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Does anybody know a reference for basic properties of tensor products of (finite) algebraic extensions of fields?

Ideally, I would like a description of $L \otimes_k K$ for arbitrary finite extensions $L, K$ of $k$ but I would settle for a reference for results such as

1) If $K / k$ is Galois with group $G$ then $K \otimes_k K \cong \oplus_{g \in G} K$.

2) If $K / k$ is purely inseparable then $K \otimes_k K$ is local with residue field $K$ and length $[K : k]$.

3) If $K / k$, $L / k$ are separable then $K \otimes_k L$ has no nilpotent elements.

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The last result is proved in Zariski and Samuel's Commutative algebra I, pgs. 195-6. They require only one of $K/k$, $L/k$ to be separable. – Adeel Dec 10 '10 at 12:51
Section 8.18 of Jacobson's basic algebra II is called "tensor products of fields" and proves several basic results (such as your 3); in fact, only one of $K/k$ or $L/k$ need be separable). Of course, he doesn't say anything about length. – Rob Harron Dec 10 '10 at 15:23
Of course ? – Georges Elencwajg Dec 10 '10 at 17:29
For question 1, see the last two pages of and for question 3, see Theorem 1.2 at – KConrad Dec 11 '10 at 12:31
I just reread the question and saw that it was focussed on algebraic extensions, rather than more general extensions, hence my suggestion of Lang was not particularly sensible, and I have deleted it. On the other hand, Todd Trimble in his answer below is completely correct: these can all be proved directly. Using the primitive element theorem, as he (implicitly) suggests, is a good tool for handling (1) and (3). In the case of (2), you can write $K$ as a succession of extensions obtained by extracting $p$th roots, and the computation is then pretty straightforward too. – Emerton Dec 12 '10 at 4:36

3 Answers 3

up vote 10 down vote accepted

Dear anon, the most complete reference might be Bourbaki's Algèbre, Chapter V.

For question 1, I suggest Bourbaki's Algèbre, Chapter V, §10, 4. Descente galoisienne, Corollaire . There the Master proves the more general result that the canonical morphism $$K\otimes_{k} K\to K^G: x\otimes y\mapsto (x \sigma (y))_{\sigma \in G} $$ is injective for any Galois extension $K/k$, finite or infinite, with Galois group $G$, and bijective if the extension is finite.

Question 3 is trivial from Bourbaki's point of view since for Him the definition of $K/k$ being a separable extension is that for any field extensions $L/k$, the $k$-algebra $K\otimes _{k} L$ has no nilpotents (neither $K$ nor $L$ is assumed finite-dimensional over $k$). As a concession to less enlightened mortals, He proves in §15, Exemple 3, that if the extension $K/k$ is algebraic ( for example finite-dimensional) this notion coincides with the one that you and I are familiar with: the minimal polynomial of any element in $K$ has simple roots .

For question 2, I cannot give you a reference which exactly answers your question. However a purely inseparable extension is a particular case of a primary extension and these are considered at the end of our reference, in §17,2. Produit d'extensions . The Corollaire there shows that the nilpotent radical $P$ of $K\otimes_{k} K$ is prime and since this algebra is finite-dimensional, it is local of dimension zero with unique prime ideal $P$ . We still must prove that its length is $[K:k]$. This is equivalent to the claim that the $K$-algebra $ (K\otimes_{k} K) /P $ is $K$ . This follows from the existence of the product map $K\otimes K \to K$ sending $x\otimes y$ to $xy$.The kernel of this map is exactly the unique prime ideal $P$ of $K\otimes K$ . (By the way, an excellent reference for the notion of "length" is Appendix A to Fulton's book Intersection Theory ; Example A.1.1 page 407 is relevant to the above discussion)

PS If you are not familiar with exotic languages, you will be relieved to know that this volume of Bourbaki exists in English translation.

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+1. Vénération touchante et justifiée ! – Chandan Singh Dalawat Dec 11 '10 at 3:23

I don't have references per se, but these can be proven hands-on. For (1) I would quote the normal basis theorem: if $w \in K$ is such that orbit of the Galois group on $w$ forms a $k$-basis, then by abstract nonsense, the functor $K \otimes_k -$ on $k$-algebras preserves the cokernel of the map $k[x] \to k[x]/(f) \cong K$ where

$$f(x) = \prod_{\sigma \in G} (x - \sigma(w))$$

so $K \otimes_k K \cong K[x]/(\prod_\sigma (x - \sigma(w)))$, which splits as $\prod_\sigma K[x]/(x - \sigma(w)) \cong \prod_\sigma K$ by the Chinese remainder theorem. This isomorphism is compatible with the Galois group action by the normal basis theorem.

The others can be handled by similar techniques. I think (3) actually reduces to (1) because if $E$ is a Galois extension of $k$ containing both $K$ and $L$, then $K \otimes_k L$ is a subalgebra of $E \otimes_k E$, and the latter contains no nilpotent elements by the previous calculation.

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I would like to second Todd's suggestion that you try to prove these yourself directly. – Emerton Dec 12 '10 at 4:37
I can and have proven all these myself by hand - evidently they are not difficult - but sometimes when you are in the middle of a proof you just want a simple statement that you can be almost sure is true without having to lose 15 minutes scribbling in a margin. Also, I like all human beings am not infallible and it is somewhat reassuring to see something in print. – anon Jul 11 '11 at 0:15
Actually the reference is more for the latter reason than the former as I have never taken a course which included Galois or field theory. – anon Jul 11 '11 at 0:19

If $K/k$ is separable, then $K=k[\alpha]\approx k[X]/(f(X))$ where $f(X)$ is the minimal polynomial of $\alpha$. Let $L$ be a field containing $k$. Then $f(X)=f_1(X)...f_r(X)$ in $L[X]$ with the $f_i$ irreducible and distinct (because $K/k$ is separable). Therefore,

$L\otimes_kK\approx L[X]/(f(X))\approx \prod L[X]/(f_i(X))$

by the Chinese remainder theorem. This describes $K\otimes L$ completely as a product of fields when $K/k$ is separable. For example, if which $f(X)$ splits in $L$, say $f(X)=(X-\alpha_{1})\cdots(X-\alpha_{n})$, then

$L\otimes_{k}K\approx L[X]/(f(X))\approx\prod_{i}L[X]/(X-\alpha_{i})\approx\prod L_{i}$

with $L_{i}=L$. The map $L\otimes_{k}K\rightarrow L_{i}$ sends $a\otimes g(\alpha)$ to $ag(\alpha_{i})$. This takes care of 1) and 3).

As for 2), if $K=k[\alpha]$ with $\alpha^{p}\in k$, then $K\otimes_{k}K=K[\epsilon]$ where $\epsilon =\alpha\otimes1-1\otimes\alpha$ and $\epsilon^{p}=\alpha^{p}\otimes 1-1\otimes\alpha^{p}=0$. That gets you started on 2).

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