Polynomial cannot be finitely generated (Elementary Algebraic Geometry by Hulek) I am reading the textbook Elementary Algebraic Geometry, by Hulek p. 23 
Or the google book: http://tinyurl.com/zjkjhvb

I cannot understand what does the author says in the part underlined.  
My understanding is if $k(t)$ is a prime polynomials, it cannot be finitely generated over $k$ by $r_i$; however, if $r_i=k(t)$, it can be finitely geberated by itself.  
I have no idea about this. Can anyone please explain this?
 A: Suppose $k(t)$ is finitely generated as a $k$-algebra by a collection of elements $r_i=p_i(t)/q_i(t)$ for $p_i(t),q_i(t)\in k[t]$ with $p_i(t)$ and $q_i(t)$ having no common factors. This means that every element $x\in k(t)$ may be written as a $k$-linear sum of products of the $r_i$, as follows:
$$x = \displaystyle\sum_{i=1}^{m} c_ir_1^{a_{i,1}}r_2^{a_{i,2}}\cdots r_n^{a_{i,n}}$$
where $a_i$ are non-negative integers and $c_i\in k$.
Pick a prime $z\in k[t]$ which is not a factor of any of $q_1,\cdots,q_n$. (We can do this as $k[t]$ has infinitely many primes, while there's a finite list of $q_i$ and each one of them is divisible by finitely many primes). Using the equation above, write $x=\frac{1}{z}$ and clear denominators. This gives
$$q_1^{b_1}\cdots q_n^{b_n} = z\left(\sum_{i=1}^m c_ip_1^{a_{i,1}}\cdots p_n^{a_{i,n}}q_1^{b_1-a_{i,1}}\cdots q_n^{b_n-a_{i,n}}\right)$$
where $b_j=\max{a_{1,j},a_{2,j},\cdots,a_{n,j}}$. The RHS is divisible by $z$, but the LHS is not, as we chose $z$ not to divide any of the $q_i$. This is a contradiction, so it's impossible for $k(t)$ to be finitely generated as a $k$-algebra. Note that the same proof would work for $k(t)$ as a $k[t]$ algebra.
On the other hand, every ring $R$ is finitely generated as an $R$-algebra over itself- the generating set is $\{1\}$. Being careful about what sense "generated" is being used in (as a module? as an algebra? over what ring?) can be a little sticky sometimes on one's first journey through this part of algebra.
