$\mathbb F_q(x)$, the field of rational functions over a finite field, is not finitely generated as a ring.

Question

I try to prove it by contradiction. If the claim is incorrect, then there exist some $f_i=p_i/q_i$, where $p_i,q_i\in \mathbb F_q[x]$, such that $$ \mathbb F_q(x)=\mathbb F_q[f_1,\cdots, f_k] $$

So $x$ is a combination of $f_i$'s. And, I don't know how to continue from this. Please help.

Answer 1

By Zariski's lemma, since the field $\mathbb F_q(x)$ is assumed to be finitely generated as $\Bbb F_q$-algebra, it follows that $\mathbb F_q(x)=\mathbb F_q[f_1,\cdots, f_k]$ is a finite extension of $\Bbb F_q$.

Therefore, the rational fractions $f_i$ are algebraic over $\Bbb F_q$, so that they are actually constants. In conclusion, we would have $\mathbb F_q(x)=\mathbb F_q$, which is clearly not possible (for instance they don't have the same dimension as $\Bbb F_q$-vector spaces, see this).

Answer 2

This is a consequence of the fact that there are infinitely many irreducible polynomials in the UFD $\Bbb{F}_q[x]$. All the elements of $\Bbb{F}_q[f_1,f_2,\ldots,f_k]$ are of the form $r(x)/s(x)$ with $r(x),s(x)\in\Bbb{F}_q[x]$ such that the irreducible factors of the denominator $s(x)$ are those of the denominators $q_i$ of the generators $f_i$.

So if we select an irreducible polynomial $s(x)$ that is not a factor of any of the polynomials $q_i$, then $1/s(x)\notin\Bbb{F}_q[f_1,\ldots,f_k]$. The claim follows.