Question

If \(L\) is ring-finite over \(K(K, L\) fields) then \(L\) is a finitely generated field extension of \(K\).

Short answer

Solution: Yes, \(L\) is a finitely generated field extension of \(K\). This is proved by the following steps: 1. Since \(L\) is a field, it is also an integral domain. 2. By the Noether Normalization Lemma, we can find a subset \(\{x_1, \dots, x_n\}\) of \(L\) which is algebraically independent over \(K\) and generates \(L\) as a module over the polynomial ring \(K[x_1, \dots, x_n]\). 3. The subset \(\{x_1, \dots, x_n\}\) forms a transcendence basis for the field extension \(L\) over \(K\). 4. Since \(L\) is an algebraic field extension of \(K(x_1, \dots, x_n)\) and the subset \(\{x_1, \dots, x_n\}\) forms a transcendence basis, we can conclude that \(L\) is a finitely generated field extension of \(K\).

Step-by-step solution

Show that \(L\) is an integral domain

Since \(L\) is a field, it is also an integral domain.

Apply Noether Normalization Lemma

By the Noether Normalization Lemma, there exists a subset \(\{x_1, \dots, x_n\}\) of \(L\) which is algebraically independent over \(K\) and such that \(L\) is finitely generated as a module over the polynomial ring \(K[x_1, \dots, x_n]\) (i.e., \(L\) is integral over \(K[x_1, \dots, x_n]\)).

Show that the subset is algebraically dependent over \(K\) and forms a transcendence basis

As \(L\) is integral over \(K[x_1, \dots, x_n]\), for each element \(y \in L\), there exists a monic polynomial \(f(T) \in K[x_1, \dots, x_n][T]\) with \(f(y)=0\). Since \(K[x_1, \dots, x_n]\) is an integral domain and \(f(T)\) is monic, this implies that each element of \(L\) is algebraic over the field of fractions of the polynomial ring \(K[x_1, \dots, x_n]\). Therefore, we can consider the field extension \(K(x_1, \dots, x_n) \subseteq L\). We can rewrite the polynomial \(f(T)\) as \(f'(T) \in K(x_1, \dots, x_n)[T]\) with \(f'(y) = 0\), which implies that \(L\) is an algebraic field extension of \(K(x_1, \dots, x_n)\). Hence, the subset \(\{x_1, \dots, x_n\}\) forms a transcendence basis for the field extension \(L\) over \(K\).

Prove that \(L\) is a finitely generated field extension of \(K\)

Since \(L\) is an algebraic field extension of \(K(x_1, \dots, x_n)\) and the subset \(\{x_1, \dots, x_n\}\) forms a transcendence basis for the field extension \(L\) over \(K\), we can conclude that \(L\) is a finitely generated field extension of \(K\).

Key concepts

Integral Domain

An integral domain represents a particular type of ring that is non-trivial, meaning it contains at least two distinct elements (usually 0 and 1), and has no divisors of zero. This absence of zero divisors means that if the product of two elements is zero, then at least one of the elements must be zero.

Within the context of field extensions, the idea of an integral domain becomes crucial. A field, by definition, is an integral domain with the additional property that every non-zero element has a multiplicative inverse. Thus, every field is an integral domain, but not every integral domain is a field. For instance, in the given exercise, since we're working with fields, we automatically avoid zero divisors, which simplifies much of our reasoning when considering extensions and algebraic dependencies.

Noether Normalization Lemma

The Noether Normalization Lemma is a foundational result in commutative algebra that facilitates the study of rings by establishing a connection to polynomial rings.

Key Takeaway from the Lemma:

Given an integral domain that is finitely generated as an algebra over a field, the lemma guarantees the existence of a set of algebraically independent elements over that field. These elements can be thought of as variables in a polynomial ring such that the original algebra is integral over this ring.

This lemma is a critical step in the solution to the exercise because it allows us to find a subset within the field extension that is algebraically independent over the base field, creating a structured framework within which we can understand the nature of the extension.

Algebraically Independent

The concept of algebraic independence is analogous to the idea of linear independence in vector spaces, but applied to elements of field extensions with respect to polynomial relationships.

Elements are algebraically independent if there's no non-zero polynomial with coefficients from the base field such that these elements, when substituted for the variables, result in zero. If such a polynomial doesn't exist, we can't 'solve' for one element in terms of the others just like we can't express a vector in terms of a linearly independent set.

In the exercise, applying the Noether Normalization Lemma helped us find a subset of algebraically independent elements that will eventually contribute to forming a transcendence basis, which is a pivotal step in proving that the field extension is finitely generated.

Transcendence Basis

A transcendence basis is a collection of algebraically independent elements in an extension field such that the field is algebraic over the field generated by these elements.

To illustrate, let’s think of the transcendence basis as a set of 'coordinates' that span an 'algebraic space' over the base field. Just as Cartesian coordinates can help us define the position of a point in Euclidean space, the transcendence basis helps to define the structure of a field extension.

The existence of a transcendence basis, as identified in the solution for our given exercise, shows that we can express any element of the field extension as an algebraic combination of this finite set and the base field, which is a major milestone in proving that the field extension is finitely generated.

Algebraic Field Extension

An algebraic field extension occurs when every element of the larger field is a root of some non-zero polynomial with coefficients in the smaller field.

This characteristic is fundamental when examining the finiteness of a field extension. For example, in our exercise, after establishing a transcendence basis, we can demonstrate that the original field extension is algebraic over the field generated by this basis.

With the transcendence basis accounting for every 'non-algebraic' aspect of the extension (since those elements are not roots of polynomials over the base field), the algebraic closing of the gap guarantees that all other elements are linked through algebraic relationships, cementing the field extension's status as finitely generated.