# A problem about an unramified prime in a Galois extension

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 $$\mathbb{Z}/p^2\mathbb{Z}$$.

My attempts: Since $$P$$ has minimal degree among the polynomials which are vanishing $$x$$ module $$\mathfrak{p}$$, it should be irreducible over the field $$\mathbb{Z}/p\mathbb{Z}$$. Therefore $$1, x, \cdots, x^{f-1}$$ are linearly independent over $$\mathbb{Z}/p$$. Also, notice that $$\dfrac{\dfrac{\mathbb{Z}}{p\mathbb{Z}}}{P(X)} \equiv \dfrac{\mathbb{Z}}{p\mathbb{Z}} \oplus x \dfrac{\mathbb{Z}}{p\mathbb{Z}} \oplus \cdots \oplus x^{f-1}\dfrac{\mathbb{Z}}{p\mathbb{Z}}$$ is a field between $$\dfrac{\mathbb{Z}}{p\mathbb{Z}}$$ and $$\dfrac{\mathcal{O}_K}{\mathfrak{p}}$$, with $$\dfrac{\mathbb{Z}}{p\mathbb{Z}}$$-degree equal to $$f=[\dfrac{\mathcal{O}_K}{\mathfrak{p}}:\dfrac{\mathbb{Z}}{p\mathbb{Z}}]$$, 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 $$\dfrac{\mathbb{Z}}{p\mathbb{Z}}$$. (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 $$\mathbb{Z}/p^2\mathbb{Z}$$?

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 $$\mathfrak{p}\setminus\mathfrak{p}^2$$, a set of representatives for $$\mathfrak{p}$$ modulo $$\mathfrak{p}^2$$ is given by $$S\cdot P(x)=\{b_0P(x)+b_1xP(x)+\dotsb+b_{f-1}x^{f-1}P(x)\ :\\ b_0,b_1,\dotsc,b_{f-1}\in\{0,1,\dotsc,p-1\}\}.$$ Therefore, a set of representatives for $$\mathcal{O}$$ modulo $$\mathfrak{p^2}$$ is given by $$S+S\cdot P(x)=\{a_0+\dotsb+a_{f-1}x^{f-1}+b_0P(x)+\dotsb+b_{f-1}x^{f-1}P(x)\ :\\ a_0,b_0,\dotsc,a_{f-1},b_{f-1}\in\{0,1,\dotsc,p-1\}\}.$$ In particular, $$\mathcal{O}=\mathbb{Z}[x]+\mathfrak{p}^2$$, and the result follows.
• This answer sounds very simple. I'm afraid my intuition is wrong. I agree with you that $S$ is a set of representatives for $\mathcal{O}_K$ modulo $\mathfrak{p}$. Also, I can see that $S\cdot P(x)$ is a set of representatives for $\mathfrak{p}$ modulo $\mathfrak{p}^2$. But why $S+S\cdot P(x)$ is a set of representatives for $\mathcal{O}_K$ modulo $\mathfrak{p}^2$? This seems intuitively true, but I can not prove it. If we have this, then you are right and the problem will be solved. – NeoTheComputer Oct 7 at 20:01
• @NeoTheComputer: Let me abbreviate $T:=S\cdot P(x)$. Start with an arbitrary $r\in\mathcal{O}$. Then there is a unique $s\in S$ such that $r-s\in\mathfrak{p}$. Hence there is also a unique $t\in T$ such that $r-s-t\in\mathfrak{p}^2$. So we proved that, for given $r\in\mathcal{O}$, there is a unique $u\in S+T$ such that $r-u\in\mathcal{p}^2$. Done. – GH from MO Oct 7 at 20:10