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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

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    $\begingroup$ 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
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    $\begingroup$ 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
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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 $f=0$.

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