If $F$ is a field, $E$ an extension field of $F$, and $u \in E$ transcendental over $F$, then $F(u) = \{f(u)g(u)^{-1}|f,g \in F[x]; g \neq 0\}$

Question

My textbook states:

If $F$ is a field, $E$ an extension field, and $u \in E$ transcendental over $F$

then $F(u) = \{f(u)g(u)^{-1}|f,g \in F[x]; g \neq 0\}$, where $F(u) = \cap \{ F \subset K|$ $K$ is a subfield of $E$ containing $u$ $\}$.

Take $x \in F(u)$, then it must be that for some polynomials $f,g \in F[x]$, we have $x = f(u)g(u)^{-1}$. Not sure how to find such polynomials.

In the opposite inclusion take any $f,g$ in $F[x]$, then $f(u)g(u)$ is contained in $F(u)$ since it is the evaluation of a polynomial in $F(u)[x]$.

Clarifications would be helpful.

Answer 1

Let $L= \{f(u)g(u)^{-1}|f,g \in F[x]; g \neq 0\}$ and $F(u) = \bigcap \{ K \supset F\,\mid \text{$K$ is a subfield of $E$ containing $u$} \}$.

Note that $L$ contains $F$ and $u$ (hint: constant map and identity map). Also it is easy to prove that $L$ is a field and is contained inside $E$. So $F(u)\subseteq L$, by definition.

Now take any element of $L$. It will be of the form $$f(u)g(u)^{-1}=\frac{f(u)}{g(u)}=\frac{a_0+a_1u+\cdots+a_nu^n}{b_0+b_1u+\cdots+b_mu^m},$$ where $a_i$'s and $b_j$'s are elements of $F$, and $g(u)\neq 0$.

Let $K$ be a subfield of $E$ containing $u$ and $F$. It is easy to see that $(a_0+a_1u+\cdots+a_nu^n),\,(b_0+b_1u+\cdots+b_mu^m)\in K$. Since $b_0+b_1u+\cdots+b_mu^m\neq 0$, it has an inverse in $K$ and hence $f(u)g(u)^{-1}\in K$. Since $K$ was arbitrary, $f(u)g(u)^{-1}\in \bigcap \{ K \supset F\,\mid \text{$K$ is a subfield of $E$ containing $u$} \}=F(u)$. So $L\subseteq F(u)$.