Exercise 12, § 3

a) With the notation and hypotheses as in Ex. 11, show that for an extension $F$ of $K$ such that $K \subset F \subset K(x)$ to be such that $F \cap K[x] \neq K$, it is necessary and sufficient that $F = K(y)$ where $y = g(x)$ for a non-constant polynomial $g \in K[T]$; then we have $K(y) \cap K[x] = K[y]$. (Write $F = K(y)$ with $y = g(x)/h(x)$, $g$ and $h$ being relatively prime polynomials of $K[T]$. If there exists a non-constant polynomial $P \in K[T]$ such that $P(x) \in F$, we can write $P(x) = Q(y)/R(y)$, where $Q$ and $R$ are two relatively prime monic polynomials of $K[T]$, not both constant. Distinguish two cases, according as $R = 1$ or $R$ is non-constant; in the second case, decompose $Q$ and $R$ into factors of the first degree in an algebraic closure $\Omega$ of $K$ and observe that if $a, b$ are distinct elements of $\Omega$, then $g - ah$ and $g - bh$ are relatively prime in $\Omega[T]$.)

b) If $F = K(y)$ with $y = g(x)$, where $g \in K[T]$ is non-constant, we may suppose that $g(T) = Tg_1(T)$, where $g_1 \in K[T]$. Show that if $K(xg(x)) \cap K[y] \neq K$, we necessarily have $g_1(T) = aT^n$ with $a \in K$ and $n > 0$. (Observe that by hypothesis and a) there exist two non-constant polynomials $P, Q$ in $K[T]$ such that $P(xg(x)) = Q(g(x)) \notin K$; deduce that there are two elements $a, b$ in $K$ and an integer $m$ such that $g(T)$ divides $a - bT^m$.)

c) Let $y, z$ be two elements of $K[x]$ such that $K(y) \cap K(z) \neq K$; show that then $K[y] \cap K[z] \neq K$.

d) Find $y \in K(x)$ such that $K(y) = K(x)$ but $K[y] \cap K[x] = K$.