Problem 22

Question

Suppose $p$ is a prime, $g \in \mathbb{Z}[X],$ and $x \in \mathbb{Z},$ such that $g(x) \equiv 0(\bmod p)$ and $\mathbf{D}(g)(x) \not \equiv 0(\bmod p)$. Show that for every positive integer $e,$ there exists an integer $\hat{x}$ such that $g(\hat{x}) \equiv 0\left(\bmod p^{e}\right),$ and give an efficient procedure to compute such an $\hat{x}$, given $p, g, x,$ and $e$. Hint: mimic the "lifting" procedure discussed in $\S 12.5 .2$

Step-by-Step Solution

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Answer

Question: Show that for any positive integer $e$, there exists an integer $\hat{x}$ such that $g(\hat{x}) \equiv 0 \left(\bmod p^e\right)$, given a prime number $p$, a polynomial $g(x)$ with integer coefficients, an integer $x$ that satisfies $g(x) \equiv 0 (\bmod p)$, and the condition $\mathbf{D}(g)(x) \not\equiv 0 (\bmod p)$, where $\mathbf{D}(g)$ denotes the derivative of the polynomial $g$. Provide an efficient procedure to compute this $\hat{x}$. Answer: To find an integer $\hat{x}$ such that $g(\hat{x}) \equiv 0 \left(\bmod p^e\right)$, we can follow the "lifting" procedure. The efficient procedure to compute $\hat{x}$ given the inputs $p, g, x,$ and $e$ is as follows: 1. Initialize $\hat{x} = x$ and $p_i = p$. 2. For each iteration $i=1,\ldots,e-1:$ a. Compute the modular inverse of $\mathbf{D}(g)(\hat{x})$ modulo $p_i$, denoted as $y$. b. Update $\hat{x}$ using the formula: $\hat{x} = \hat{x} - \left(\frac{g(\hat{x})}{p_i}\right) \times \left(\frac{\mathbf{D}(g)(\hat{x})}{y}\right)$. c. Update $p_i = p_i \times p$. 3. Return the final value of $\hat{x}$. By following this procedure, we can efficiently compute $\hat{x}$ for any positive integer $e$.

1Step 1: Understand the lifting procedure

We first need to understand the lifting procedure, which is a technique to find the roots of a polynomial modulo a prime power. Given a root of a polynomial $g(\tilde{x}) \equiv 0 (\bmod p)$, and its derivative $\mathbf{D}(g)(\tilde{x}) \not\equiv 0 (\bmod p)$, we can compute a new root $\hat{x}$ satisfying $g(\hat{x}) \equiv 0 (\bmod p^e)$ by using the formula: $$\hat{x} = \tilde{x} - \left(\frac{g(\tilde{x})}{p}\right) \times \left(\frac{\mathbf{D}(g)(\tilde{x})}{\mathbf{D}(g)(\tilde{x})^{-1} (\bmod p)}\right)$$ where $\mathbf{D}(g)(\tilde{x})^{-1} (\bmod p)$ represents the modular inverse of $\mathbf{D}(g)(\tilde{x})$ modulo $p$.

2Step 2: Iterate the lifting procedure

We will now use the lifting procedure repeatedly for $e-1$ times to get the desired $\hat{x}$. Let's start with the given integer $x$ such that $g(x) \equiv 0 (\bmod p)$ and its derivative $\mathbf{D}(g)(x) \not\equiv 0 (\bmod p)$. For each iteration $i=1,\ldots,e-1:$ 1. Compute the modular inverse $\mathbf{D}(g)(x)^{-1} (\bmod p)$. 2. Apply the lifting procedure to get a new root $\hat{x}$ by using the formula given in Step 1. 3. Update $x := \hat{x}$ and $p := p^2$.

3Step 3: Verify that $g(\hat{x}) \equiv 0 (\bmod p^e)$

After completing the $e-1$ iterations of the lifting procedure, we will get an integer $\hat{x}$. We can verify that $g(\hat{x}) \equiv 0 (\bmod p^e)$ by directly evaluating the polynomial $g$ at $\hat{x}$ and checking the divisibility by $p^e$. If the result is a multiple of $p^e$, then we have successfully found the desired integer $\hat{x}$.

