Primitive $p^n$-th root of unity in $\bar{\mathbb{Q}}_p$.

Question

I am trying to solve the following exercise in Koblitz's "$p$-adic Numbers, $p$-adic analysis, and Zeta-Functions".

Let $p$ be a prime. Let $a$ be a primitive $p^n$-th root of unity in $\bar{\mathbb{Q}}_p$. Find $|a-1|_p$.

Can you give me some hints? I was thinking about using the relation given by the irreducible polynomial of $a$, but this does not seem useful.

Answer 1

Here’s the way an old (really old) hand looks at this:

I like to use the additive valuation $v_p$, $v_p(p)=1$. You’ve already seen from @JyrkiLahtonen’s comment that $v(\zeta_p-1)=1/(p-1)$. Let’s write $\zeta_p=1+\epsilon_1$, $v(\epsilon_1)=1/(p-1)$. Now look at $\zeta_{p^2}=1+\epsilon_2$. You have $$ (1+\epsilon_2)^p=1+p(\text{stuff}) + \epsilon_2^p=1+\epsilon_1\,, $$ which tells you that $\epsilon_2^p=\epsilon_1-p(\text{stuff})$. It follows that $v(\epsilon_2)=v(\epsilon_1)/p$. You can see the rest of the argument.

Answer 2

Show $\mathbf Q_p(a)/\mathbf Q_p$ is totally ramified because $a-1$ is the root of an Eisenstein polynomial (this is how you prove the $p^n$-th cyclotomic polynomial is irreducible over $\mathbf Q$ or $\mathbf Q_p$). Koblitz shows that in a totally ramified extension of degree $n$, a uniformizer $\pi$ has $|\pi|_p = (1/p)^{1/n}$. So now it remains to show that the root of an Eisenstein polynomial in $\mathbf Z_p[x]$ is a uniformizer in the extension it generates over $\mathbf Q_p$. Since $a-1$ is the root of an Eisenstein polynomial over $\mathbf Q_p$, you therefore can get $|a-1|_p$ once you know the degree $[\mathbf Q_p(a-1):\mathbf Q_p] = [\mathbf Q_p(a):\mathbf Q_p]$.

Here are more details. Say $\alpha$ is a root of an Eisenstein polynomial $x^n + c_{n-1}x^{n-1} + \cdots + c_1x + c_0$: $c_i \in p\mathbf Z_p$ for $i = 0,\ldots,n-1$ and $c_0 \not\in p^2\mathbf Z_p$. Since Eisenstein polynomials with respect to $p$ are irreducible over $\mathbf Q_p$, $[\mathbf Q_p(\alpha):\mathbf Q_p] = n$. Let $e$ be the ramification index of $\mathbf Q_p(\alpha)/\mathbf Q_p$, so $e \leq n$. By the formula for the $p$-adic absolute value of a number algebraic over $\mathbf Q_p$ in terms of the constant term of its minimal polynomial over $\mathbf Q_p$, we have $|\alpha|_p = \sqrt[n]{|c_0|_p} = (1/p)^{1/n} < 1$. The biggest absolute value less than $1$ on $\mathbf Q_p(\alpha)$ is $(1/p)^{1/e}$, so $(1/p)^{1/n} \leq (1/p)^{1/e}$, which implies $1/n \geq 1/e$, and hence $e \geq n$. Together with $e \leq n$ we get $e = n$. Thus the extension $\mathbf Q_p(\alpha)/\mathbf Q_p$ is totally ramified, and from $|\alpha|_p = (1/p)^{1/n} = (1/p)^{1/e}$ we see that $\alpha$ is a uniformizer of $\mathbf Q_p(\alpha)$.