Let $m \geq 2$ be an integer. Consider the polynomial $f(x)=x^{2m}-p^{2(m-1)} \in \mathbb{Q}_p[x]$.

I want to study the field extension by the zeros of $f(x)$ over $\mathbb{Q}_p$.

What is the degree extension? Is it totally ramified?

My approach is not complete but I have tried as follows.

$m=2$ case: The polynomial becomes $f(x)=x^{4}-p^{2} \in \mathbb{Q}_p[x]$.

We know that $x^2 \equiv -1 \pmod 4$ has a solution if $p \equiv 1 \pmod 4$.

  Thus,

• if $p \equiv 1 \pmod 4$, then $\sqrt p$ exists in $\mathbb{Q}_p$ and thus $[\mathbb{Q}_p(\alpha): \mathbb{Q}_p]=1$, where $f(\alpha)=0$. Not interesting.

• $p \not\equiv 1 \pmod 4$, then $\sqrt p$ doesn't exist in $\mathbb{Q}_p$. Write $f(x)=f_1(x)f_2(x)=(x^2+p)(x^2-p)$; its zeros are $\pm \sqrt{p}, \pm \sqrt{-p}$. It is a degree-two extension. Both $f_1$ and $f_2$ are Eisenstein polynomials, so they define a total ramified extension. Each individual ramification index is $2$. Thus, if $\pi_1, \pi_2$ be the ramification index of $\mathbb{Q}_p(f_1(x)=0)$ and $\mathbb{Q}_p(f_1(x)=0)$ respectively, then $v_p(\pi_1)=\frac{1}{2}=v_p(\pi_2)$. Thus, we can take $\pi_1=\pi_2=u \sqrt{p}$, where $u$ is a unit. Since the uniformizers are equal, the compositum is also totally ramified, I think. That is, the zeros of $f(x)$ give a degree-two totally ramified extension.

General case: Write $f(x)=(x^m+p^{m-1})(x^m+p^{m-1})=f_1(x)f_2(x)$. Since $\gcd(m,m-1)=1$, both $f_1(x)$ and $f_2(x)$ are irreducible over $\mathbb{Q}_p$, by simple argument of Newton polygon. Now consider the field $\mathbb{Q}_p(f_1(x)=0)$ i.e., $\mathbb{Q}_p(\sqrt[m-1]{p^m})=\mathbb{Q}_p(\sqrt[m-1]{a}),~a=p^m$. Then $v(a)=m$. Since $ m-1 \nmid v_p(a)$, by the answer of Emerton in this post, it is a "tamely ramified" extension. Same for the extension $\mathbb{Q}_p(f_2(x)=0)$. Thus, the degree of extension $\mathbb{Q}_p(f(x)=0)$ is $m$. I am unable to say anything about the ramification of $\mathbb{Q}_p(f(x)=0)$.

I appreciate to improve my approach

Let $m \geq 2$ be an integer. Consider the polynomial $f(x)=x^{2m}-p^{2(m-1)} \in \mathbb{Q}_p[x]$.

I want to study the field extension by the zeros of $f(x)$ over $\mathbb{Q}_p$.

What is the degree extension? Is it totally ramified?

My approach is not complete but I have tried as follows.

$m=2$ case: The polynomial becomes $f(x)=x^{4}-p^{2} \in \mathbb{Q}_p[x]$.

Note $\sqrt p$ doesn't exist in $\mathbb{Q}_p$. Write $f(x)=f_1(x)f_2(x)=(x^2+p)(x^2-p)$; its zeros are $\pm \sqrt{p}, \pm \sqrt{-p}$. It is a degree-two extension. Both $f_1$ and $f_2$ are Eisenstein polynomials, so they define a total ramified extension. Each individual ramification index is $2$. Thus, if $\pi_1, \pi_2$ be the ramification index of $\mathbb{Q}_p(f_1(x)=0)$ and $\mathbb{Q}_p(f_1(x)=0)$ respectively, then $v_p(\pi_1)=\frac{1}{2}=v_p(\pi_2)$. Thus, we can take $\pi_1=\pi_2=u \sqrt{p}$, where $u$ is a unit. Since the uniformizers are equal, the compositum is also totally ramified, I think. That is, the zeros of $f(x)$ give a degree-two totally ramified extension.

General case: Write $f(x)=(x^m+p^{m-1})(x^m+p^{m-1})=f_1(x)f_2(x)$. Since $\gcd(m,m-1)=1$, both $f_1(x)$ and $f_2(x)$ are irreducible over $\mathbb{Q}_p$, by simple argument of Newton polygon. Now consider the field $\mathbb{Q}_p(f_1(x)=0)$ i.e., $\mathbb{Q}_p(\sqrt[m-1]{p^m})=\mathbb{Q}_p(\sqrt[m-1]{a}),~a=p^m$. Then $v(a)=m$. Since $ m-1 \nmid v_p(a)$, by the answer of Emerton in this post, it is a "tamely ramified" extension. Same for the extension $\mathbb{Q}_p(f_2(x)=0)$. Thus, the degree of extension $\mathbb{Q}_p(f(x)=0)$ is $m$. I am unable to say anything about the ramification of $\mathbb{Q}_p(f(x)=0)$.

I appreciate to improve my approach

Let $m \geq 2$ be an integer. Consider the polynomial $f(x)=x^{2m}-p^{2(m-1)} \in \mathbb{Q}_p[x]$.

