Question

Let \(F\) be a field of characteristic zero, \(f \in F[x]\) nonconstant, and \(k \in \mathbb{N}\). Prove that $$ \underbrace{\operatorname{gcdE}(\operatorname{gcdE}(\cdots \operatorname{gcdE}}_{\text {ktimes }}(f) \cdots))=\operatorname{gcd}\left(f, E f, \ldots, E^k f\right) $$

Short answer

The statement is true by induction. The repeated "gcdE" is equivalent to gcd of polynomial and its shifts.

Step-by-step solution

Understanding Operators

We need to distinguish between the two operators used: 1. \( \operatorname{gcd} \) stands for the greatest common divisor of two or more polynomials.2. \( E \) is the shift operator that acts on any polynomial \( f \) as \( E(f(x)) = f(x+1) \). The exercise is asking us to show that iterating the "\( \operatorname{gcdE} \)" operation \( k \) times on a polynomial \( f \) is equivalent to calculating the gcd of \( f \), \( Ef \), ..., \( E^k f \).

Initial Setup for Proof Induction

To prove the statement, we will use induction on \( k \). When \( k = 1 \), the statement is simply \( \operatorname{gcdE}(f) = \operatorname{gcd}(f, Ef) \), which holds by the definition of the \( \operatorname{gcdE} \) operation since \( \operatorname{gcdE}(f) = \operatorname{gcd}(f, E(f)) \).

Inductive Hypothesis

Assume that the statement holds for some \( k = n \), i.e.,\[ \underbrace{\operatorname{gcdE}(\operatorname{gcdE}(\cdots\operatorname{gcdE}}_{n \text{ times }}(f)\cdots)) = \operatorname{gcd}(f, Ef, \ldots, E^n f). \]We will prove that the statement holds for \( k = n + 1 \).

Inductive Step

Based on the inductive hypothesis, \[ \operatorname{gcdE}^{(n+1)}(f) = \operatorname{gcdE}(\operatorname{gcd}(f, Ef, ..., E^n f)). \]Apply the \( \operatorname{gcdE} \) operator:\[ \operatorname{gcd}(\operatorname{gcd}(f, Ef, ..., E^n f), E^{n+1}f) = \operatorname{gcd}(f, Ef, ..., E^{n+1}f) \].The right side remains the gcd of all polynomials transformed by \( E \) from 0 to \( n+1 \), validate the hypothesis continues to hold for \( n+1 \).

Conclusion

By induction, we have shown that for all \( k \), \[ \underbrace{\operatorname{gcdE}(\operatorname{gcdE}(\cdots\operatorname{gcdE}}_{k \text{ times }}(f)\cdots)) = \operatorname{gcd}(f, Ef, \ldots, E^k f) \]. This completes the proof.

Key concepts

Shift Operator in Algebra

In algebra, shifts can transform functions by offering a unique method of evaluating changes without altering the original structure. The shift operator is denoted as \( E \) and acts on polynomials. Applying it to a polynomial \( f(x) \) gives us \( f(x+1) \). This means each term in the polynomial is shifted to the right by one unit.

The significance of the shift operator in algebra involves how it allows us to evaluate polynomials at consecutive integers. Consider any polynomial \( f(x) = ax^2 + bx + c \). Applying the shift operator, you'll get \( E(f(x)) = a(x+1)^2 + b(x+1) + c \). After expanding, you'll observe how each term’s structure relates closely to the original.

Using the shift operator helps simplify advanced polynomial operations, particularly when seeking common structures or behaviors across multiple shifted instances of a polynomial.

Inductive Proof

Inductive proof is a powerful mathematical technique to demonstrate the truth of an infinite number of cases using two essential steps: the base case and the inductive step. It often feels like setting a line of dominoes in motion—prove one, and the rest follow.

To use induction, start with a base case, usually where the least value of \( k \) (often \( k = 1 \)) holds true. In our problem, it shows that for \( k = 1 \), \( \operatorname{gcdE}(f) = \operatorname{gcd}(f, E(f)) \). Then, assume that the statement holds for \( k = n \). Next comes the inductive step: prove it also holds for \( k = n + 1 \). Each step builds upon the previous, helping to construct a full proof.

In this exercise, induction allows us to demonstrate a polynomial property over many iterative operations. Thus, it is invaluable for handling operations repeated extensively across algebraic structures.

Field of Characteristic Zero

Fields are algebraic structures that equip mathematicians with a framework to study elements and operations. These fields contain no zero divisors and possess unique properties that simplify complex algebraic expressions. In particular, a 'field of characteristic zero' ensures integers can be directly managed without transformation.

In our context, we're working within an abstract field \( F \) where denoted operations like addition and multiplication behave just as you'd expect with real numbers. This specific characteristic means there is no smallest positive number of times you can add \( 1 \) to zero to achieve zero again.

The field's characteristic affects how polynomial roots are treated, guaranteeing certain algebraic properties hold. It simplifies computations, ensuring the gcd evaluations behave predictably, no matter how often \( E \) is applied. The certainty and straightforward nature of fields of characteristic zero are fundamental to simplifying proofs and polynomial evaluations.