4Step 4: Efficient procedure to compute $\hat{x}$

Based on the aforementioned steps, here is the efficient procedure to compute $\hat{x}$ given the inputs $p, g, x,$ and $e$: 1. Initialize $\hat{x} = x$ and $p_i = p$. 2. For each iteration $i=1,\ldots,e-1:$ a. Compute the modular inverse of $\mathbf{D}(g)(\hat{x})$ modulo $p_i$, denoted as $y$. b. Update $\hat{x}$ using the formula: $\hat{x} = \hat{x} - \left(\frac{g(\hat{x})}{p_i}\right) \times \left(\frac{\mathbf{D}(g)(\hat{x})}{y}\right)$. c. Update $p_i = p_i \times p$. 3. Return the final value of $\hat{x}$.

Key Concepts

Lifting ProcedurePolynomial Roots ModuloNewton's Method

Lifting Procedure

The lifting procedure is a powerful concept used to find solutions to polynomial congruences. Let's say we have a root $ \tilde{x} $ such that $ g(\tilde{x}) \equiv 0 \pmod{p} $, where $ g $ is a polynomial. The goal is to "lift" this solution to a higher modulus, specifically $ p^e $, where $ p $ is a prime and $ e $ is a positive integer.

This is achieved by utilizing Hensel's Lemma, which provides that if $ \mathbf{D}(g)(\tilde{x}) ot\equiv 0 \pmod{p} $, there exists a unique $ \hat{x} \equiv \tilde{x} \pmod{p} $ such that $ g(\hat{x}) \equiv 0 \pmod{p^2} $.

The essence of the lifting process is iterative, enhancing the solution step by step, ultimately meeting the modulus requirement $ p^e $.

During each step of the iteration, the root is refined using:

Finding the modular inverse of the polynomial's derivative (ensures accurate adjustments)
Applying the formula to progressively "lift" the solution by adjusting for higher power moduli

This procedure grants us the desired accuracy and specificity needed for complex mathematical and computational tasks.

Polynomial Roots Modulo

Finding polynomial roots modulo a prime is about discovering values that satisfy a polynomial equation under modular constraints. In our context, the polynomial equation is $ g(x) \equiv 0 \pmod{p} $.

Initially, you might have some value $ x $ that satisfies $ g(x) \equiv 0 \pmod{p} $, meaning when $ x $ is substituted into $ g(x) $, it yields a multiple of the prime $ p $. This is the foundation on which the lifting procedure builds. To deal with higher moduli like $ p^2, p^3, \ldots, p^e $, understanding initial roots becomes critical.

In these modular systems:

Roots must fit within the constraints of congruences, meaning simplistic rules of division don't apply.
The accuracy of roots modulo increasing powers of $ p $ requires iterative refinement.

Ultimately, the task is to "lift" these roots to be compatible with larger congruence classes, leveraging criteria stated in Hensel's Lemma. This makes roots modulo a stepping stone in solving broader polynomial congruences.

Newton's Method

Newton's Method is a mathematical technique typically used for finding successively better approximations to the roots (or zeroes) of a real-valued function. In the realm of modular arithmetic, it parallels in function by refining approximations within a modular system.

Just as Newton's Method iteratively hones in on a real solution by evaluating function derivatives, the lifting procedure borrows this principle under the constraints of modular arithmetic.

Here's how it aligns:

In standard settings, the function's derivative plays a key role in creating the next approximation.
In our modular context, the derivative $ \mathbf{D}(g)(x) $ functions similarly, but must consider the inverse within the prime modulus.
The adaptation to modular arithmetic involves similar iterative refinement, ensuring congruence at each step.

This strategy is central to making minor adjustments that respect the modulus, eventually achieving a solution that holds for higher powers. The synergy between Newton's Method's classic approach and modular arithmetic demonstrates mathematical techniques' adaptability across various settings.

Problem 21

Problem 23

Other exercises in this chapter

Problem 20

Let $g \in R[X],$ and let $x \in R$ be a root of $g .$ Show that $x$ is a multiple root of $g$ if and only if $x$ is also a root of \(\mathbf{D}(g)\

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Problem 21

Let $g \in R[X]$ with $\operatorname{deg}(g)=k \geq 0,$ and let $x \in R .$ Show that if we evaluate $g$ at $X+x,$ writing $$ g(X+x)=\sum_{i=0}^{k} b_

View solution

Problem 23

Let $F$ be a field. Show that every non-zero ideal of $F \llbracket X \rrbracket$ is of the form $\left(X^{m}\right)$ for some uniquely determined integer

View solution

Problem 24

Let $F$ be a field. Show that $F((X))$ is the field of fractions of $F \llbracket X \rrbracket ;$ that is, there is no subfield $E \subsetneq F((X))$ th

View solution