2
$\begingroup$

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


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.

$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – NeoTheComputer Oct 7 at 20:01
  • 1
    $\begingroup$ @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. $\endgroup$ – GH from MO Oct 7 at 20:10
  • 1
    $\begingroup$ Thank you so much for your clarification. $\endgroup$ – NeoTheComputer Oct 7 at 20:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.