I want to study the field extension by the zeros of $f(x)$ over $\mathbb{Q}_p$.

What is the degree extension? Is it totally ramified?

$m=2$ case: The polynomial becomes $f(x)=x^{4}-p^{2} \in \mathbb{Q}_p[x]$.

We know that $x^2 \equiv -1 \pmod 4$ has a solution if $p \equiv 1 \pmod 4$.

Thus,

• if $p \equiv 1 \pmod 4$, then $\sqrt p$ exists in $\mathbb{Q}_p$ and thus $[\mathbb{Q}_p(\alpha): \mathbb{Q}_p]=1$, where $f(\alpha)=0$. Not interesting.

• $p \not\equiv 1 \pmod 4$, then $\sqrt p$ doesn't exist in $\mathbb{Q}_p$. Write $f(x)=f_1(x)f_2(x)=(x^2+p)(x^2-p)$; its zeros are $\pm \sqrt{p}, \pm \sqrt{-p}$. It is a degree-two extension. Both $f_1$ and $f_2$ are Eisenstein polynomials, so they define a total ramified extension. Each individual ramification index is $2$. Thus, if $\pi_1, \pi_2$ be the ramification index of $\mathbb{Q}_p(f_1(x)=0)$ and $\mathbb{Q}_p(f_1(x)=0)$ respectively, then $v_p(\pi_1)=\frac{1}{2}=v_p(\pi_2)$. Thus, we can take $\pi_1=\pi_2=u \sqrt{p}$, where $u$ is a unit. Since the uniformizers are equal, the compositum is also totally ramified, I think. That is, the zeros of $f(x)$ give a degree-two totally ramified extension.

General case: Write $f(x)=(x^m+p^{m-1})(x^m+p^{m-1})=f_1(x)f_2(x)$. Since $\gcd(m,m-1)=1$, both $f_1(x)$ and $f_2(x)$ are irreducible over $\mathbb{Q}_p$, by simple argument of Newton polygon. Now consider the field $\mathbb{Q}_p(f_1(x)=0)$ i.e., $\mathbb{Q}_p(\sqrt[m-1]{p^m})=\mathbb{Q}_p(\sqrt[m-1]{a}),~a=p^m$. Then $v(a)=m$. Since $ m-1 \nmid v_p(a)$, by this post, it is a "tamely ramified" extension. Same for the extension $\mathbb{Q}_p(f_2(x)=0)$. Thus, the degree of extension $\mathbb{Q}_p(f(x)=0)$ is $m$. I am unable to say anything about the ramification of $\mathbb{Q}_p(f(x)=0)$.

Let $m \geq 2$ be an integer. Consider the polynomial $f(x)=x^{2m}-p^{2(m-1)} \in \mathbb{Q}_p[x]$.

I want to study the field extension by the zeros of $f(x)$ over $\mathbb{Q}_p$.

What is the degree extension? Is it totally ramified?

My approach is not complete but I have tried as follows.

$m=2$ case: The polynomial becomes $f(x)=x^{4}-p^{2} \in \mathbb{Q}_p[x]$.

We know that $x^2 \equiv -1 \pmod 4$ has a solution if $p \equiv 1 \pmod 4$.

Thus,

• if $p \equiv 1 \pmod 4$, then $\sqrt p$ exists in $\mathbb{Q}_p$ and thus $[\mathbb{Q}_p(\alpha): \mathbb{Q}_p]=1$, where $f(\alpha)=0$. Not interesting.

• $p \not\equiv 1 \pmod 4$, then $\sqrt p$ doesn't exist in $\mathbb{Q}_p$. Write $f(x)=f_1(x)f_2(x)=(x^2+p)(x^2-p)$; its zeros are $\pm \sqrt{p}, \pm \sqrt{-p}$. It is a degree-two extension. Both $f_1$ and $f_2$ are Eisenstein polynomials, so they define a total ramified extension. Each individual ramification index is $2$. Thus, if $\pi_1, \pi_2$ be the ramification index of $\mathbb{Q}_p(f_1(x)=0)$ and $\mathbb{Q}_p(f_1(x)=0)$ respectively, then $v_p(\pi_1)=\frac{1}{2}=v_p(\pi_2)$. Thus, we can take $\pi_1=\pi_2=u \sqrt{p}$, where $u$ is a unit. Since the uniformizers are equal, the compositum is also totally ramified, I think. That is, the zeros of $f(x)$ give a degree-two totally ramified extension.

General case: Write $f(x)=(x^m+p^{m-1})(x^m+p^{m-1})=f_1(x)f_2(x)$. Since $\gcd(m,m-1)=1$, both $f_1(x)$ and $f_2(x)$ are irreducible over $\mathbb{Q}_p$, by simple argument of Newton polygon. Now consider the field $\mathbb{Q}_p(f_1(x)=0)$ i.e., $\mathbb{Q}_p(\sqrt[m-1]{p^m})=\mathbb{Q}_p(\sqrt[m-1]{a}),~a=p^m$. Then $v(a)=m$. Since $ m-1 \nmid v_p(a)$, by the answer of Emerton in this post, it is a "tamely ramified" extension. Same for the extension $\mathbb{Q}_p(f_2(x)=0)$. Thus, the degree of extension $\mathbb{Q}_p(f(x)=0)$ is $m$. I am unable to say anything about the ramification of $\mathbb{Q}_p(f(x)=0)$.

I appreciate to improve my approach