Field Theory - Nicholson - Section 6.2 - Exercise 31

[Math Amateur]

In Section 6.2 of Nicholson: Introduction to Abstract Algebra, Exercise 31 reads as follows:

Let $$E \supseteq F$$ be fields and let $$u \in E$$ be transcendental over F.

(a) Show that $$F(u) = \{ f(u){g(u)}^{-1} \ | \ f,g \in F[x] ; g(x) \ne 0 \}$$

(b) Show that $$F(u) \cong F(x)$$ where F(x) is the field of quotients of the integral domain F[x].

(c) Show that every element $$w \in F(u), w \notin F$$, is transcendental over F.

Can someone help me approach this problem.

Peter

[This has also been posted on MHF]

 

[jvicens]

Hi Peter,

Sure, I'd be happy to help you approach this problem. Let's break it down step by step:

(a) To show that F(u) = \{ f(u){g(u)}^{-1} \ | \ f,g \in F[x] ; g(x) \ne 0 \} , we need to show that every element in F(u) can be written in the form f(u){g(u)}^{-1} for some f,g \in F[x], g(x) \ne 0, and that every such expression is in F(u).

First, let's consider an arbitrary element in F(u). By definition, this element can be written as a quotient of two polynomials in u with coefficients in F. So, let's say we have an element a/b \in F(u), where a,b \in F[x] and b \ne 0. We can rewrite this as a{b}^{-1}, where b^{-1} is the inverse of the polynomial b. Since a and b are both polynomials in u with coefficients in F, a/b can be written as f(u){g(u)}^{-1} for some f,g \in F[x] and g(x) \ne 0.

Next, we need to show that every expression in the form f(u){g(u)}^{-1} is in F(u). This is true because both f(u) and g(u) are polynomials in u with coefficients in F, and we know that F(u) is the field of fractions of F[x]. This means that any expression of the form f(u)/g(u) is in F(u), and we can rewrite this as f(u){g(u)}^{-1}.

Therefore, we have shown that every element in F(u) can be written in the form f(u){g(u)}^{-1} for some f,g \in F[x], g(x) \ne 0, and every such expression is in F(u). So, we can conclude that F(u) = \{ f(u){g(u)}^{-1} \ | \ f,g \in F[x] ; g(x) \ne 0 \} .

(b) To show that F(u) \cong F(x), we need to find an isomorphism between these two fields. An isomorphism is a bijective homomorphism, which means that it is a function that