non-constant 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 non-constant 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 non-zero 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.

Title	non-constant element of rational function field
Canonical name	NonconstantElementOfRationalFunctionField
Date of creation	2013-03-22 15:02:50
Last modified on	2013-03-22 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