Proof. Let $\pi \in O_K$ be a uniformiser and let $r=v(f^{\prime }(a))$, with $v$ the normalised valuation ($v(\pi )=1$). We construct a sequence ${(x_n)}_{n=1}^{\infty }$ in $O_K$ such that:

• (i) $f(x_n)\equiv 0(mod{\pi }^{n+2r})$
• (ii) $x_{n+1}\equiv x_n(mod{\pi }^{n+r})$

Take $x_1=a$: then $f(x_1)\equiv 0(mod\pi 1+2r)$.

Now we suppose we have constructed $x_1,\dots ,x_n$ satisfying (i) and (ii). Define

Since $x_n\equiv x_1(mod{\pi }^{r+1})$, we have

and hence

by (i).

It follows that $x_{n+1}\equiv x_n(mod{\pi }^{n+r})$, so (ii) holds. Note that letting $X,Y$ be indeterminates, we have

where $f_i(X)\in O_K[X]$ and $f_0(X)=f(X)$, $f_1(X)=f^{\prime }(X)$. Thus

where $c=-\frac{f(x_n)}{f^{\prime }(x_n)}$.

Since $c\equiv 0(mod{\pi }^{n+r})$ and $v(f_i(x_n))\ge 0$ we have

so (i) holds.

Property (ii) implies that ${(x_n)}_{n=1}^{\infty }$ is Cauchy, so let $x\in O_K$ such that $x_n\to x$. Then $f(x)=\underset{n\to \infty }{\lim }f(x_n)=0$ by (i).

Moreover, (ii) impies that

This proves existence.

Uniqueness: suppose $x^{\prime }$ also satisfies $f^{\prime }(x)=0$, $|x^{\prime }-a|<|f^{\prime }(a)|$. Set $\delta =x^{\prime }-x\ne 0$. Then

and the ultrametric inequality implies

But

Hence $|f^{\prime }(x)\delta |\le |\delta {|}^2$, so $|f^{\prime }(x)|<|\delta |$, a contradiction. □

Proof. Apply Theorem 4.1 to a lift $c\in O_K$ of $\overline{c}$. Then $|f(c)|<1=|f^{\prime }(c){|}^2$ since $\overline{c}$ is a simple root. □

Proof. Case $p>2$: Let $b\in {{\mathbb{Z}}_p}^{\times }$. Applying to $f(X)=X^2-b$, we find that $b\in {({{\mathbb{Z}}_p}^{\times })}^2$ if and only if $b\in {({𝔽}_p^{\times })}^2$. Thus ${{\mathbb{Z}}_p}^{\times }∕{({{\mathbb{Z}}_p}^{\times })}^2\cong {𝔽}_p^{\times }∕{({𝔽}_p^{\times })}^2\cong \mathbb{Z}∕2\mathbb{Z}$ (${𝔽}_p^{\times }\cong \mathbb{Z}∕(p-1)\mathbb{Z}$).

We have an isomorphism

given by $(u,n)\mapsto u{p}^n$. Thus

Case $p=2$: Let $b\in {\mathbb{Z}}_2^{\times }$. Consider $f(X)=X^2-b$. Note $f^{\prime }(X)=2X\equiv 0(mod2)$. Let $b\equiv 1(mod8)$. Then

Hensel’s Lemma version 1 gives

Then

Again using ${\mathbb{Q}}_2^{\times }\equiv {\mathbb{Z}}_2^{\times }\times \mathbb{Z}$, we find that ${\mathbb{Q}}_2^{\times }\cong {(\mathbb{Z}∕2\mathbb{Z})}^3$. □

Proof. Example Sheet 1. □

Proof. Upon scaling, we may assume $f(X)\in O_K[X]$ with $\underset{i}{\max }(|a_i|)=1$. Thus we need to show that $\max (|a_0|,|a_n|)=1$. If not, let $r$ minimal such that $|a_r|=1$, then $0<r<n$. Thus we have

Then Theorem 4.4 implies $f(X)=g(X)h(X)$ with $0<\mathrm{deg}<n$. □