Let $F$ ba an extension extension of the field $K$. An element $a$ in $F$ is said to be algebraic if $a$ is the root of some non-zero polynomial $f \in K[x]$.

Using only the definition, how do we prove that for a field of rational functions, $K(x_1, x_2, ... x_n)$, over a field $K$,

(a) $x_i$ is transcendental over K

(b) Every non-constant rational function is transcendental over K

  • 4
    $\begingroup$ For (a) you can use the fact that the powers of $x_i$ are linearly independent, which can be checked in your favorite concrete construction of what the polynomial and rational functions are. (I am familiar with one in which you take a big cartesian product and call the indeterminate $x$ to be one of the unit vectors.) For (b) same idea except you need to clear the denominator. Then you get a polynomial in $x_1$ up to $x_n$ being $0$. Again, use linear independence of the monomials. $\endgroup$ – Jeff Dec 21 '13 at 8:01
  • $\begingroup$ If $ X_{i} $ were algebraic over $ K $, then $ \displaystyle \sum_{k = 0}^{m} a_{k} X_{i}^{k} = 0_{K(X_{1},\ldots,X_{n})} $ for some $ m \in \mathbb{N} $ and scalars $ a_{0},\ldots,a_{m} $, where $ a_{m} \neq 0_{K} $. However, this is not possible because $ \displaystyle \sum_{k = 0}^{m} a_{k} X_{i}^{k} $ is a non-zero polynomial. $\endgroup$ – Berrick Caleb Fillmore Dec 21 '13 at 8:14
  • 1
    $\begingroup$ What is your definition of "a field of rational functions", or even a polynomial ring? The exact form of a full argument would depend on the definition. $\endgroup$ – ronno Dec 21 '13 at 8:23

If a fraction $\frac{p}{q}$ were algebraic over $K$, then there would exist a polynomial $f\in K[T]$ of degree $n>0$ such that $f(\frac pq)=0$. Then $q^nf(\frac pq)$ is a polynomial in $K[x_1,...,x_n]$ that evaluates to 0. So it has to be $0$. Conclude from considering the highest degree components this is only possible if $f=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.