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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 $L\otimes_k k(x_1,\dots,x_n)=(k[x_1,\dots,x_n]-(0))^{-1}L[x_1,\dots,x_n]$ ?

brunoh
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    Let $\Delta: k(x)\otimes_k k(x) \rightarrow k(x)$ be the ring homomorphism induced by multiplication. In other words $\Delta(a \otimes b)=ab$ and extend linearly. Now $x\otimes x^{-1} - 1\otimes 1$ is a nontrivial element of the kernel of $\Delta$, so $k(x) \otimes_k k(x)$ has a non-trivial ideal. – PVAL-inactive Mar 24 '15 at 00:14
  • @PVAL thank you for comment which is clear. So it means Eisenbud's proof is wrong. And is my correction of it correct ? Ps: I upvoted your comment. – brunoh Mar 24 '15 at 00:25
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    I do not have Eisenbud handy or otherwise. It might help others to know what the result of Maclane being proven was, so they could give other references for it. Perhaps in context, it might be clear that the statement for $L/k$ algebraic is enough, but I really do not know. – PVAL-inactive Mar 24 '15 at 00:34
  • @PVAL And by the way how does the tensorial product $k(x)\otimes k(x)$ looks like ? Is there a simple way to describe it without the tensor product (e.g. using localization ...) ? Something like "polynomials in $y$ with coefficients in $k(x)$ divided by polynomials in $y$ with coefficients in $k$" ? – brunoh Mar 24 '15 at 00:36
  • @PVAL nope. I have checked that. – brunoh Mar 24 '15 at 00:36
  • If I recall correctly it is not even finitely generated as a $k$-algebra. I certainly do not know a "nice" description for it or what it looks like. Perhaps someone else here does. – PVAL-inactive Mar 24 '15 at 00:51
  • Your question in the title doesn't really match your post completely. – Pedro Mar 24 '15 at 17:36
  • @PedroTamaroff Yes, you are right unfortunately. But the question did arise from a mistake I tried to understand in Eisenbud's and a discussion with PVAL. So after some edits listed in the question, I thought this title was more catchy and summarize everything well. Sorry again. – brunoh Mar 24 '15 at 17:39
  • The map $k(x) \otimes k(y) \to k(x,y)$ induced by $a \otimes b \mapsto ab$ isn't surjective, because I'm pretty sure $\frac 1{x+y}$ isn't in its image. – mercio Mar 24 '15 at 17:45
  • @mercio yes you are right. IMO, $k(x)\otimes k(x)$ looks like polynomials in $y$ with coefficients in $k(x)$ divided by polynomials in $y$ with coefficients in $k$, but that is the whole point of my question. – brunoh Mar 24 '15 at 18:01
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    What you wrote down doesn't look symmetric in $x$ and $y$. I think it's ${ f/gh ; f \in k[x,y], g \in k[x] - {0}, h \in k[y] - {0} }$ – mercio Mar 24 '15 at 18:09
  • @mercio I agree with you ... but it is also actually the same than what I wrote ! Just way more elegant by making it symmetric. If you prove it and make it like answer I will considered it as such. Meanwhile I upvoted your comment. – brunoh Mar 24 '15 at 18:24

2 Answers2

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@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.

Duchamp Gérard H. E.
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  • @Duchamp_Gérard I like your answer, except for the proof using basis (which seems to be correct), but thank you very much, I will accept it as an answer and upvote it. As another proof is it okay to write that $L\otimes_k k(x)=L\otimes_{k[x]} (k[x]\otimes_k k(x))= (L\otimes_{k[x]} k[x]_{(0)})\otimes_k k[x]=(k[x]-{0})^{-1}L\otimes k[x]=(k[x]-{0})^{-1}L[x]$ ?? In the case of $L=k(y)$, it gives $(k[x]-{0})^{-1}(k[y]-{0})^{-1}k[x,y]$. – brunoh Mar 25 '15 at 16:32
  • @brunoh Thanks a lot. What is $k[x]_{(0)}$ ? – Duchamp Gérard H. E. Mar 26 '15 at 06:57
  • @Duchamp_Gérard_H.E. $k[x]_{(0)}$ is the localization at (0) of $k[x]$, i.e. its fraction field, i.e. $k(x)$. – brunoh Mar 26 '15 at 08:14
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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 $$L\otimes_k k(x)=L\otimes_k (k[x]\otimes_{k[x]} k(x))$$ $$=(L\otimes_k k[x])\otimes_{k[x]} k[x]_{(0)}$$ $$=L[x]\otimes_{k[x]} k[x]_{(0)}$$ $$=(k[x]-{0})^{-1}L[x]$$

brunoh
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    It is correct ! You can check it using the universal properties (1) of the tensor product in one way and (2) of $S^{-1}A$ in the other way and constructing two inverse arrows. +1 – Duchamp Gérard H. E. Mar 26 '15 at 20:01
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    [except for the proof using basis]---> by the way, using universal properties, one can prove very easily that $$A[S^{-1}]\otimes_R B[T^{-1}]\simeq A\otimes_R B[(S\otimes T)^{-1}]$$ and one recovers the result without bases. – Duchamp Gérard H. E. Mar 27 '15 at 14:24
  • @Duchamp_Gérard thank you very much. I like it ! +1 – brunoh Mar 27 '15 at 14:44