Proof that $K\otimes_F L$ is not noetherian

Let $F$ be a field and $K$ and $L$ be extension fields of $F$ such that $\mathrm{tr.deg}_F(K) = \infty$ and $\mathrm{tr.deg}_F(L) = \infty$.

It seems to be proved that $K\otimes_F L$ is not noetherian by the following paper.

P. Vámos, On the minimal prime ideals of a tensor product of two fields, Mathematical Proceedings of the Cambridge Philosophical Society, 84 (1978), pp. 25-35.

However, I don't have an easy access to a university library. I would appreciate if someone would post the proof (which needs not be that of Vámos).

Solution 1:

Here is a slight generalization of the relevant section in Vámos' paper.

Lemma: Let $F \subseteq E$ be a field extension and $A,B$ be two $E$-algebras $\neq 0$. If there is a strictly ascending chain of intermediate fields between $F$ and $E$, then $A \otimes_F B$ is not Noetherian.

Proof: Let $\{F_n\}$ be a strictly ascending chain of intermediate fields between $F$ and $F'$. For every $n \leq m$ we have a commutative diagram $$\begin{array}{c} & A \otimes_F B & \\ & \swarrow ~~~~~~~ \searrow \\ A \otimes_{F_n} B & \rightarrow & A \otimes_{F_m} B. \end{array}$$ Hence, if $I_n$ denotes the kernel of $A \otimes_F B \to A \otimes_{F_n} B$, we see that $\{I_n\}$ is an ascending chain of ideals. It is strictly ascending: Choose $f \in F_{n+1} \setminus F_n$. Then $f \otimes 1 - 1 \otimes f \in I_{n+1}$, but not $\in I_n$ since $f \otimes 1 \neq 1 \otimes f$ in $A \otimes_{F_n} B$ since $\{1,f\}$ is $F_n$-linearly independent in $A$ resp. $B$. $\square$

Corollary: Let $K,L$ be extension fields of $F$ such that $\mathrm{tr.deg}_F(K)$ and $\mathrm{tr.deg}_F(L)$ are infinite. Then $K \otimes_F L$ is not Noetherian.

Proof. Let $E$ be a rational function field over $F$ in countably many variables. Then $E$ embeds into $K$ and $L$, and there is a strictly ascending chain of intermediate fields between $F$ and $E$. Thus the Lemma applies. $\square$