Problem 5
Question
If \(A\) is an integral domain, we have seen that in \(A[x]\), $$ \operatorname{deg} a(x) b(x)=\operatorname{deg} a(x)+\operatorname{deg} b(x) $$ Show that if \(A\) is not an integral domain, we can always find polynomials \(a(x)\) and \(b(x)\) such that \(\operatorname{deg} a(x) b(x)<\operatorname{deg} a(x)+\operatorname{deg} b(x)\)
Step-by-Step Solution
Verified Answer
If \(A\) has zero-divisors, polynomials can produce a zero product with lower degree.
1Step 1: Understand the Definition of Integral Domain
An integral domain is a commutative ring with unity and no zero-divisors. This means for any non-zero elements \(a\) and \(b\) in \(A\), \(a \cdot b eq 0\). The absence of zero-divisors ensures that the product of non-zero polynomials does not lose degree.
2Step 2: Consider a Non-Integral Domain
In a non-integral domain \(A\), there are zero-divisors. This means there exist non-zero elements \(c\) and \(d\) in \(A\) such that \(c \cdot d = 0\).
3Step 3: Construct Polynomials in \(A[x]\)
Let's consider two polynomials \(a(x) = cx\) and \(b(x) = dx\) where \(c, d \in A\) and are non-zero, but \(c \cdot d = 0\). The degree of \(a(x)\) is 1 and the degree of \(b(x)\) is 1.
4Step 4: Calculate the Product of the Polynomials
The product is given by \(a(x) \cdot b(x) = (cx) \cdot (dx) = (c \cdot d)x^2 = 0 \cdot x^2 = 0.\) Therefore, the degree of \(a(x) \cdot b(x)\) is \(-\infty\) when considering the zero polynomial has undefined degree, typically treated as \(-\infty\).
5Step 5: Compare the Degrees
Since \(a(x)\) and \(b(x)\) each have degree 1, their sum is \(1 + 1 = 2\). However, the degree of their product is \(-\infty\) because the product is the zero polynomial. Hence, \(\deg(a(x) \cdot b(x)) < \deg(a(x)) + \deg(b(x))\) when zero-divisors exist in \(A\).
Key Concepts
PolynomialsZero-DivisorsCommutative RingDegree of Polynomials
Polynomials
Polynomials play a fundamental role in algebra and mathematics in general. They are expressions composed of variables and coefficients involving operations of addition, subtraction, and multiplication. Imagine them as functions where you input a number for the variable, and the polynomial gives you an output based on that number.
- Form: A polynomial is of the form \(a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\), where \(a_n, a_{n-1}, \ldots, a_0\) are coefficients and \(n\) is a non-negative integer known as the degree of the polynomial.
- Operations: You can add, subtract, and multiply polynomials, and these operations follow the usual algebraic rules. It's important to realize, though, that the outcome of these operations depends on the coefficients, which can be part of various algebraic structures such as fields or rings.
Zero-Divisors
Zero-divisors are intriguing elements found within some algebraic structures, such as rings. In simple terms, a zero-divisor is a non-zero element that can multiply another non-zero element to yield zero as a product. This concept is critical when working with polynomials as it affects how they behave under multiplication.
- Example: Imagine integers modulo 4, for instance, where 2 multiplied by 2 results in 0. Here, 2 is a zero-divisor because it is not zero, yet its product with itself is zero.
- Impact: When zero-divisors appear in the coefficients of polynomials, usual expectations regarding the polynomial's degree can change. Specifically, the product of two polynomials might not have a degree equal to the sum of their individual degrees when zero-divisors are involved.
Commutative Ring
A commutative ring is a foundational concept in algebra. It is a set equipped with two operations, usually called addition and multiplication, which mimic the familiar arithmetic operations.
- Basic Properties: In a commutative ring, the addition and multiplication of elements are commutative, meaning \(a + b = b + a\) and \(a \cdot b = b \cdot a\) for all elements \(a\) and \(b\) in the ring.
- Prominent Features: These rings have an additive identity (often 0) and a multiplicative identity (often 1). There are no divisors of zero in the case of integral domains, but in general commutative rings, zero-divisors may exist.
- Relevance: Commutative rings are vital in understanding polynomial behavior since polynomials themselves form a commutative ring when their coefficients come from a commutative ring.
Degree of Polynomials
The degree of a polynomial gives us valuable information about its behavior and characterizes many of its fundamental properties. It is defined as the highest power of the variable with a non-zero coefficient in the polynomial.
- Significance: The degree of polynomials dictates much about their shape and behavior, such as the number of potential solutions an equation involving that polynomial might have.
- Rules: In integral domains, the degree of the product of two polynomials is simply the sum of their degrees. However, in rings with zero-divisors, this rule may not apply, as seen when two non-zero polynomials multiply to produce a polynomial with degree -∞, which happens if their product is the zero polynomial.
- Special Cases: This highlights a distinction between integral domains and other types of rings. In a non-integral domain, the product’s degree might not align with expectations, demonstrating how zero-divisors can alter typical outcomes.
Other exercises in this chapter
Problem 4
Let \(A\) be an integral domain; prove the following: If \((x+1)^{2}=x^{2}+1\) in \(A[x]\), then \(A\) must have characteristic \(2 .\) If \((x+1)^{4}=x^{4}+1\)
View solution Problem 4
Show that the following is true in \(A[x]\) for any ring \(A:\) For any odd \(n\), (a) \(x+1\) is a factor of \(x^{n}+1\) (b) \(x+1\) is a factor of \(x^{n}+x^{
View solution Problem 5
Find an example of each of the following in \(\mathbb{Z}_{8}[x]:\) a divisor of zero, an invertible element, an idempotent element.
View solution Problem 5
Prove the following: In \(\mathbb{Z}_{3}[x], x+2\) is a factor of \(x^{m}+2\), for all \(m .\) In \(\mathbb{Z}_{n}[x]\), \(x+(n-1)\) is a factor of \(x^{m}+(n-1
View solution