Finitely generated as a field vs. as an algebra?

Question

Let $K$ be a field, $L$ a $K$-algebra. For $x_1, \ldots, x_n \in L$, denote $K[x_1, \ldots, x_n]$ for the smallest subalgebra of $L$ containing $K$ and $x_1, \ldots, x_n$. If $L$ is a field, denote $K(x_1, \ldots, x_n)$ for the smallest subfield of $L$ containing $K$ and $x_1, \ldots, x_n$. Clearly $K[x_1, \ldots, x_n] \subseteq K(x_1, \ldots, x_n)$.

If $L$ is a field and $L = K(x_1, \ldots, x_n)$ for some $x_1, \ldots, x_n \in L$, can we conclude that $L = K[y_1, \ldots, y_m]$ for some $y_1, \ldots, y_m \in L$? Both hints and full answers are appreciated.

Answer 1

There is an easy way to fix this, and it's actually related to a well-known theorem:

Zariski's lemma: If $L/K$ is a field extension such that $L$ is a finitely generated $K$-algebra, then $L$ is a ($K$-vector space) finite extension of $K$.

The theorem this is related to, and morally equivalent to, is Hilbert's nullstellensatz.

So, if you know that $L=K[x_1,\ldots,x_n]$ then in fact, $L/K$ is finite. From that we derive the following easy (to state) solution to your problem:

Corollary: Let $L/K$ be an extension with $L=K(x_1,\ldots,x_n)$ with $x_i\in L$. Then, the following are equivalent:

1. $L=K[y_1,\ldots,y_m]$ for some $y_j\in L$.
2. $L=K[x_1,\ldots,x_n]$.
3. $L/K$ is finite.

So, one easily sees how 'DonAntonio' found their answer--one needs to adjoin to $K$ a transcendental.

Answer 2

No. For example, $\;L=\Bbb Q(\pi)\;$ is a field, but it can't be that $\;L=\Bbb Q[\alpha]\;$ , as:

(1) If $\;\alpha\;$ is algebraic over $\;\Bbb Q\;$ then $\;\Bbb Q[\alpha]=\Bbb Q(\alpha)\neq\Bbb Q(\pi)\;$ , as the last one is infinite dimensional over $\;\Bbb Q\;$ whereas the first one is finite dimensional, and

(2) if $\;\alpha\;$ is transcendental over $\;\Bbb Q\;$ then $\;\Bbb Q[\alpha]\;$ is not even a field and in fact $\;\Bbb Q[\alpha]\cong\Bbb Q[x]=$ the "usual" polynomial ring of rational polynomials