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Let \(F\) be a field. Let \(z=s / t,\) where \(s, t \in F[X], \operatorname{deg}(s)<\operatorname{deg}(t)\) and \(\operatorname{gcd}(s, t)=1\). (a) Show that if \(F\) is finite, there exist integers \(k, k^{\prime}\) such that \(0 \leq k

Short Answer

Expert verified
Question: Show that if $F$ is finite, there exist integers $k, k'$ such that $0 \leq k < k'$ and $sX^{k} ≡ sX^{k'}(\bmod t)$. Answer: If $F$ is finite, there are a finite number of elements, say $q$. Then for any $k$ ranging from $0$ to $q-1$, consider the residue classes of $sX^k \pmod{t}$. By the Pigeonhole Principle, for at least one pair of distinct integers $k$ and $k'$ within this range, we have $sX^k \equiv sX^{k'} (\bmod t)$.

Step by step solution

01

Understanding the problem

When \(F\) is finite, its order, which is the number of elements, is a finite number, say \(q\). We need to show that there exist integers \(k, k'\) such that \(0 \leq k<k'\) and \(s X^{𝑘} \equiv s X^{𝑘^{\prime}}(\bmod t)\).
02

Using the Pigeonhole Principle

For \(k=0,1, \ldots, q-1\), consider the residue classes of \(sX^k \pmod{t}\). Since there are only \(q\) different residue classes \(\bmod {t}\), by the Pigeonhole Principle, there must be two distinct integers \(k\) and \(k'\) within this range such that \(sX^k \equiv sX^{k'} (\bmod t)\). (b) Show that for integers \(k, k^{\prime}\) with \(0 \leq k<k^{\prime},\) the sequence of coefficients of the reversed Laurent series representing \(z\) is \(\left(k, k^{\prime}-k\right)\) -periodic if and only if \(s X^{k} \equiv s X^{k^{\prime}}(\bmod t)\)
03

Understanding the problem

We are given the condition \(s X^{k} \equiv s X^{k^{\prime}}(\bmod t)\) and want to show that this implies that the coefficients of the reversed Laurent series representing \(z\) are \((k, k'-k)\)-periodic.
04

Expanding the reversed Laurent series

Let \(\tau(z)\) denote the reversed Laurent series of \(z\). From \(sX^{k} \equiv sX^{k'} \pmod{t}\), we have \(z X^{k} - z X^{k^{\prime}} \equiv 0 \pmod{t}\). By definition of the reversed Laurent series, we get \(\tau(z X^k)-\tau(z X^{k'}) = \tau\left(\frac{sX^k}{t}\right) - \tau\left(\frac{sX^{k'}}{t}\right)\). The coefficients of \(\tau\left(\frac{sX^k}{t}\right)\) and \(\tau\left(\frac{sX^{k'}}{t}\right)\) are periodic with lengths \(k\) and \(k'-k\), which implies that the difference is periodic with combined length \(\operatorname{lcm}(k, k'-k)\). Since both lengths are relatively prime, their lcm is equal to their product, \((k, k'-k)\). So, the coefficients of \(\tau(z)\) are \((k, k'-k)\)-periodic. The reverse implication is fairly similar: if the coefficients are \((k, k'-k)\)-periodic, then \(s X^k \equiv s X^{k'} \pmod{t}\). (c) Show that if \(F\) is finite and \(X \nmid t,\) then the reversed Laurent series representing \(z\) is purely periodic with period equal to the multiplicative order of \([X]_{t} \in(F[X] /(t))^{*}\)
05

Understanding the problem

We consider the case when \(X \nmid t\), and want to show that the reversed Laurent series is purely periodic.
06

Finding the period

First, we note that \([X]_t\) has multiplicative order \(p \in \mathbb{N}\) if \((X^p-1) \equiv 0 \pmod{t}\). By definition of the reversed Laurent series, we have \(\tau(z) = \sum_{i \geq 0} a_i X^{-d_{z} - i}\). Next, we note that the reversed Laurent series of \(zX^p\) has the same period as \(\tau\left(\frac{sX^p}{t}\right)\) by our analysis in part (b), which is equal to the multiplicative order of \([X]_t\). Therefore, the reversed Laurent series is purely periodic. (d) More generally, show that if \(F\) is finite and \(t=X^{k} t^{\prime},\) with \(X \nmid t^{\prime},\) then the reversed Laurent series representing \(z\) is ultimately periodic with pre-period \(k\) and period equal to the multiplicative order of \([X]_{t^{\prime}} \in\left(F[X] /\left(t^{\prime}\right)\right)^{*}\)
07

