Around a little mistake in Eisenbud's Commutative Algebra: What does $k(x)\otimes k(x)$ look like? In Eisenbud's Commutative Algebra with a view towards Algebraic Geometry book, in Appendix A1 p. 564 in his proof of a result of Maclane, Prof. Eisenbud said that if $L$ is an extension of a field $k$, we have $L\otimes_k k(x_1,\dots,x_n)=L(x_1,\dots,x_n)$ which is the field of rational functions over $L$ in $n$ variables. Nevertheless, a result of A. Grothendieck is that $$\dim(L\otimes_k K)=\min (\text{trdeg}_k(K),\text{trdeg}_k(L))$$
Therefore, if $L$ is a transcendental extension of $k$, $L\otimes_k k(x_1,\cdots,x_n)$ cannot be a field. 
There is something I do not understand here. What is it ?
Edit: to save the proof of the result of MacLane's it seems to me it is sufficient to say that $L\otimes_k k(x_1,\dots,x_n)$ is a localization of  $L\otimes_k k[x_1,\dots,x_n]=L[x_1,\dots,x_n]$ which is clearly reduced. Am I right ?
Edit2: Following discussions in the comment, I am now curious to know the structure of $k(x)\otimes_k k(x)$. Can someone describe it simply to me without the tensor product ? It should at least be a domain ...
Edit3 : Am I right to write that $L\otimes_k k(x_1,\dots,x_n)=(k[x_1,\dots,x_n]-(0))^{-1}L[x_1,\dots,x_n]$ ?
 A: @brunoh (I post this as a - partial - answer as it is too long for a comment) [And by the way how does the tensorial product $k(x)\otimes_k k(x)$ look like ?]--> You have an easy combinatorial description of $k(x)\otimes_k k(x)$: it is, through the arrow $\varphi:\ k(x)\otimes_k k(x)\rightarrow k(x,y)$ (which turns out to be an embedding), the sub-algebra $k[x,y]S^{-1}$, where $S$ is the multiplicative semigroup generated by the irreducible polynomials in $k[x]$ and the irreducible polynomials in $k[y]$ or, equivalently, fractions of the form 
$$\frac{P(x,y)}{Q_1(x)Q_2(y)}\ ;\ Q_i\not\equiv 0\ .$$
  When $k$ is algebraically closed this is easy to see as the standard partial fraction decomposition basis reads 
$$\{x^n\}_{n\geq 0}\sqcup \{\frac{1}{(x-a)^n}\}_{a\in k\atop n\geq 1}\ .$$
Calling $B$ this basis, one has just to check that the image by $\varphi$ of $B\otimes B$ is linearly independent which seems to be a routine.
When $k$ is not algebraically closed, you have to combine elements of the basis $B$ to get a basis of $k(x)$ which reads, in this case 
$$\{x^n\}_{n\geq 0}\sqcup \{\frac{x^m}{P^n}\}_{P\in Irr(k[x])\atop m<deq(P),\ n\geq 1}\ .$$
and proceed as above.
A: To complement the answer of Gérard Duchamp, I would like to give a bit more general proof to my own query, correcting the one I gave in my last comment
We have $$L\otimes_k k(x)=L\otimes_k (k[x]\otimes_{k[x]} k(x))$$
$$=(L\otimes_k k[x])\otimes_{k[x]} k[x]_{(0)}$$
$$=L[x]\otimes_{k[x]} k[x]_{(0)}$$
$$=(k[x]-{0})^{-1}L[x]$$
