# 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

 $\displaystyle\varrho=\frac{f(X)}{g(X)}$ (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

 $\displaystyle\varrho\,g(X)\!-\!f(X)=0,$ (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.

## References

• 1 B. L. van der Waerden: .  Siebte Auflage der Modernen Algebra.  Erster Teil.
— Springer-Verlag. Berlin, Heidelberg (1966).
Title non-constant element of rational function field NonconstantElementOfRationalFunctionField 2013-03-22 15:02:50 2013-03-22 15:02:50 pahio (2872) pahio (2872) 17 pahio (2872) Theorem msc 12F99 field of rational functions rational function field