Question

Let \(K\) be a field, \(L=K(X)\) the field of rational functions in one variable over \(K\). (a) Show that any element of \(L\) that is integral over \(K[X]\) is already in \(K[X] .\) (Hint: If \(z^{n}+a_{1} z^{n-1}+\cdots=0\), write \(z=F / G, F, G\) relatively prime. Then \(F^{n}+a_{1} F^{n-1} G+\cdots=0\) so \(G\) divides \(F\).) (b) Show that there is no nonzero element \(F \in K[X]\) such that for every \(z \in L, F^{n} z\) is integral over \(K[X]\) for some \(n>0 .\) (Hint: See Problem 1.44.)

Short answer

2. How do we clear the denominators in the equation involving z, F, and G? 3. How do we conclude that G must be a unit in K[X]? 4. What contradiction do we reach if we assume there exists an F that makes F^n z integral over K[X] for all z in L?

Step-by-step solution

(a) Proving any integral elements of \(L\) are in \(K[X]\)

We want to show that any element \(z\) of \(L\) that is integral over \(K[X]\) is already in \(K[X]\). Let's assume that \(z\) is integral over \(K[X]\). According to the hint, we can represent \(z\) as the fraction \(F/G\), where \(F\) and \(G\) are relatively prime polynomials in \(K[X]\). Since \(z\) is integral over \(K[X]\), there exists a monic polynomial \(m(X) \in K[X]\) such that $$m(z) = z^n + a_1 z^{n-1} + \cdots + a_n=0$$ Since \(z=F/G\), we can substitute this into the equation to get $$\left(\frac{F}{G}\right)^n + a_1 \left(\frac{F}{G}\right)^{n-1} + \cdots + a_n=0$$ Now, we want to clear the denominators. To do that, we can multiply both sides of the equation by \(G^n\): $$F^n + a_1F^{n-1}G + \cdots + a_nG^n=0$$ The left side of the equation is a polynomial in \(F\) and \(G\). Now, observe that \(G\) divides all terms except the first one. This means that \(G\) must divide \(F^n\). Since \(F\) and \(G\) are relatively prime, \(G\) must be a unit in \(K[X]\), which essentially means that \(G\) is a nonzero constant from the field \(K\). Thus, we can write \(z=F/G\) as a single polynomial in \(K[X]\), and so any element of \(L\) that is integral over \(K[X]\) is already in \(K[X]\).

(b) Proving there is no such \(F\) that makes \(F^n z\) integral over \(K[X]\)

We want to show that there is no nonzero element \(F \in K[X]\) such that for every \(z \in L\), \(F^n z\) is integral over \(K[X]\) for some \(n > 0\). To do this, let's assume there is such an element \(F\) and try to reach a contradiction. Let \(z = X\) and let \(A = F^n X\), where A is integral over \(K[X]\). So, there exists a monic polynomial \(m_A(X) \in K[X]\) such that $$m_A(A) = A^m + a_1 A^{m-1} + \cdots + a_m = 0$$ Substituting \(A = F^n X\), we get: $$\left(F^n X\right)^m + a_1 \left(F^n X\right)^{m-1} + \cdots + a_m = 0$$ Now, observe that each term on the left side of this equation has at least an \(X\) term. When we add these terms together, there must still be an \(X\) term since we cannot cancel any terms. However, the right side of the equation is \(0\), which has no \(X\) term. This is a contradiction, and therefore, we conclude there is no such \(F\) that makes \(F^n z\) integral over \(K[X]\) for all \(z \in L\).

Key concepts

Integral Extensions

Integral extensions in algebraic geometry refer to elements of a larger ring or field that satisfy a polynomial equation with coefficients in a smaller ring or field. For example, consider a ring extension where we have two rings, say ring A and ring B, with ring A being a subring of ring B. An element from ring B is said to be integral over ring A if it satisfies a polynomial equation

• with coefficients in ring A
• and the leading coefficient equal to 1 (monic polynomial).

Integral elements are significant because they exhibit a notion of 'closure' under algebraic operations, maintaining the relationships set by polynomial equations.
In the context of the exercise, a rational function in the field of rational functions over a field is integral over the polynomial ring if it can be expressed as a polynomial in that ring. This implies that if a rational function is integral over the polynomial ring, it is just a polynomial itself.

Polynomial Rings

Polynomial rings are foundational structures in algebraic geometry and abstract algebra. A polynomial ring, typically denoted as \( K[X] \), consists of polynomials where the coefficients come from a field or a ring \( K \) and \( X \) represents a variable. These rings form the building blocks in algebra as they allow us to study and construct more complex algebraic structures.

• The elements of a polynomial ring can be added or multiplied following the familiar rules of polynomial arithmetic.
• They are characterized by their ability to serve as the domain for rational functions and to define curves and shapes in algebraic geometry.

In algebra, polynomial rings enable the creation of other mathematical entities such as ideals. They are also crucial when considering the notion of integrality because any polynomial equation, used in determining integrality, experiences its coefficients originating from a polynomial ring.
The exercise explores polynomial rings as it analyzes the nature of elements in the field of rational functions being integral over \( K[X] \), which constitutes a polynomial ring.

Fields in Algebra

In algebra, a field is a set equipped with two operations that abide by the laws similar to arithmetic operations on numbers. Fields serve as the essential playing ground for algebraic tricks, allowing addition, subtraction, multiplication, and division (except by zero), much like the familiar set of rational numbers.

• Fields form the basic structural underpinning of many algebraic entities.
• Each element in a field has an inverse for both addition and multiplication, essentially forming a very symmetric system.

A field, with its characteristic properties, allows us to understand more advanced structures such as vector spaces and algebraic extensions.
In algebraic geometry and in these exercises, fields provide the coefficients for rings, including polynomial rings, and host rational functions. The field \( K \) plays a central role in understanding the elements that are integral over polynomial structures like \( K[X] \). It ensures that every rational function can interact seamlessly with polynomial rings when analyzing integrality in algebra.

Rational Functions

Rational functions are a key concept in algebra that involve ratios of polynomials. A rational function can generally be expressed as \( \frac{F(X)}{G(X)} \), where \( F(X) \) and \( G(X) \) are polynomials and \( G(X) \) is not zero. Essentially, these functions extend polynomial functions by enabling division.

• Rational functions are highly adapted for describing and working with mappings such as curves and surfaces in algebraic geometry.
• They provide a broader toolkit for solving equations and modeling complex algebraic structures based on polynomial ratios.

In the context of integral extensions, rational functions are examined to determine whether they are simply polynomials when confined to satisfying certain polynomial equations with coefficients from a polynomial ring.
This exercise demonstrates such an approach by analyzing whether rational functions in a particular field can still be seen as polynomials within a designated polynomial ring when considered for their integrality over the ring.