Root of power $p$ from $1$ in the field of $p$-adic numbers

How to prove that there is no root of power $p$ from $1$ other than $1$ for $p$ not equal to $2$?

Solution 1:

You need to know the Eisenstein criterion for irreducibility: if $R$ is a unique factorization domain (like $\Bbb Z$ or $\Bbb Z_p$), and if $f(X)\in R[X]$ is monic with all lower coefficients divisible by the prime element $p$, and if $p^2$ does not divide the constant term of $f$, then $f$ is irreducible in $R[X]$.

Now consider $g(X)=X^p-1=(X-1)(X^{p-1}+X^{p-2}+\cdots+X+1)$. You want to show that the second factor, call it $h$, is $\Bbb Z_p$-irreducible, and in particular has no roots in $\Bbb Z_p$. Note that the degree of $h$ is $p-1$, and its roots are all the $p$-th roots of unity other than $1$.

Look now at $G(X)=g(X+1)=X\bigl((X+1)^{p-1}+\cdots+(X+1)+1\bigr)$, in which the second factor, call it $H(X)$, clearly has constant term equal to $p$. Now look at this equation modulo $p$: as an equation with coefficients in $\Bbb Z/(p)$, the field with $p$ elements, it becomes $G(X)=(X+1)^p-1=X\cdot X^{p-1}$. This shows that as a $\Bbb Z/(p)$-polynomial, $H(X)=X^{p-1}$, in other words, $H$ is monic with all lower coefficients divisible by $p$. Thus $H$ is irreducible, by Eisenstein, and the corresponding $h$ is also irreducible.

Solution 2:

Here is a proof which seems to rely only on the fact that every element $x \in \Bbb Q_p$ has an absolute value $\vert x\vert_p \in p^\Bbb Z$ (or additive valuation $v_p(x) \in \Bbb Z$), and basic properties of ultrametrics. However, underlying are the same polynomial manipulations as in Lubin's answer.

Namely, assume $\zeta \in \Bbb Q_p$ with $\zeta^p =1$ but $\zeta \neq 1$. Call $x := \zeta -1$. Then

$$1 = (x+1)^p = x^p+px^{p-1}+\binom{p}{2}x^{p-2} +...+px+1$$ so $$0= x^p+px^{p-1}+\binom{p}{2}x^{p-2} +...+px =x\cdot(x^{p-1}+px^{p-2}+\binom{p}{2}x^{p-3} +...+\binom{p}{p-2}x+p)$$ and since $x\neq 0$, the second factor is zero, meaning $$x^{p-1}+px^{p-2}+\binom{p}{2}x^{p-3} +...+\binom{p}{p-2}x = -p$$

The right hand side has absolute value $p^{-1}$ (or: additive valuation $v_p(RHS) =1$). However, if $\vert x\vert_p \ge 1$ (i.e $v_p(x)\le 0$), the left hand side has absolute value $\ge 1$ (i.e. $v_p(LHS)\le 0$). If, on the other hand, $\vert x\vert_p \le p^{-1}$ (i.e. $v_p(x)\ge 1$), then the left hand side has absolute value $\le p^{-2}$ (i.e. $v_p(LHS)\ge 2$) because each term has such value (because the binomial coefficients are all divisible by $p$, and $p-1 \ge 2$ -- here we use that $p\neq 2$, otherwise indeed $x=-2 \in \Bbb Q \subset \Bbb Q_p$). Contradiction.

(Indeed, the correct absolute value of $x$ would be $p^{-\frac{1}{p-1}}$ (i.e. $v_p(x) =\frac{1}{p-1}$) but no elements of $\Bbb Q_p$ have such value.)

I think one can also prove by elementary means that for every sequence of integers (or rationals) $(a_n)_n$ such that $a_n^p \rightarrow 1$ ($p$-adically), necessarily $a_n \rightarrow 1$ ($p$-adically). I leave that to you. Update: The argument I had in mind for that is basically the one that makes up the answer by user reuns now.

Solution 3:

$\quad$ Otherwise I think there is that elementary argument :

$p\ne 2$.

$\mathbb{Z}/(p^k)^\times$ has $(p-1)p^{k-1}$ elements.

$H_k= \{ c \in \mathbb{Z}/(p^k)^\times, c \equiv 1 \bmod p \}$ is a subgroup of order $p^{k-1}$.

$(1+p)^{p^{k-2}} \equiv 1+p^{k-1} \bmod p^k$ so $1+p \bmod p^k$ is of order $p^{k-1}$ and it is a generator of $H_k$ cyclic (note the special case $(1+2)^{2^{k-2}} \equiv 1 +2.2^{k-2}+2^2 \frac{2^{k-2}(2^{k-2}-1)}{2}\equiv 1\bmod 2^k$)

The elements of $H_k$ and $\mathbb{Z}/(p^k)^\times$ whose order divides $ p$ are of the form $(1+p)^{e p^{k-2}}$.

If $\zeta^p = 1 \in \mathbb{Z}_p$ then $\zeta \equiv (1+p)^{e_k p^{k-2}} \equiv 1+e_k p^{k-1}\bmod p^k$ so that $\zeta \equiv 1 \bmod p^{k-1}$ and $\zeta = 1 $.