How to show this tensor product is regular?

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Let $L,K$ be field extensions of a field $k$, and let $K$ be finitely generated ($K=k(x_1,\cdots,x_n)$). I need to prove $K\otimes_k L$ is regular in the following two cases:

  1. $K$ is separated.
  2. $L$ is separated.

Here the definition of "separated" is : the tensor product of it with any field extension of $k$ is reduced.

I can't handle any of these cases.

In the first case, I have done some calculation:

We can assume $K=k(x_1,\cdots,x_n,y)$. Here $x_1,\cdots,x_n$ is a transcendental basis, and $y$ is separated over $k':=k(x_1,\cdots,x_n)$. Denote the minimal polynomial of $y$ over $k'$ by $f$. Set $S=k'\setminus\{0\}$, $A=k[x_1,\cdots,x_n]$, then $k'=S^{-1}A$. We know:

$$K\otimes_k L=k'(y)\otimes_{k'}(k' \otimes_k L)$$

But

$$k' \otimes_k L=S^{-1}A\otimes_k L=S^{-1}A\otimes_A (A\otimes_k L)=S^{-1}A\otimes_A L[x_1,\cdots,x_n]=S^{-1}L[x_1,\cdots,x_n]\triangleq B$$

Hence we have:

$$K\otimes_k L=k'(y)\otimes_{k'}B=B[T]/(f)$$

But I don't know how to prove this ring is regular...

In the second case, I have no idea.

Could you help me finish the solution or give other methods? Thanks!

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Question: "Let $L,K$ be field extensions of a field $k$, and let $K$ be finitely generated $(K=k(x1,⋯,xn))$. I need to prove $K⊗_k L$ is regular in the following two cases: $K$ is separated. $L$ is separated. Here the definition of "separated" is : the tensor product of it with any field extension of k is reduced."

Answer: The usual terminology is (you find this in Matsumuras book on commutative ring theory) that $k \rightarrow K$ is "separable" iff for any field extension $k \subseteq l$ it follows $l \otimes_k K$ is reduced.

If $k \subseteq K$ is separable it follows in particular that $L\otimes_k K$ is reduced. If $dim_k(L)=n$ it follows $dim_K(L\otimes_k K)=n$ and hence $L\otimes_k K\cong K_1 \oplus \cdots \oplus K_d$ is a direct sum of fields. It follows $L\otimes_k K$ is regular. Hence this proves the claim in this particular case.