The Stacks project

Lemma 15.113.1 (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. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 15.113: Krasner's lemma

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 09EI. Beware of the difference between the letter 'O' and the digit '0'.