Consider the extension $\mathbb{Q}_{p}(\zeta)/\mathbb{Q}_{p}$, where $\zeta$ is a $n$-th primitive root of unity and $(n,p)=1$.

I want to show that $\mathbb{Q}_{p}(\zeta)/\mathbb{Q}_{p}$ is an unramified extension.

My attempt of understanding proof

Let $f\in\mathbb{Q}_{p}[x]$ be the minimal polynomial of $\zeta$ over $\mathbb{Q}_{p}$.

Question $(1)$. Is $f$ a polynomial with coefficientes actually in $\mathbb{Z}_{p}$? Why?

If the above is true, then the polynomial $\overline{f}$obtained by reducing coeficients modulo $p\mathbb{Z}_{p}$ is monic and vanishes at the image $\overline{\zeta}$ of the element $\zeta$ by the canonical homomorphism $\mathcal{O}\to\mathcal{O}/\beta$, where $\mathcal{O}$ (resp. $\beta$) is the valuation ring (resp. maximal ideal) of $\mathbb{Q}_{p}(\zeta)$. By the way, no doubt that $\zeta\in\mathcal{O}$.

If we show that $\overline{f}$ is irreducible, then it is the minimal polynomial of $\overline{\zeta}$ over $\mathbb{F}_{p}$. We know that $f$ is a primitive polynomial (since it is monic) and hence we can use Hensel's Lemma as a tool for that.

Namely, if $\overline{f}$ is a product of coprime polynomials $\overline{g}$, $\overline{h}\in\mathbb{F}_{p}[x]$, each of degree $>1$, then $f$ would be a product of polynomials $g$, $h\in\mathbb{Z}_{p}[x]$, each of degree >1, a contradiction.

However, we still need to prove that $\overline{f}$ is not a product of noncoprime polynomials.

Neukirch says that since $\overline{f}$ divides the coefficient-reduced modulo $p\mathbb{Z}_{p}$ separable polynomial $x^{n}-\overline{1}\in\mathbb{F}_{p}[x]$, this case cannot happen, but I don't understand this.

Question $(2)$. Why is $x^{n}-\overline{1}\in\mathbb{F}_{p}[x]$ a separable polynomial?

Question $(3)$. If $\overline{f}$ is separable, then can't it be a product of noncoprime polynomials?

If the above is cleared, then we actually confirmed that $\overline{f}$ is the minimal polynomial of $\overline{\zeta}$.

In particular, being irreducible gives us that $\operatorname{deg}(f)=\operatorname{deg}(\overline{f})$, which means

\begin{equation} [\mathbb{Q}_{p}(\zeta)\colon\mathbb{Q}_{p}]=\operatorname{deg}(f)=\operatorname{deg}(\overline{f})=[\mathbb{F}_{p}(\overline{\zeta})\colon\mathbb{F}_{p}]. \end{equation}

On the other hand, we have that

\begin{equation} [\mathbb{Q}_{p}(\zeta)\colon\mathbb{Q}_{p}]\geq [\beta\colon\mathbb{F}_{p}] \end{equation}

Putting these two together gives $\mathbb{F}_{p}(\overline{\zeta})=\beta$ and hence the extension is unramified.

Is this correct? Thank you advance for answering my above questions

EDIT. Can someone point out where we used that (n,p)=1?


2 Answers 2


Question (1). Yes.

Recall that $\zeta$ satisfies monic cyclotomic polynomial $\Phi(x) \in \mathbb{Z}[x] \subseteq \mathbb{Z}_p[x]$. This means $\zeta$ is integral in $\mathbb{Q}_p(\zeta)$ over $\mathbb{Z}_p$. So its minimal polynomial must not only have coefficients in $\mathbb{Z}_p$ but be monic as well.

We can prove that if $f$ is monic and $\overline{f}$ is separable [this assumption means $\overline{f}$ makes sense which implies $f$ has integral coefficients] then $f$ must be separable as well. Suppose not, say $$f = (x - \alpha)^2 g$$ then $\alpha$ must be integral and $g$ must also have integral coefficients. So we have factorization $$\overline{f} = (x - \overline{\alpha})^2 \overline{g}$$ which contradicts separability of $\overline{f}$. Note that this reduction only makes sense because $\alpha$ is integral and $g$ have integral coefficient.

Question (2). Use $\gcd(n,p) = 1$ to deduce that the derivative is co-prime with the polynomial.

Question (3). Obviously. If $f = g h$ with $g, h$ non co-prime then roots of $\gcd(g, h)$ will have multiplicities $\geq 2$ in $f$ so $f$ cannot be separable.


This follows from a more general result about unramified extensions: The finite extension $F/\mathbb{Q}_p$ is unramified when you can write $F=\mathbb{Q}_p(\alpha)$ so that $\alpha$ is a root of a monic polynomial $f(x) \in \mathbb{Z}_p[x]$ and $\overline{\alpha}$ is a simple root of $\overline{f}(x) \in \mathbb{F}_p[x]$.

You can find this theorem on any text about algebraic number theory. The point is that $\zeta$ satisfies $x^n-1$ and when $(n,p)=1$ the polynomial $x^n-1$ has no repeated root modulo $p$: because $n t^{n-1} \neq 0$ when $t^n=1$.

  • $\begingroup$ If $f(x)\in\mathbb{Q}_{p}[x]$, then how come $f\in\mathbb{F}_{p}[x]$? I believe you meant that this result is valid only for monic polynomials $f(x)\in\mathbb{Z}_{p}[x]$, right? of course, those that satisfy what you said. Isn't irreducibility needed? If we had a polynomial $f(x)\in\mathbb{Q}_{p}[x]$ monic and irreducible, it would have coefficients in $\mathbb{Z}_{p}$ or not necessarily? This is my question $(1)$ above. $\endgroup$
    – Shoutre
    Nov 6, 2015 at 15:50
  • $\begingroup$ Thanks, I edited my answer. The irreducibility is not needed. $\endgroup$ Nov 6, 2015 at 16:12

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