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# non-constant element of rational function field

Let $K$ be a field. Every simple 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: Algebra. Siebte Auflage der Modernen Algebra. Erster Teil.

— Springer-Verlag. Berlin, Heidelberg (1966).

## Mathematics Subject Classification

12F99*no label found*

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## Recent Activity

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

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