Understanding the problem

In this case, we have \(t=X^k t'\), and we want to show that the reversed Laurent series becomes periodic after a certain pre-period.
08

Decompose the Laurent series

Consider the sequence of coefficients of the reversed Laurent series representing \(z\), and denote it as \((a_i)_{i \geq 0}\). We can naturally decompose this series into two parts: the first \(k\) coefficients corresponding to the pre-period, and the remaining coefficients corresponding to the periodic part. Now, consider the sequence of coefficients of the reversed Laurent series representing \(z \cdot X^k\). By the periodicity of the reversed Laurent series, the coefficients of this series should be equal to the periodic part of the sequence \((a_i)_{i \geq 0}\). By our analysis in parts (b) and (c), this holds true if we can show that the coefficients are equal to the multiplicative order of \([X]_{t'}\).
09

Proving periodicity and pre-period

To prove this, note that since \(t = X^k t'\) and \(X \nmid t'\), we know that \(z \cdot X^k = s / t'\). Rewriting this as a reversed Laurent series, we have \(\tau(z\cdot X^k) = \tau\left(\frac{sX^k}{t'}\right)\). Using the properties shown in parts (b) and (c), we conclude that the reversed Laurent series representing \(z \cdot X^k\) has a period equal to the multiplicative order of \([X]_{t'}\). As a result, this implies that the reversed Laurent series representing \(z\) is ultimately periodic with pre-period \(k\) and period equal to the multiplicative order of \([X]_{t'}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Finite Fields
A finite field, also known as a Galois field, is a field that contains a finite number of elements. In such a field, the arithmetic of numbers, polynomials, or even matrices follows particular rules that are sometimes counterintuitive when compared to the infinite fields like the real numbers or complex numbers we are more familiar with. For instance, in a finite field, every element other than zero has a multiplicative inverse, and the set of non-zero elements forms a group under multiplication.

Finite fields are especially important in areas like coding theory, cryptography, and in our particular case, understanding the periodicity in reversed Laurent series. A key feature of finite fields that we exploit in the exercise is that the number of different polynomial residue classes modulo some polynomial is limited by the size of the field. This feature interacts with the Pigeonhole Principle to ensure the existence of periodic patterns in sequences or series derived from polynomials over finite fields.
Polynomial Residue Classes
The notion of polynomial residue classes is analogous to the concept of integer residues when performing modular arithmetic with integers. Essentially, we're looking at the set of remainders when polynomials are divided by a given polynomial, similar to how we look at remainders of integers when divided by an integer modulus.

Formally, if we have a polynomial t over a finite field, the residue class of a polynomial s modulo t, denoted as [s]_t, consists of all polynomials that differ from s by a multiple of t. This is important because it implies that, in these residue classes, polynomials can be considered equivalent or 'congruent' to each other, just as we consider numbers congruent in integer modular arithmetic. In our exercise, understanding these residue classes is core to grasping why certain series can be periodic and why they exhibit repetitive patterns.
Multiplicative Order
Multiplicative order is an important concept in number theory and abstract algebra that also applies to elements within a finite field. In essence, it refers to the number of times one must multiply an element by itself to get the multiplicative identity (which is 1 in the context of fields).

For an element g in a finite field, its multiplicative order is the smallest positive integer n such that g^n = 1, where the operations are carried out in the field. This concept is directly related to the periodicity discussed in the exercise, where the multiplicative order of the residue class [X]_t determines the periodic behavior of the reversed Laurent series derived from it. Understanding the multiplicative order allows students to calculate the periodicity of such sequences, which is a crucial part of solving problems involving cyclical patterns in finite fields.
Pigeonhole Principle
The Pigeonhole Principle is a simple yet powerful tool in combinatorics and probability theory. The principle states that if you have more pigeons than pigeonholes, at least one pigeonhole must contain more than one pigeon. This might sound like common sense, but it has profound implications in mathematics.

For our exercise, we use the Pigeonhole Principle to reason that since there are a finite number of residue classes for the polynomial sX^k modulo t, and we can form more power multiples of sX^k than there are distinct residue classes, two of these power multiples must be congruent, i.e., fall into the same 'pigeonhole.' This congruence is the basis for establishing the periodicity of the reversed Laurent series for a given polynomial fraction, making the Pigeonhole Principle an indispensable concept for these types of proofs in finite field arithmetic.

