# A problem about an unramified prime in a Galois extension

MathOverflow Asked by NeoTheComputer on December 9, 2020

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.

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

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