Reference for tensor products of fields - MathOverflow

Reference for tensor products of fields

2010-12-10T12:18:44Z

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.

Answer by Todd Trimble for Reference for tensor products of fields

2010-12-10T13:23:20Z

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.

Answer by Georges Elencwajg for Reference for tensor products of fields

2010-12-10T14:03:39Z

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.

Answer by Mephisto for Reference for tensor products of fields

2010-12-12T07:38:59Z

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).