Field of rational functions

Remarks

Example: $\frac{x^2 - 1}{x - 1} = \frac{x + 1}{1}$ in $\Q(x)$, since $(x^2 - 1) \cdot 1 = (x + 1)(x - 1)$. Non-example: no element $p/q$ of $\Q(x)$ satisfies $(p/q)^2 = x$, since $p^2 = x q^2$ would give $2 \deg p = 1 + 2 \deg q$, which is impossible. So $K(x) \supset K$ is a field extension in which $x$ is transcendental over $K$. An element of $K(x)$ is a formal fraction, not a function on $K$: over $\F_2$, the fraction $\frac{1}{x^2 + x}$ is a nonzero element, although $x^2 + x$ vanishes at every point of $\F_2$. Over $\R$ or $\C$, the elements correspond to rational functions in the sense of analysis.