Question

Let \(R\) be a ring (commutative, with 1), \(f, g \in R[x]\), and \(n \in \mathbb{N}\). Prove that \(f=g \bmod x^{n+1}\) implies \(f^{\prime} \equiv g^{\prime} \bmod x^{n}\), and give an example where \(f^{\prime} \equiv g^{\prime} \bmod x^{n+1}\) does not hold.

Short answer

\( f=g \mod x^{n+1} \) implies \( f' \equiv g' \mod x^n \), but not necessarily \( \mod x^{n+1} \).

Step-by-step solution

Identify the given relationship

We start by understanding the given modulo relationship, which is \( f = g \mod x^{n+1} \). This means that the difference \( f - g \) is divisible by \( x^{n+1} \), or equivalently \( f - g = h \cdot x^{n+1} \) for some polynomial \( h \in R[x] \).

Differentiate both sides

Differentiate the relationship \( f - g = h \cdot x^{n+1} \). The derivative of a constant equivalent to 0, here \( h \cdot x^{n+1} \), is \( h' \cdot x^{n+1} + h \cdot (n+1)x^n \).

Check derivative coefficients

Since \( h' \cdot x^{n+1} + h \cdot (n+1)x^n \) is derived from \( x^{n+1} \), the lowest degree term that can remain is a multiple of \( x^n \). Thus, for \( f' - g' = h' \cdot x^{n+1} + h \cdot (n+1)x^n \), it must be divisible by \( x^n \).

Verify divisibility by \( x^n \)

Considering the derivative, \( (f') - (g') = h' \cdot x^{n+1} + h \cdot (n+1)x^n \), note that the smallest term degree is \( h \cdot (n+1)x^n \), confirming divisibility by \( x^n \). Therefore, \( f' \equiv g' \mod x^n \).

Provide a counterexample

To illustrate \( f' \equiv g' \mod x^{n+1} \) doesn't necessarily hold, consider the polynomials \( f = x^{n+1} \) and \( g = 0 \) in the ring \( R[x] \). Clearly, \( f \equiv g \mod x^{n+1} \), but \( f' = (n+1)x^n \) and \( g' = 0 \), hence \( f' ot\equiv g' \mod x^{n+1} \).

Key concepts

Polynomial Derivatives

In mathematics, a polynomial is an expression consisting of variables and coefficients, involving operations like addition, subtraction, and multiplication. A derivative of a polynomial, much like derivatives in calculus, represents the polynomial's rate of change. For polynomial expression:\[ f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \]The derivative, denoted as \( f'(x) \), can be calculated using power rules to find:\[ f'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \ldots + a_1 \]The power rule involves multiplying the coefficient of each term by its power and reducing the power by one. For polynomials in ring theory, a similar operation applies, ensuring that the ring's properties are maintained when computing derivatives. Differentiation shifts polynomial degrees downward, which plays a fundamental role in our original exercise because it impacts divisibility by lower powers of \( x \). Understanding derivatives in this context helps in manipulating and proving equalities like \( f' \equiv g' \mod x^n \), because the differentiation shifts powers, aiding in checking divisibility.

Modular Arithmetic

Modular arithmetic is a system of arithmetic for integers, where numbers wrap around after reaching a certain value, known as the modulus. An analogy frequently used is that of a clock, which "resets" after 12 hours. When applied to polynomials, this concept involves considering equivalence classes where two polynomials are equivalent if their difference is divisible by a specified polynomial.For instance, in the context of our exercise, if we have:\[ f \equiv g \mod x^{n+1} \]This implies that the polynomial \( f - g \) is a multiple of \( x^{n+1} \). This means:\[ f = g + h \cdot x^{n+1} \]for some polynomial \( h \). This concept of equivalence is central when analyzing the properties of derivatives within ring theory. Polynomial congruences like these are essential for understanding how certain properties (like divisibility and equality) behave in more complex structures, particularly in commutative rings. Grasping this is key to understanding why a polynomial's derivative will retain some properties when modular arithmetic comes into play.

Commutative Algebra

Commutative algebra is a branch of algebra that deals with commutative rings and their ideals. These structures form the backbone of many constructs in algebraic geometry, number theory, and beyond. A commutative ring is a set equipped with two operations: addition and multiplication, where multiplication is commutative. This means \( a \times b = b \times a \) for any elements \( a \) and \( b \) in the ring.Rings like \( R[x] \), used in our exercise, involve polynomials whose coefficients belong to a ring \( R \). The concept of modularity within these rings leads to powerful methods for solving polynomial equations and manipulating their derivatives, as we've seen. Commutative algebra provides the framework for understanding the behavior of modules and ideals which in turn explain how polynomial properties transform under operations like differentiation and congruence.Understanding commutative algebra's role illuminates why concepts of divisibility and equality in polynomial expressions become more flexible and intricate within rings. Such comprehension helps one grasp the mechanics behind results like \( f' \equiv g' \mod x^n \), fostering a deeper understanding of the interplay between algebraic structures.