Example 2

Let $K$ be a field and $F$ the field $K(X)$ of rational functions in one indeterminate over $K$. Let $E$ be the subfield $K(X^3)$ of $F$; we have $F = E(X)$ and $X$ is algebraic over $E$, since it is a root of the polynomial $Y^3 - X^3$ of the ring $E[Y]$; this polynomial is irreducible in $E[Y]$, for in the contrary case it would have at least one factor of the first degree, and there would then exist two non-zero polynomials $u(X)$, $v(X)$ of $K[X]$ such that $(u(X^3))^3 = X^3(v(X^3))^3$ which is absurd, for if $m$ and $n$ are the degrees of $u$ and $v$, this would imply $9m = 9n + 3$, or $3m = 3n + 1$. The field $F$ is thus of degree 3 over $E$, and every element of $F$ may be written in just one way as a linear combination $f(X^3) + Xg(X^3) + X^2 h(X^3)$, where $f, g, h$ are three rational fractions of $K(X)$.