Given a surjection of rings with nilpotent kernel, the root of a polynomial lifts to a unique root. Let $ \phi : R_1 \rightarrow R_2$ be a surjection of rings such that the kernel is nilpotent, let $f \in R_1[T]$ be such that its image under the map induced by $\phi, \ $ namely $g \in R_2[T], \ $  has a root, $\overline{\alpha}$, and $g'(\overline{\alpha}) \in R_2^\times$. I'm trying to show that there's a unique $\alpha \in R_1$ such that: $\phi(\alpha)=\overline\alpha \ $ and $ \ f(\alpha)=0.$

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*Uniqueness
if $ \ \alpha, \beta \in R_1$ are such that $f(\alpha)=f(\beta)=0 \ $ and $ \ \phi(\alpha)=\phi(\beta)=\overline{\alpha},\ $ then $g$ has a double root at $\overline{\alpha}$, thus $g'(\overline{\alpha}) \notin R_2^\times \ \  \unicode{x21af}$.

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*Existence
I was thinking that we should show that if there's a solution in $\frac{R_1}{(ker \phi)^k}, \ $ then it lifts to a solution in $\frac{R_1}{(ker \phi)^{k+1}},$ and since $$ \exists \ n \text{ with } (\ ker \ \phi \ )^n= (0), \text{ we would get a solution in } \frac{R_1}{(0)} \cong R_1 \text{ as requested}$$  but I've got no clue on how to prove the existence of a lift to $ \frac{R_1}{(ker \phi)^{k+1}}, \  $ any help?
 A: To start with, we can make an (unimportant) reduction to simplify things. Namely, with $R=R_1$, $I=\ker \phi$, we have a polynomial $f\in R[T]$ whose reduction $\bar f\in (R/I)[T]$ has a root $\bar \alpha$, with $\bar f'(\bar \alpha)$ invertible. The goal is to show that, given that $I$ is nilpotent, that $\bar \alpha$ has a unique lift to a root of $f$. If $I^n = 0$, we can consider the filtration $$R = R/I^n \to R/I^{n-1}\to \cdots \to R/I$$ so it suffices to consider the case where $n=2$, since this allows us to lift at each stage.
So suppose $I^2=0$ and $\bar f$ has root $\bar \alpha$ with $\bar f'(\bar \alpha)$ invertible. Let $\alpha_0$ be any lift of $\bar \alpha$; we want to find a unique $a\in I$ for which $f(\alpha_0+a) = 0$. Let $f(x) = a_nx^n+\cdots+a_0$. Since $a^2=0$, we see that for any $m$, we have $(\alpha_0+a)^m = \alpha_0^m + m\alpha_0^{m-1}a$, so in particular we have $f(\alpha_0+a) = f(\alpha_0)+f'(\alpha_0)a$. Since $\bar f'(\bar \alpha)$ is invertible in $R/I$, there is some $b\in R$ and $y\in I$ for which $f'(\alpha_0)b = 1+y$. But then $f'(\alpha_0)$ is invertible in $R$, since $$f'(\alpha_0)b\cdot (1-y) = 1-y^2 = 1.$$ In particular we are forced to choose $a = -f(\alpha_0)f'(\alpha_0)^{-1}$, giving both existence and uniqueness of a lift.
This strategy allows you to even lift a root of $R/I$ uniquely to the $I$-adic completion of $R$, which, in fancier terms, is saying that for any ring $R$ and ideal $I$, the pair $(\hat R_I,I)$ is a henselian pair. This, sometimes in more or less general forms, is called "Hensel's lemma." Your particular question is simply a special case, where $\hat R_I = R$ since $I$ is nilpotent.
