nonconstant element of rational function field
Let $K$ be a field. Every simple (http://planetmath.org/SimpleFieldExtension) transcendent field extension $K(\alpha )/K$ may be represented by the extension $K(X)/K$, where $K(X)$ is the field of fractions^{} of the polynomial ring^{} $K[X]$ in one indeterminate^{} $X$. The elements of $K(X)$ are rational functions, i.e. rational expressions
$\varrho ={\displaystyle \frac{f(X)}{g(X)}}$  (1) 
with $f(X)$ and $g(X)$ polynomials^{} in $K[X]$.
Theorem.
Let the nonconstant rational function (1) be reduced to lowest terms and let the greater of the degrees of its numerator and denominator be $n$. This element $\varrho $ is transcendental^{} with respect to the base field^{} $K$. The field extension $K(X)/K(\varrho )$ is algebraic and of degree $n$.
Proof. The element $X$ satisfies the equation
$\varrho g(X)f(X)=0,$  (2) 
the coefficients of which are in the field $K(\varrho )$, actually in the ring $K[\varrho ]$. If all these coefficients were zero, we could take one nonzero coefficient ${b}_{\nu}$ in $g(X)$ and the coefficient ${a}_{\nu}$ of the same power of $X$ in $f(X)$, and then we would have especially $\varrho {b}_{\nu}{a}_{\nu}=0$; this would mean that $\varrho =\frac{{a}_{\nu}}{{b}_{\nu}}$ = constant, contrary to the supposition. Thus at least one coefficient in (2) differs from zero, and we conclude that $X$ is algebraic with respect to $K(\varrho )$. If $K(\varrho )$ were algebraic with respect to $K$, then also $X$ should be algebraic with respect to $K$. This is not true, and therefore we see that $K(\varrho )$ is transcendental, Q.E.D.
Further, $X$ is a zero of the ${n}^{\mathrm{th}}$ degree polynomial
$$h(Y)=\varrho g(Y)f(Y)$$ 
of the ring $K(\varrho )[Y]$, actually of the ring $K[\varrho ][Y]$, i.e. of $K[\varrho $, Y]. The polynomial is irreducible in this ring, since otherwise it would have there two factors, and because $h(Y)$ is linear in $\varrho $, the other factor should depend only on $Y$; but there can not be such a factor, for the polynomials $f(Z)$ and $g(Z)$ are relatively prime. The conclusion is that $X$ is an algebraic element over $K(\varrho )$ of degree $n$ and therefore also
$$(K(X):K(\varrho ))=n,$$ 
Q.E.D.
References

1
B. L. van der Waerden: Algebra^{}. Siebte Auflage der Modernen Algebra. Erster Teil.
— SpringerVerlag. Berlin, Heidelberg (1966).
Title  nonconstant element of rational function field 

Canonical name  NonconstantElementOfRationalFunctionField 
Date of creation  20130322 15:02:50 
Last modified on  20130322 15:02:50 
Owner  pahio (2872) 
Last modified by  pahio (2872) 
Numerical id  17 
Author  pahio (2872) 
Entry type  Theorem 
Classification  msc 12F99 
Synonym  field of rational functions 
Synonym  rational function field 