Decompose tensor product of fields into direct product of domains Let $k$ be a perfect field and let $K/k$ be a finitely generated extension of transcendence degree $1$ as well as $L/k$ any extension. I would like to prove that $K\otimes_k L$ is isomorphic to a direct product of domains (as $L$-algebras), where the fraction field of each factor is finitely generated of transcendence degree $1$ over $L$.
So far I have figured out that since $k$ is perfect, $K\otimes_k L$ is reduced. I also know that $K\otimes_k L$ is module-finite over $k(T)\otimes_k L$, where $T\in K$ is transcendental over $k$. Also, $k(T)\otimes_k L$ is a domain with fraction field $L(T)$ (I believe). So far I have failed to decompose $K\otimes_k L$ into a direct product. I know that if $k(T)\otimes_k L$ were a field, then such a product decomposition would be easy to achieve and then each factor would already be a field, more specifically a finite field extension of $k(T)\otimes_k L$. I would like to do something analogous in the case of $k(T)\otimes_k L$ not being a field, i.e. I need a decomposition $K\otimes_k L\cong \prod_i A_i$ such that $k(T)\otimes_k L\hookrightarrow A_i$ is a module-finite extension.
Unfortunately, I have not been able to find a good reference for these things, so any tips on a good source are also highly appreciated.
 A: Let $S= k[x] \setminus \{0\}$. The case that $S^{-1}L[x]$ is a field is clear (and can also be done with the methods described below, it's even simpler in fact), so assume that $S^{-1}L[x]$ is not a field.
Choose a separating transcendence basis for $K/k$ (cf. Thm. 9.27, p. 116 in these field theory notes by Milne ), which must consist of one element, call it $x$. So that $x$ is transcendental over $k$ and $K/k(x)$ is finite separable. Choose a primitive element $\alpha$ for $K/k(x)$, such that $K\cong k(x)[y]/(f)$ for some irreducible separable monic polynomial $f \in k(x)[y]$. 
We get
$K \otimes_k L\\  \cong k(x)[y]/(f) \otimes_k L \\\cong k(x)[y]/(f) \otimes_{k(x)[y]} k(x)[y] \otimes_{k(x)} k(x) \otimes_{k[x]} k[x] \otimes_k L \\ \cong k(x)[y]/(f) \otimes_{k(x)[y]} k(x)[y] \otimes_{k(x)} k(x) \otimes_{k[x]} L[x] \\ \cong k(x)[y]/(f) \otimes_{k(x)[y]} k(x)[y] \otimes_{k(x)} S^{-1}L[x] \\ \cong k(x)[y]/(f) \otimes_{k(x)[y]} (S^{-1}L[x])[y] \\ \cong (S^{-1}L[x])[y]/(f)$.
Let $f=f_1 \cdot \ldots \cdot f_n$ be the factorization of $f$ over $L(x) = \operatorname{Frac}(S^{-1}L[x])$ into monic irreducibles. Since $k(x) \subset S^{-1}L[x]$ and $f$ is monic and $S^{-1}L[x]$ is a PID, all the factors $f_i$ have coefficients in $S^{-1}L[x]$. Since all the $f_i$ are distinct (as $f$ is separable) and irreducible and $S^{-1}L[x][y]$ is a UFD by Gauss' lemma, we have as ideals in $(S^{-1}L[x])[y]$ $$(f)=(f_1) \cdot (f_2) \cdot \ldots \cdot (f_n) = (f_1) \cap (f_2) \cap\ldots \cap (f_n)$$ so that the natural map
$$\varphi:(S^{-1}L[x])[y]/(f) \to \prod_{k=1}^n (S^{-1}L[x])[y]/(f_k)$$ is injective. (If $S^{-1}L[x]$ is a field, then it is also bijective and we're done.) Call the LHS $R$ and the RHS $A$. $A / R$ is an integral extension, since $A$ is already integral over $S^{-1}L[x]$. Both $A$ and $R$ are one-dimensional rings, being integral extensions of the PID $S^{-1}L[x]$. Furthermore, $A$ is equidimensional: Since it is integral over $S^{-1}L[x]$ (note that $S^{-1}L[x] \to A$ is injective), it follows that a prime ideal can't be both maximal and minimal, since we could get a contradiction with going-up and incomparability. (As an alternative argument for equidimensionality, note that maximal ideals in the polynomial ring in one variable over a PID which is not a field are never principal, you can prove this by explicit calculations of the form of maximal ideals or by Krull's principal ideal theorem.)
So it suffices to show the following lemma:

If $R \hookrightarrow A$ is an integral extension of one-dimensional Noetherian rings and $A$ is equidimensional, then if $A$ is a finite product of domains, so is $R$.

The main ingredient is Prop 2.20 in Eisenbud's "Commutative Algebra": A Noetherian ring $R$ is a finite product of domains iff for every maximal ideal $P$ of $R$, the local ring $R_P$ is a domain.   
For $P$ a maximal ideal of $R$, choose a maximal ideal $Q$ in $A$ above $P$, then we get an induced map $g: R_P \to A_Q$. Suppose that $g$ is not injective. Then $\operatorname{ker}(g) = PR_P$, because $R_P$ is one-dimensional and $\operatorname{ker}(g)$ must be a prime ideal, as $A_Q$ is a domain. Thus $A_Q$ is an integral domain and an integral extension of $R_P/PR_P \cong R/P$, which is a field, so $A_Q$ must be a field, thus there is no prime ideal properly contained in $Q$, so $Q$ is a minimal prime ideal. But since $A$ is equidimensional, a maximal prime ideal can't be minimal, thus $g$ is injective and $R_P$ is a domain, because $A_Q$ is a domain.
This finishes the proof.
