(p-1)th Roots of the p-Adic Integers (7.6.11e)

Dummit and Foote Abstract Algebra, section 7.6, exercise 11e:

Show that if $a_1 \not ≡ 0 \mod{p}$, then there is an element $a=(a_i)$ in the inverse limit $\mathbb{Z}_p$ satisfying $a_j^{p-1} ≡ 1 \mod{p^j}$ and $\mu_{j1}(a_j)=a_1$ for all $j$. Deduce that $\mathbb{Z}_p$ contains $p-1$ distinct $(p-1)$th roots of 1.

Proof: We claim that for any $a_1 \not ≡ 0 \mod{p}$, the coordinates of $a$ are uniquely determined and satisfy the given criteria. We proceed by induction on the $n$th coordinate: The case holds true for $n=1$, so observe the $n+1$th coordinate. $a_{n+1}^{p-1} ≡ 1 \mod{p^{n+1}}$ if and only if its associated polynomial representation $(b_0+b_1p+...+b_np^n)^{p-1} ≡ 1 \mod{p^{n+1}}$ holds true, so this is equivalent to solving for $b_n$ since $a_n=b_0+b_1p+...+b_{n-1}p^{n-1}$ has been uniquely determined. $$(b_0+b_1p+...+b_np^n)^{p-1} ≡ 1 \mod{p^{n+1}}⇔$$ $$((b_0+b_1p+...+b_{n-1}p^{n-1})+(b_np^n))^{p-1} ≡ 1 \mod{p^{n+1}}⇔$$ $$\sum_{k=0}^{p-1} {p-1\choose k}(b_0+b_1p+...+b_{n-1}p^{n-1})^k(b_np^{n})^{p-1-k}≡1 \mod{p^{n+1}}⇔$$ $$(b_0+b_1p+...+b_{n-1}p^{n-1})^{p-1}+$$ $$(p-1)(b_0+b_1p+...+b_{n-1}p^{n-1})^{p-2}(b_np^n) ≡ 1 \mod{p^{n+1}}$$ Since $(b_0+b_1p+...+b_{n-1}p^{n-1})^{p-1}≡ 1 \mod{p^n}$ has been uniquely determined, we have $(b_0+b_1p+...+b_{n-1}p^{n-1})^{p-1}=1+vp^n$ for some integer $v$. $$1+vp^n+(p-1)(b_0+b_1p+...+b_{n-1}p^{n-1})^{p-2}(b_np^n) ≡ 1 \mod{p^{n+1}}⇔$$ $$1+vp^n-b_0^{p-2}b_np^n ≡ 1 \mod{p^{n+1}}⇔$$ $$1+(v-a_1^{p-2}b_n)p^n ≡ 1 \mod{p^{n+1}}⇔$$ $$v-a_1^{p-2}b_n ≡ 0 \mod{p}⇔$$ $$b_n≡a_1^{2-p}v \mod{p}~\square$$