Question

Consider a field \(F\) and its field of rational functions \(F(X) .\) Let \(\alpha \in F(X) \backslash F\). Show that \(X\) is algebraic over \(F(\alpha),\) and that \(\alpha\) is transcendental over \(F\)

Short answer

Question: Show that for any element α in F(X) that is not in F, X is algebraic over F(α), and α is transcendental over F. Answer: To show that X is algebraic over F(α), we can consider the rational function α = p(X)/q(X) and the polynomial P(x) = xq(X) - p(X) in F(α)[X]. Evaluating P(x) at X gives P(X) = 0, so X is algebraic over F(α). To show that α is transcendental over F, we can assume the contrary and suppose that there exists a non-zero polynomial R(x) ∈ F[X] such that R(α) = 0. Clearing the denominators by multiplying both sides with a power of q(X) results in a polynomial equation with F[X] coefficients. This would mean that p(X)/q(X) is a root of this polynomial, which contradicts the fact that p(X) and q(X) have no common factors. Thus, α is transcendental over F.

Step-by-step solution

Define algebraic and transcendental elements.

An element γ is algebraic over a field E if there exists a non-zero polynomial P(x) with coefficients in E such that P(γ) = 0. If no such polynomial exists, γ is called transcendental over E.

Define the field of rational functions F(X).

The field of rational functions F(X) is the field of fractions of the polynomial ring F[X]. In other words, it is the set of all rational expressions of the form \(\frac{p(X)}{q(X)}\), where p(X) and q(X) are polynomials with coefficients in F and q(X) is not the zero polynomial.

Prove X is algebraic over F(α).

Consider the rational function α = \(\frac{p(X)}{q(X)}\) with p(X), q(X) ∈ F[X] and q(X) ≠ 0. Then, we have p(X) - αq(X) = 0. Consider the polynomial P(x) = xq(X) - p(X) in F(α)[X]. Its coefficients are from F(α), and it is non-zero. Evaluating P(x) at X, we have P(X) = Xq(X) - p(X) = 0. Thus, X is algebraic over F(α).

Prove α is transcendental over F.

Suppose, for contradiction, α is algebraic over F. Then there exists a non-zero polynomial R(x) ∈ F[X] such that R(α) = 0. R(α) can be expressed as \(R\left(\frac{p(X)}{q(X)}\right) = 0\). Clearing the denominators by multiplying both sides with a power of q(X), we get a polynomial equation with F[X] coefficients. However, this would mean that p(X)/q(X) is a root of this polynomial, which contradicts the fact that p(X) and q(X) have no common factors, as they are elements of the field of fractions F(X). Thus, there is no such polynomial R(x) ∈ F[X], and α must be transcendental over F.

Key concepts

Algebraic elements

In mathematics, when working within the framework of field theory, understanding the concept of algebraic elements is crucial. An element is said to be algebraic over a field if it satisfies a particular type of polynomial equation.

Imagine, for instance, you have a field, let's call it \(E\), and an element \(\gamma\). This element \(\gamma\) is algebraic over \(E\) if and only if there exists a non-zero polynomial \(P(x)\) with coefficients in \(E\) for which \(P(\gamma) = 0\).

It's like finding a special polynomial key that unlocks the door to \(\gamma\). Finding such a polynomial implies that \(\gamma\) isn't "free" but is instead bound by this special relationship with the field.

• Polynomial P(x) is the "key" that shows the dependency of \(\gamma\) on field \(E\).
• If no such key exists, \(\gamma\) is not algebraic; it's something entirely else: transcendental.

In the problem at hand, the element \(X\) is shown to be algebraic over \(F(\alpha)\) because a polynomial \(P(x) = xq(X) - p(X)\) exists with coefficients in \(F(\alpha)\) such that \(P(X) = 0\). This underlines its algebraic nature within the specified field structure.

Transcendental elements

If algebraic elements are tied to their field by a polynomial bond, transcendental elements are the elusive creatures that defy such chains.

An element is considered transcendental over a field \(E\) if there is no non-zero polynomial \(P(x)\) with coefficients in \(E\) for which the element is a root. In simpler terms, no matter how hard you try, you can't find a polynomial with roots pointing back to this element.

Transcendental elements are like free spirits in the mathematical realm, unconstrained by algebraic expressions rooted in a given field.

• They remain independent of any polynomial relationship within their specified field.
• This freedom from algebraic constraints makes them uniquely interesting in mathematics.

In the context given, \(\alpha\) is highlighted as transcendental over \(F\). Attempting to assume otherwise leads us to paradoxes, ultimately re-emphasizing its transcendental status. There is no polynomial \(R(x) \in F[X]\) that makes \(\alpha\) a root, thus confirming its transcendental nature.

Rational functions

Rational functions are an essential part of field theory and algebra, stitching together polynomials to form expressions of greater complexity.

Essentially, a rational function is the ratio of two polynomials. You can think of it like a sophisticated fraction where both the numerator and the denominator are polynomials from a specified field. Specifically, within a field \(F\), the rational functions form the field \(F(X)\). This field includes all expressions of the form \(\frac{p(X)}{q(X)}\), where \(q(X) eq 0\).

This inclusion of all such expressions results in an expansive field, extending beyond simple polynomials.

• Rational functions allow algebraists to extend and use field operations on polynomial expressions efficiently.
• They are central to defining and exploring other key concepts such as algebraic and transcendental elements.

When \(\alpha\) in our problem is expressed as a rational function \(\frac{p(X)}{q(X)}\), it lives within the space \(F(X)\), positioning \(\alpha\) and \(F(X)\) as vital players in understanding both algebraic and transcendental considerations.