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Most popular questions from this chapter

Assume \(2_{R} \in R^{*}\). Suppose that we are given two polynomials \(g, h \in R[X]\) of length at most \(\ell,\) along with a primitive \(2^{n}\) th root of unity \(\omega \in R,\) where \(2 \ell \leq 2^{n}<4 \ell .\) Let us "pad" \(g\) and \(h,\) writing \(g=\sum_{i=0}^{2^{n}-1} a_{i} X^{i}\) and \(h=\sum_{i=0}^{2^{n}-1} b_{i} X^{i},\) where \(a_{i}\) and \(b_{i}\) are zero for \(i \geq \ell\). Show that the following algorithm correctly computes the product of \(g\) and \(h\) using \(O(\ell \operatorname{len}(\ell))\) operations

This exercise develops a fast algorithm, called the fast Fourier transform or FFT, for computing the function \(\mathcal{E}_{n, \omega} .\) This is a recursive algorithm 17.6 Faster polynomial arithmetic (*) FFT \(\left(n, \omega ; a_{0}, \ldots, a_{2^{n}-1}\right)\) that takes as input an integer \(n \geq 0,\) a primitive \(2^{n}\) th root of unity \(\omega \in R,\) and elements \(a_{0}, \ldots, a_{2^{n}-1} \in R,\) and runs as follows: if \(n=0\) then return \(a_{0}\) else $$ \begin{array}{l} \left(\alpha_{0}, \ldots, \alpha_{2^{n-1}-1}\right) \leftarrow \mathrm{FFT}\left(n-1, \omega^{2} ; a_{0}, a_{2}, \ldots, a_{2^{n}-2}\right) \\\ \left(\beta_{0}, \ldots, \beta_{2^{n-1}-1}\right) \leftarrow \mathrm{FFT}\left(n-1, \omega^{2} ; a_{1}, a_{3}, \ldots, a_{2^{n}-1}\right) \\\ \text { for } i \leftarrow 0 \text { to } 2^{n-1}-1 \mathrm{do} \\ \quad \gamma_{i} \leftarrow \alpha_{i}+\beta_{i} \omega^{i}, \gamma_{i+2^{n-1}} \leftarrow \alpha_{i}-\beta_{i} \omega^{i} \\ \text { return }\left(\gamma_{0}, \ldots, \gamma_{2^{n}-1}\right) \end{array} $$

Let \(n\) be a large, positive integer. We can factor \(n\) using trial division in time \(n^{1 / 2+o(1)} ;\) however, using fast polynomial arithmetic in \(\mathbb{Z}_{n}[X],\) one can get a simple, deterministic, and rigorous algorithm that factors \(n\) in time \(n^{1 / 4+o(1)} .\) Note that all of the factoring algorithms discussed in Chapter \(15,\) while faster, are either probabilistic, or deterministic but heuristic. Assume that we can multiply polynomials in \(\mathbb{Z}_{n}[X]\) of length at most \(\ell\) using \(M(\ell)\) operations in \(\mathbb{Z}_{n},\) where \(M\) is a well-behaved complexity function, and \(M(\ell)=\ell^{1+o(1)}\) (the algorithm from Exercise 17.24 would suffice). (a) Let \(\ell\) be a positive integer, and for \(i=1, \ldots, \ell,\) let $$ a_{i}:=\prod_{j=0}^{\ell-1}(i \ell-j) \bmod n $$ Using fast polynomial arithmetic, show how to compute \(\left(a_{1}, \ldots, a_{\ell}\right)\) in time \(\ell^{1+o(1)} \operatorname{len}(n)^{O(1)}\) (b) Using the result of part (a), show how to factor \(n\) in time \(n^{1 / 4+o(1)}\) using a deterministic algorithm.

Let \(g, h \in R[X, Y]\) with \(g=\sum_{i=0}^{m-1} g_{i} Y^{i}\) and \(h=\sum_{i=0}^{m-1} h_{i} Y^{i},\) where each \(g_{i}\) and \(h_{i}\) is a polynomial in \(X\) of degree less than \(k .\) The product \(f:=g h \in R[X, Y]\) may be written \(f=\sum_{i=0}^{2 m-2} f_{i} Y^{i},\) where each \(f_{i}\) is a polynomial in \(X .\) Show how to compute \(f,\) given \(g\) and \(h,\) using \(O(M(k m))\) operations in \(R\). Hint: for an appropriately chosen integer \(t>0,\) first convert \(g, h\) to \(\tilde{g}, \tilde{h} \in R[X],\) where \(\tilde{g}:=\sum_{i=0}^{m-1} g_{i} X^{t i}\) and \(\tilde{h}:=\sum_{i=0}^{m-1} h_{i} X^{t i} ;\) next, compute \(\tilde{f}:=\tilde{g} \tilde{h} \in R[X] ;\) finally, "read off" the \(f_{i}\) 's from the coefficients of \(\tilde{f}\).

Let \(x, a_{0}, \ldots, a_{\ell-1} \in R,\) and let \(k\) be an integer with \(0

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