STATEMENT: Suppose that $F\subseteq E$ is a field extension of $F$. And assume $u\in E$ is transcendental over $E$. Then it readily follows that $F(u)\cong F(x)$, where $F(x)$ is the quotient field of $F[X]$, the ring of polynomials over $F$.
QUESTION: How do they arrive that the conclusion that $F(u)\cong F(x)$. Does it just stem from the fact that $F[u]\cong F[x]$ then their quotient fields must also be isomorphic?