Question

Let \(L\) be a field, \(k\) an algebraically closed subfield of \(L\). (a) Show that any element of \(L\) that is algebraic over \(k\) is already in \(k\). (b) An algebraically closed field has no module-finite field extensions except itself. 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 \(K[X \mid\) for some \(n>0 .\) (Hint: See Problem 1.44.)

Short answer

Question: Prove that (a) any element of L that is algebraic over k is already in k, and (b) an algebraically closed field has no module-finite field extensions except itself, where L is a field and k is an algebraically closed subfield of L. Answer: (a) If x is an element of L that is algebraic over k, then there exists a nonzero polynomial f(X) in \(k[X]\) with f(x) = 0. Since k is algebraically closed, x must be a root of f(X) in k, and therefore, x is in k. (b) If L were a module-finite field extension over K, it would contain elements that are algebraic over K. However, every element in L is already in K, which implies L and K are the same field. Therefore, there are no module-finite field extensions of L except itself.

Step-by-step solution

Define an element algebraic over k

An element x in L is said to be algebraic over k if there exists a nonzero polynomial in the ring \(k[X]\) (polynomials with coefficients in k) such that f(x) = 0.

Show that algebraic element is in k

Let x be an element of L that is algebraic over k. Then, there exists a nonzero polynomial f(X) in \(k[X]\) such that f(x) = 0. Since k is algebraically closed, x must be a root of f(X) that is in the subfield k. Therefore, x is already in k. (b) An algebraically closed field has no module-finite field extensions except itself

Define module-finite field extensions

A field extension L/K is said to be module-finite if L can be expressed as a finitely generated K-module.

Show that every element is integral over K

Let \(z \in L\). Then there exists a nonzero polynomial F(X) in \(K[X]\) such that \(F^{n}(z)\) is integral over \(K[X]\) for some \(n > 0\). Since \(K[X]\) is algebraically closed, there exists an element of K that is a root of F(X), say x. Then, \(F^{n}(x)\) is integral over \(K[X]\) as well.

Showing the contradiction

If L were a module-finite field extension over K, then it would contain elements that are algebraic over K. However, we have just shown that every element in L is already in K. This implies that L and K are the same field, and there are no module-finite field extensions of L except itself. Hence, we have proven that any element of L that is algebraic over k is already in k (part a) and that an algebraically closed field has no module-finite field extensions except itself (part b).

Key concepts

Algebraic Element

When we dive into the world of algebra, there's a fundamental concept known as an algebraic element. This term pops up when we deal with elements of fields. Simply put, if we have a field, and you find an element in it that happens to be the solution of a polynomial equation with coefficients from a smaller subfield, you've found yourself an algebraic element.

Consider this like a scavenger hunt. You're given a polynomial equation like a riddle, and your mission is to find the numbers (or elements) that satisfy this riddle. If the element is within your smaller field (the one you got the coefficients from), it's like finding the treasure right under the starting point—it's algebraic!

For example, if we are dealing with the field of complex numbers, which is algebraically closed, any algebraic number found there would also be a complex number. This is due to the fundamental property of algebraically closed fields, they contain all the solutions (or roots) to polynomials that you can cook up using elements from within the field.

Module-Finite Field Extensions

The phrase module-finite field extension might sound like heavy-duty terminology, but it's an accessible idea once broken down. Think of a field extension as an extended family. It's a larger field that includes a smaller 'parent' field. Now, if the larger family can be managed by a few key members, it's considered module-finite—meaning it doesn't need an endless number of elements to describe it.

This concept becomes fascinating when you remark that an algebraically closed field—like that all-knowing grandparent—doesn't really allow for these vast but finitely managed extensions, except for itself. Why? Because if an element can be described as a finite combination of aspects from the grandparent field, it's not really extending the family; it's just revealing what's already been there.

Polynomial Equations

Crossing the bridge from the abstract to the applicable, we encounter polynomial equations. These mathematical expressions form the backbone of algebra and bring a structured way to understand the symphony of numbers. At their core, polynomial equations are like recipes that require specific ingredients—the coefficients—and ask you to find the missing secret spice—the variable that satisfies the equation.

Given a polynomial equation, the roots or solutions are the values that make the equation true when you replace the variable with them. It's a game of mix and match until everything balances out. And in an algebraically closed field, you will always find these roots within the field itself. No matter how complex the polynomial, the solutions are guaranteed to be 'in-house'.

Field Theory

Moving on to the grand umbrella that covers all these concepts, field theory, which is a rich and elegant tapestry within mathematics. This is where mathematicians explore various fields—no pun intended! These are sets equipped with two operations that seem familiar—they're like addition and multiplication, but these operations follow specific rules, allowing for division and subtraction to happen smoothly, minus the zero division debacle.

In this playground, we discuss the relationships between different fields. Discovering how they extend or sit within each other, much like our earlier family analogy, gives us incredible insights into the structure of our mathematical universe. Field theory also explains why certain equations have solutions and others don't, guiding us on how to navigate these numerical oceans with theoretical compasses.