Question

Show that \(L=K(X)\) (the field of rational functions in one variable) is a finitely generated field extension of \(K\), but \(L\) is not ring-finite over \(K\). (Hint: If \(L\) were ringfinite over \(K\), a common denominator of ring generators would be an element \(b \in\) \(K[X]\) such that for all \(z \in L, b^{n} z \in K[X]\) for some \(n\); but let \(z=1 / c\), where \(c\) doesn't divide \(b\) (Problem 1.5).)

Short answer

Question: Show that the field of rational functions in one variable, \(L=K(X)\), is a finitely generated field extension of \(K\), but \(L\) is not ring-finite over \(K\). Answer: \(L=K(X)\) is a finitely generated field extension of \(K\) with a single generator \(x\); however, it is not ring-finite over \(K\) because assuming a common denominator for ring generators leads to a contradiction.

Step-by-step solution

Show that \(L=K(X)\) is a finitely generated field extension of \(K\).

The field of rational functions \(L=K(X)\) consists of all fractions of the form \(f(X)/g(X)\), where \(f(X)\) and \(g(X)\) are polynomials with coefficients in \(K\) and \(g(X)\neq 0\). \(L\) can be written as a simple field extension \(K(X)=K(x)\), where \(x\) is an indeterminate over \(K\). Then, every element in \(L\) can be expressed as a \(K\)-linear combination of powers of \(x\). This implies that \(L\) is a finitely generated field extension of \(K\) with a single generator \(x\).

Assume \(L\) is ring-finite over \(K\) and derive properties of common denominator

Now, let's assume that \(L\) is ring-finite over \(K\). That means that there exists a set of finitely many ring generators \(z_1, z_2, ..., z_n \in L\) that generate \(L\) as a \(K\)-algebra. Let's also assume there exists a common denominator \(b \in K[X]\) such that for all \(z \in L\), we have \(b^n z \in K[X]\) for some positive integer \(n\).

Using the hint and Problem 1.5 to show that the conditions cannot hold

Let \(z = 1/c\), where \(c \in K[X]\) and \(c\) doesn't divide \(b\). According to Problem 1.5, we can always find such a \(c\). Now, for our \(z\), we have: \(b^n \frac{1}{c} \in K[X]\). Multiplying both sides by \(c^n\), we get: \(b^n \in K[X]c^n\) Since \(c\) doesn't divide \(b\), the left-hand side is not divisible by \(c\). However, the right-hand side is divisible by \(c\), which is a contradiction. So, our assumption that a common denominator \(b\) exists is false.

Conclusion: \(L\) is not ring-finite over \(K\)

We've shown that \(L\) is a finitely generated field extension of \(K\) with generator \(x\). However, we have also shown, using the hint and Problem 1.5, that \(L\) cannot be ring-finite over \(K\), as it would require a common denominator that leads to a contradiction. Therefore, \(L\) is not ring-finite over \(K\).

Key concepts

Rational Functions

Rational functions are expressions that can be written as the quotient of two polynomials. Given two polynomials, say \( f(X) \) and \( g(X) \), a rational function is of the form \( \frac{f(X)}{g(X)} \). Here, \( g(X) \), the denominator, should not be zero to avoid undefined expressions.

These functions extend the idea of fractions from numbers to algebraic expressions. If \( f(X) \) and \( g(X) \) have coefficients in a field \( K \), together they describe elements of the field of rational functions over \( K \), often symbolized as \( K(X) \).

An example of a rational function is \( \frac{1}{X} \). Though expressions may look like simple fractions, their behavior and properties resemble that of polynomials mixed with proportional properties of divisions. These functions are crucial for studying behaviors of curves, especially in algebraic geometry.

Finitely Generated Fields

A field is said to be finitely generated if there is a finite set of elements within the field from which every other element of the field can be derived using field operations (addition, subtraction, multiplication, and division).

In the context of our original problem, the statement \( L = K(X) \) indicates that \( K(X) \), the field of rational functions, is a field extension of \( K \). This is a simple extension created by adding just one new element—typically denoted as \( X \) or \( x \).

This simple extension implies that \( L \) is a finitely generated field because every rational function in \( K(X) \) can be expressed using the single element \( X \) and the operation rules of \( K \). The finiteness comes from the finite generation, which begins with basic field elements much like building blocks.

Understanding finitely generated extensions is vital for exploring more complex field structures and algebraic concepts.

K-Algebra

A \( K \)-algebra is a ring that includes the field \( K \) as a subring, enriched with the properties of vector spaces over \( K \). An important concept in field theory, \( K \)-algebras allow us to explore how fields can combine with other algebraic structures.

In our exercise, we consider \( L = K(X) \) as a \( K \)-algebra because \( L \) is a field containing \( K \) and is closed under the operations that make it both a vector space and a ring. This closely ties to how algebraic structures can interact within this context.

Specifically, the K-algebra \( L \) is called a field extension, and it's important to note that not all K-algebras are field extensions. The exercise distinguishes between being finitely generated and being ring-finite. In this context, being finitely generated refers to how the field (or algebra) can be built from a few elements, while being ring-finite involves more specific constraints on the structure, like having finite dimensions as a vector space, which is not the case for \( L \).

This distinction helps in understanding different types of algebraic relationships and how one can classify various algebraic objects based on generation rules and constraints.