,dotsc,p-1}}.$$ Therefore, a set of representatives for $mathcal{O}$ modulo $mathfrak{p^2}$ is given by $$S+Scdot P(x)={a_0+dotsb+a_{f-1}x^{f-1}+b_0P(x)+dotsb+b_{f-1}x^{f-1}P(x) :\ a_0,b_0,dotsc,a_{f-1},b_{f-1}in{0,1,dotsc,p-1}}.$$ In particular, $mathcal{O}=mathbb{Z}[x]+mathfrak{p}^2$, and the result follows.

Correct answer by GH from MO on December 9, 2020

Add your own answers!

Related Questions

Applications of Set theory vs. model theory in mathematics

0  Asked on November 3, 2021 by mohammad-golshani

     

Circular track riddle

1  Asked on November 3, 2021

   

Ask a Question

Get help from others!

© 2022 AnswerBun.com. All rights reserved. Sites we Love: PCI Database, MenuIva, UKBizDB, Menu Kuliner, Sharing RPP