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Eisenbud states the following corollary to Hensel's lemma:

Given a polynomial $f(t,x)$ over a field $k$, with $x=a$ a simple root of $f(0,x)$, then there exists a unique power series $x(t) \in k[[t]]$ with $x(0) = a$ and $f(t,x(t)) = 0$ identically. His hint is to use the complete local ring $R = k[[t]]$ for $R[x]$.

Since $a$ is a simple root of $f(0,a)$, we know that $f'(t,a) \neq 0$. But, I'm not sure why $$ \begin{split} f(t,a) &\equiv 0 \pmod{f'(a)\cdot (t)} \\ &= 0 \pmod{\alpha\cdot(t)} \end{split} $$ This would hold of course if $f(t,a) \equiv f(0,a)$, but I'm not sure this is the case.

Zach Teitler
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54321user
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    But $f'(a)$ is a nonzero element of the field, and therefore $f'(a)\cdot(t)=(t)$. I think it will all be clear if you write $f(t,x)=\sum_{i=0}^\infty g_i(x)t^i$ – Lubin Aug 27 '17 at 05:38

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We will apply Hensel's Lemma to $$\tilde{f}(x)=f(t,x)=g_0(t)+g_1(t)x+\cdots +g_m(t)x^m=h_0(x)+h_1(x)t+\cdots +h_n(x)t^n\in k[\![t]\!][x].$$ (It might be more natural to write the first expansion of $f(t,x)$ as a polynomial in $x$, but the latter expansion is helpful. This is where we use the assumption that $f(t,x)$ is a polynomial, instead of a general element in $k[\![t]\!][x]$.)

By assumption, $f(0,a)= h_0(a)=0$ and $\tilde f'(a)\in k[\![t]\!]$ is nonzero, moreover, $$h_0'(a)=\left.\tilde{f}'(a)\right\vert_{t=0}=\partial_t f(0,a)\ne 0\in k.$$ Thus $\tilde f'(a)$, as a unit plus something in the Jacobson radical, is a unit. Then $$\tilde f(a)=h_0(a)+(\text{something}) t\equiv 0\pmod{\tilde f'(a)(t)=(t)}.$$

Peter Griffin
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