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[parent] non-constant element of rational function field (Theorem)

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 1   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

$\displaystyle h(Y) = \varrho\,g(Y)\!-\!f(Y)$
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
$\displaystyle (K(X):K(\varrho)) = n,$
Q.E.D.

Bibliography

1
B. L. van der Waerden: Algebra. Siebte Auflage der Modernen Algebra. Erster Teil.
-- Springer-Verlag. Berlin, Heidelberg (1966).



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Other names:  field of rational functions, rational function field

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Cross-references: conclusion, relatively prime, factors, irreducible, mean, power, ring, coefficients, equation, proof, algebraic, base field, transcendental, denominator, numerator, degrees, lowest terms, reduced, polynomials, rational functions, indeterminate, polynomial ring, field of fractions, extension, field extension, field
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This is version 14 of non-constant element of rational function field, born on 2005-02-16, modified 2005-08-26.
Object id is 6762, canonical name is NonConstantElementOfRationalFunctionField.
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Classification:
AMS MSC12F99 (Field theory and polynomials :: Field extensions :: Miscellaneous)

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