Reference for tensor products of fields - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:53:07Z http://mathoverflow.net/feeds/question/48912 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48912/reference-for-tensor-products-of-fields Reference for tensor products of fields anon 2010-12-10T12:18:44Z 2010-12-13T17:59:30Z <p>Does anybody know a reference for basic properties of tensor products of (finite) algebraic extensions of fields?</p> <p>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</p> <p>1) If $K / k$ is Galois with group $G$ then $K \otimes_k K \cong \oplus_{g \in G} K$.</p> <p>2) If $K / k$ is purely inseparable then $K \otimes_k K$ is local with residue field $K$ and length $[K : k]$.</p> <p>3) If $K / k$, $L / k$ are separable then $K \otimes_k L$ has no nilpotent elements.</p> http://mathoverflow.net/questions/48912/reference-for-tensor-products-of-fields/48921#48921 Answer by Todd Trimble for Reference for tensor products of fields Todd Trimble 2010-12-10T13:23:20Z 2010-12-10T13:23:20Z <p>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 </p> <p>$$f(x) = \prod_{\sigma \in G} (x - \sigma(w))$$ </p> <p>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. </p> <p>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. </p> http://mathoverflow.net/questions/48912/reference-for-tensor-products-of-fields/48924#48924 Answer by Georges Elencwajg for Reference for tensor products of fields Georges Elencwajg 2010-12-10T14:03:39Z 2010-12-12T00:59:44Z <p>Dear anon, the most complete reference might be Bourbaki's Algèbre, Chapter V.</p> <p><strong>For question 1</strong>, 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 <em>or infinite</em>, with Galois group $G$, and bijective if the extension is finite.</p> <p><strong>Question 3</strong> 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 . </p> <p><strong>For question 2</strong>, 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 <em>Corollaire</em> 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 <em>Intersection Theory</em> ; Example A.1.1 page 407 is relevant to the above discussion)</p> <p><strong>PS</strong> If you are not familiar with exotic languages, you will be relieved to know that this volume of Bourbaki exists in English translation. </p> http://mathoverflow.net/questions/48912/reference-for-tensor-products-of-fields/49128#49128 Answer by Mephisto for Reference for tensor products of fields Mephisto 2010-12-12T07:38:59Z 2010-12-13T17:59:30Z <p>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,</p> <p>$L\otimes_kK\approx L[X]/(f(X))\approx \prod L[X]/(f_i(X))$</p> <p>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</p> <p>$L\otimes_{k}K\approx L[X]/(f(X))\approx\prod_{i}L[X]/(X-\alpha_{i})\approx\prod L_{i}$</p> <p>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). </p> <p>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).</p>