# Thread: Field Theory - Nicholson - Section 6.2 - Exercise 31

1. ## Field Theory - Nicholson - Section 6.2 - Exercise 31

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

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

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

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

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

Can someone help me approach this problem.

Peter

2. ## Re: Field Theory - Nicholson - Section 6.2 - Exercise 31

Since $\displaystyle E$ is a field extension of $\displaystyle F$, there exists a function $\displaystyle f_0(x) \in F[x]$ such that $\displaystyle f_0(u) = 0$. Let $\displaystyle f_1(x) = f_0(x)+1$. Let $\displaystyle f_u(x) = x$. Let $\displaystyle f_k(x) = kf_1(x)$ for any $\displaystyle k \in F$. Essentially, what I am trying to do is find all elements of the form a+bu. I would think this is the way to start.

3. ## Re: Field Theory - Nicholson - Section 6.2 - Exercise 31

Thanks SlipEternal

One quick question:

You write: "Since E is a field extension of F, there exists a function $\displaystyle f_0(x) \in F[x]$ such that $\displaystyle f_0(u) = 0$."

But $\displaystyle u \in E$ is transcendental over F, so doesn't this mean that there is no polynomial $\displaystyle f_0(x)$ in F[x] such that $\displaystyle f_0(u) = 0$.

Can you clarify?

Thanks.

Peter

4. ## Re: Field Theory - Nicholson - Section 6.2 - Exercise 31

Right, I am getting tired. This should be much easier than that. So $\displaystyle f_{a+bu}(x) = a+bx$ for any $\displaystyle a,b \in F$. Now, $\displaystyle f_{a+bu}(u) = a+bu$. Now through products of functions, you should be able to get any element of $\displaystyle F(u)$