Lemma 15.101.1 (Krasner's lemma). Let $A$ be a complete local domain of dimension $1$. Let $P(t) \in A[t]$ be a polynomial with coefficients in $A$. Let $\alpha \in A$ be a root of $P$ but not a root of the derivative $P' = \text{d}P/\text{d}t$. For every $c \geq 0$ there exists an integer $n$ such that for any $Q \in A[t]$ whose coefficients are in $\mathfrak m_ A^ n$ the polynomial $P + Q$ has a root $\beta \in A$ with $\beta - \alpha \in \mathfrak m_ A^ c$.

Proof. Choose a nonzero $\pi \in \mathfrak m$. Since the dimension of $A$ is $1$ we have $\mathfrak m = \sqrt{(\pi )}$. By assumption we may write $P'(\alpha )^{-1} = \pi ^{-m} a$ for some $m \geq 0$ and $a \in A$. We may and do assume that $c \geq m + 1$. Pick $n$ such that $\mathfrak m_ A^ n \subset (\pi ^{c + m})$. Pick any $Q$ as in the statement. For later use we observe that we can write

$P(x + y) = P(x) + P'(x)y + R(x, y)y^2$

for some $R(x, y) \in A[x, y]$. We will show by induction that we can find a sequence $\alpha _ m, \alpha _{m + 1}, \alpha _{m + 2}, \ldots$ such that

1. $\alpha _ k \equiv \alpha \bmod \pi ^ c$,

2. $\alpha _{k + 1} - \alpha _ k \in (\pi ^ k)$, and

3. $(P + Q)(\alpha _ k) \in (\pi ^{m + k})$.

Setting $\beta = \mathop{\mathrm{lim}}\nolimits \alpha _ k$ will finish the proof.

Base case. Since the coefficients of $Q$ are in $(\pi ^{c + m})$ we have $(P + Q)(\alpha ) \in (\pi ^{c + m})$. Hence $\alpha _ m = \alpha$ works. This choice guarantees that $\alpha _ k \equiv \alpha \bmod \pi ^ c$ for all $k \geq m$.

Induction step. Given $\alpha _ k$ we write $\alpha _{k + 1} = \alpha _ k + \delta$ for some $\delta \in (\pi ^ k)$. Then we have

$(P + Q)(\alpha _{k + 1}) = P(\alpha _ k + \delta ) + Q(\alpha _ k + \delta )$

Because the coefficients of $Q$ are in $(\pi ^{c + m})$ we see that $Q(\alpha _ k + \delta ) \equiv Q(\alpha _ k) \bmod \pi ^{c + m + k}$. On the other hand we have

$P(\alpha _ k + \delta ) = P(\alpha _ k) + P'(\alpha _ k)\delta + R(\alpha _ k, \delta )\delta ^2$

Note that $P'(\alpha _ k) \equiv P'(\alpha ) \bmod (\pi ^{m + 1})$ as $\alpha _ k \equiv \alpha \bmod \pi ^{m + 1}$. Hence we obtain

$P(\alpha _ k + \delta ) \equiv P(\alpha _ k) + P'(\alpha ) \delta \bmod \pi ^{k + m + 1}$

Recombining the two terms we see that

$(P + Q)(\alpha _{k + 1}) \equiv (P + Q)(\alpha _ k) + P'(\alpha ) \delta \bmod \pi ^{k + m + 1}$

Thus a solution is to take $\delta = -P'(\alpha )^{-1} (P + Q)(\alpha _ k) = - \pi ^{-m} a (P + Q)(\alpha _ k)$ which is contained in $(\pi ^ k)$ by induction assumption. $\square$

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