Around a little mistake in Eisenbud's Commutative Algebra: What does $k(x)\otimes k(x)$ look like?

@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.