# Is this field extension finite?

Let $\mathbb{F}, \mathbb{F'}$ be fields, $\mathbb{F} \subset \mathbb{F'}$, and $\alpha \in \mathbb{F'}$. Assume $\mathbb{F(\alpha)}$ is finitely generated algebra over $\mathbb{F}$. Is this true: $(\mathbb{F}(\alpha):\mathbb{F}) < \infty$ ?

First i thought it's a consequence of this theorem: Finitely Generated Algebraic Extension is Finite

But we don't know if $\alpha$ is algebraic or not.

If $\alpha$ is transcendental (hence basically a free variable) over $F$, then one may show that $F(\alpha)$ is not finitely generated as follows: Notice that $F[\alpha]$ is a Euclidean ring and hence factorial. A standard argument shows that $F[\alpha]$ contains infinitely many prime elements.
Let now $\frac{f_1}{g_1},\dotsc,\frac{f_n}{g_n}\in F(\alpha)$ be arbitrary elements and denote by $S$ the subalgebra generated by these. It suffices to show $S\neq F(\alpha)$. Let $\{p_1,\dotsc,p_N\}$ be the set of distinct primes dividing $g_1\dotsm g_n$. If $p$ is a prime element with $p\notin \{p_1,\dotsc,p_N\}$ then it is clear that $\frac 1p\notin S$ (as the denominator of each element of $S$ has to be a product of primes from $\{p_1,\dotsc,p_N\}$). This proves $S\neq F(\alpha)$ and as $S$ was an arbitrary finitely generated subalgebra of $F(\alpha)$, it follows that $F(\alpha)$ itself cannot be finitely generated.
By contraposition we have shown that if $F(\alpha)$ is a finitely generated algebra over $F$, then $\alpha$ is algebraic over $F$ and hence $[F(\alpha) : F] < \infty$.
• As I understood, this proof relies on fact that there are infinitely many primes, so we can choose such $p$ that $p\notin \{p_1,\dotsc,p_N\}$. But if $\alpha$ is algebraic, why this doesn't hold true? – Khan Mar 16 '17 at 6:33
• If $\alpha$ is algebraic, then $F(\alpha) = F[\alpha]$ is a field. Hence there are no prime elements. We really need that we are working with a polynomial ring. – Claudius Mar 16 '17 at 8:13