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The following was an exercise in some MIT course notes on p-adic numbers and Hensel's lemma.

Let $f\in \Bbb Z_p[X]$. Suppose that $b$ is a simple root of $f$. Prove that for any $a\in \Bbb Z_p$: if $|a-b|_p<|f'(b)|_p$ then $|f(a)|_p<|f'(a)|_p^2$. Conclude that if no $a\in \Bbb Z_p$ satisfies $|f(a)|_p<|f'(a)|_p^2$, then $f$ does not have a simple root in $\Bbb Z_p$.

I am having trouble with the first part, which should be easy.

We have $$f(x) = f(a) + f'(a)(x-a) + g(x)(x-a)^2$$ for some polynomial $g$.

Plugging in $x=b$ gives $f(a)=f'(a)(a-b)-g(b)(a-b)^2$ so taking valuations gives $|f(a)|_p=|f'(a)(a-b)-g(b)(a-b)^2|_p$.

Now, $$ |f'(a)(a-b)|=|f'(a)|\cdot |a-b|<|f'(a)|\cdot |f'(b)| $$

How should I continue ? I guess I should use somewhere the ultrametric inequality, but I don't see where.

Dr. Heinz Doofenshmirtz
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    Use a Taylor expansion of $f'(x)$ to relate $f'(a)$ and $f'(b)$, that should be the last piece of the puzzle. – Merosity Feb 23 '21 at 16:13
  • Could you elaborate on that? We have $f'(x)=f'(b)+f''(b)(x-b)+h(b)(x-b)^2$. Now $x=a$ gives $f'(a)=f'(b)+f''(b)(a-b)+h(b)(a-b)^2$ ? – Dr. Heinz Doofenshmirtz Feb 23 '21 at 22:26
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    We don't have to go so far, $|f'(a)|_p = |f'(b)+h(b)(a-b)|_p$ since we know $|f'(b)|_p>|a-b|_p$ and $|h(b)|_p \le 1$ we have by the ultrametric inequality's strong property that $|f'(a)|_p = |f'(b)|_p$ – Merosity Feb 24 '21 at 02:02
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    @Merosity of course, thank you !! I could now finish, thank you. – Dr. Heinz Doofenshmirtz Feb 24 '21 at 11:18
  • See Theorem 7.1 in https://kconrad.math.uconn.edu/blurbs/gradnumthy/hensel.pdf with $\alpha$ there being $b$ in your post. – KCd Feb 26 '21 at 05:36

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