Problem 3
Question
If \(c\) is transcendental over \(F\), so are \(c+1, k c\) (where \(k \in F)\), and \(c^{2}\).
Step-by-Step Solution
Verified Answer
If \( c \) is transcendental over \( F \), so are \( c+1, kc, \) and \( c^2 \).
1Step 1: Understand Transcendental Numbers
A number \( c \) is transcendental over a field \( F \) if it is not a root of any non-zero polynomial with coefficients in \( F \). This means that no polynomial equation with coefficients from \( F \) evaluated at \( c \) equals zero.
2Step 2: Analyze \( c+1 \)
Suppose \( c+1 \) is algebraic over \( F \). Then there exists a polynomial \( f(x) \) with coefficients in \( F \) such that \( f(c+1) = 0 \). However, by substituting \( x = c+1 \), we would also have a polynomial where \( c \) is a root, contradicting \( c \)'s transcendental nature. Thus, \( c+1 \) must also be transcendental.
3Step 3: Consider \( k c \)
Suppose \( kc \) is algebraic over \( F \). There would be a non-zero polynomial \( g(x) \) with coefficients from \( F \) such that \( g(kc) = 0 \). Since \( k \in F \), this would imply that \( c \) is a root of another polynomial, contradicting its transcendental nature. Therefore, \( k c \) is also transcendental.
4Step 4: Evaluate \( c^2 \)
Assume \( c^2 \) is algebraic over \( F \). Then a polynomial exists with coefficients in \( F \) such that \( h(c^2) = 0 \). This would imply \( c \) is a root of another polynomial, which is impossible if \( c \) is transcendental. Thus, \( c^2 \) is transcendental.
Key Concepts
Field TheoryAlgebraic NumbersPolynomial Equations
Field Theory
Field Theory provides the foundation for understanding various mathematical structures, especially in algebra. A "field" is a set equipped with two operations typically called addition and multiplication, subject to certain conditions.
These include the existence of additive and multiplicative identities and inverses.
Fields are a crucial concept in abstract algebra because they extend many arithmetic properties from familiar number systems, like rational numbers and real numbers.
These include the existence of additive and multiplicative identities and inverses.
Fields are a crucial concept in abstract algebra because they extend many arithmetic properties from familiar number systems, like rational numbers and real numbers.
- **Additive Identity**: There exists a number 0 such that for any number a in the field, \( a + 0 = a \).
- **Multiplicative Identity**: There exists a number 1 such that for any number a in the field, \( a \times 1 = a \).
- **Inverses**: For every nonzero number a, there exists a number \( a^{-1} \) such that \( a \times a^{-1} = 1 \).
- **Commutative Properties**: Addition and multiplication are commutative, meaning for any numbers a and b, \( a + b = b + a \) and \( a \times b = b \times a \).
Algebraic Numbers
Algebraic Numbers are among the central concepts of algebra. A number is deemed algebraic over a field if there exists a non-zero polynomial with coefficients within that field for which the number is a root. Essentially, if a number is algebraic, it can "solve" a polynomial equation with those coefficients.
For example, consider the square root of 2, denoted as \(\sqrt{2}\). This is an algebraic number over the rational number field \(\mathbb{Q}\) because it is a root of the polynomial \(x^2 - 2 = 0\). Here’s how algebraic numbers are characterized:
For example, consider the square root of 2, denoted as \(\sqrt{2}\). This is an algebraic number over the rational number field \(\mathbb{Q}\) because it is a root of the polynomial \(x^2 - 2 = 0\). Here’s how algebraic numbers are characterized:
- Numbers like integers, rational numbers, and certain roots are often algebraic over the field of rational numbers \(\mathbb{Q}\).
- Algebraic numbers provide the building blocks for more complex number sets, such as algebraic closures, which are fields where every non-zero polynomial has a root.
- However, not every number is algebraic. Numbers that do not satisfy any such polynomial equation (like \(\pi\) or \(e\)) are termed 'transcendental'.
Polynomial Equations
Polynomial Equations are mathematical expressions that involve sums of powers of a variable, combined using coefficients from a given field. They play a crucial role in algebra, serving as the foundational components for describing relationships and solving problems.
Formally, a polynomial over a field \( F \) is an expression of the form:\[ a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \]where each coefficient \( a_i \) belongs to the field \( F \).
Polynomial equations can be classified based on their degree, which is the highest power of the variable in the expression. The degree gives insight into the behavior and properties of the polynomial, such as the number of possible roots:
Formally, a polynomial over a field \( F \) is an expression of the form:\[ a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \]where each coefficient \( a_i \) belongs to the field \( F \).
Polynomial equations can be classified based on their degree, which is the highest power of the variable in the expression. The degree gives insight into the behavior and properties of the polynomial, such as the number of possible roots:
- **Linear Equations** have the form \( ax + b = 0 \) and generally have one root.
- **Quadratic Equations** contain terms up to \( x^2 \) and can have two roots. They follow \( ax^2 + bx + c = 0 \).
- **Higher-degree Polynomials** may have more complex behavior with potentially more roots, following the Fundamental Theorem of Algebra, which states that a polynomial of degree \( n \) in the complex numbers has exactly \( n \) roots.
Other exercises in this chapter
Problem 1
Prove that each of the following numbers is algebraic over \(\mathbb{Q}\) : (a) \(i\) (b) \(\sqrt{2}\) (c) \(2+3 i\) (d) \(\sqrt{1+\sqrt[3]{2}}\) (e) \(\sqrt{i-
View solution Problem 2
\([a(x) b(x)]^{\prime}=a^{\prime}(x) b(x)+a(x) b^{\prime}(x)\)
View solution Problem 3
Find the minimum polynomial of the following numbers over the indicated fields: $$ \begin{array}{ll} \sqrt{3}+i & \text { over } \mathbb{R} ; \text { over } \ma
View solution Problem 4
If \(F\) has characteristic 0 and \(a^{\prime}(x)=0\), then \(a(x)\) is a constant polynomial. Why is is conclusion not necessarily true if \(F\) has characteri
View solution