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Let $k$ be a field, $L$, $H$ extension fields of $k$, and $G=L\otimes_k H$. I wonder why (I want to know the proof but I can't find) the prime ideal of $G$ must be maximal, and its properties:

a) if $L$ is separable over $k$, then $G$ is reduced.

b) if $L$ is algebraic and purely inseparable over $k$, then $G$ has a unique prime ideal.

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When $L$ is of finite degree, (a) is a straightforward exercise using the primitive element theorem, the CRT, and some basic facts about tensor products. As such, I don't think it's appropriate for the site. For arbitrary algebraic $L$, the same result follows because $L\otimes_kH$ is the union of the reduced rings $F\otimes_kH$ with $F$ a finite subextension of $L/k$ (this follows from flatness and what it means to be algebraic). The statement about prime ideals, at least in the case $L/k$ finite, is a fact about artinian rings. – Keenan Kidwell Jun 29 '10 at 18:56
Thank you !And what about the prime ideal in the case L/k inifinite – TOM Jun 30 '10 at 8:00
up vote 6 down vote accepted

For (a): a field extension $L/k$ is separable is $L$ is a separable algebra, that is, if $L\otimes K$ is a semisimple algebra for all field extensions $K/k$. In particular, if $L$ is separable over $k$ and $K$ is an extension of $k$, then $L\otimes K$ will have no nilpotent elements because it is semisimple.

For details, see for example, Pierce's beautiful book Associative algebras.

For (b), for which the argument is more elaborate, see Jacobson's Lectures in Abstract Algebra, vol. 3, where the result is part 3 of Theorem 21; he does without algebraicity, by the way.

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Dear TOM,

Ad (a) This is the definition of separable! An (arbitrarily big) extension $L/k$ is said to be separable if for any field extension $H/k$ the algebra $G=L\otimes_k H$ is reduced i.e, without non zero nilpotents. This is the definition given by Bourbaki, in Algebra Chapter V, §15. Clearly this answer to your question would be idiotic if I didn't give you criteria for being separable. Here they are. Let $L/k$ be a field extension.

I) if $char (k)=0$, then $L$ is always separable

II) If $char(k)=p> 0$ , then the following are equivalent:

$\quad$ a) the extension $L/k$ is separable

$\quad$ b) there exists a $k$-basis $(a_i)_{i\in I}$ of $L$ seen as a $k$ vector space such that $(a_i^p)_{i\in I}$ is $k$- linearly free

$\quad$ c) in an algebraic closure $L^{alg}$ of $L$, the subextensions $L$ and $k^{\frac{1}{p^\infty}}$ (purely inseparable closure of $k$) are linearly disjoint

$\quad$ d) for every subextension $M$ of $L$ ($k\subset M\subset L$) which is of finite type over $k$ ( as a field extension ! ) there exists a finite transcendence basis $m_1, m_2,...,m_s$ of $M/k$ such that the extension $M/k(m_1, m_2,...,m_s)$ is finite and separable in the elementary sense of galois theory (the minimal polynomial of every element has simple zeroes)

Ad (b) An extension $L/k$ is said to be primary if it is separably closed in the sense that every element in $L$ which is algebraic and separable over $k$ (minimal polynomial has simple zeroes ) already is in $k$. This is the case in your question: a purely inseparable extension is primary. The main result on primary extensions is that the tensor product of a primary extension with any extension has irreducible spectrum i.e. that its nilpotent radical $N=\sqrt{0}$ is prime. To conclude that $N$ is the only prime ideal of $G=L\otimes_k H$ it suffices to show that $G$ has dimension zero.This follows from a result to be found in a surprising place: the $Errata$ to EGA IV, Quatrième partie, Remarque, page 349

Grothendieck's best hidden result If $L$ and $H$ are field extensions of the field $k$, the Krull dimension of their tensor product is given by

$$\operatorname{dim}(L\otimes_k H )= \min (\operatorname{tr.deg} L, \operatorname{tr.deg} H) $$

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Dear Georges, I took the liberty of changing max to min: if $H=k$ then $L\otimes_k H\simeq L$ is a field, so it has Krull dimension 0, and similarly for $H/k$ algebraic (unless there is an error in the erratum, EGA confirms max). Also, huge +++ for "Grothendieck's best hidden result"! In my ignorance, I $\textit{falsely believed}$ (cf…) that the tensor product of fields always has Krull dimension 0, which falls apart immediately upon inspection. – Victor Protsak Jun 30 '10 at 2:21
Dear Victor, thank you for the correction: you are absolutely right, of course. Incidentally, I am quite moved by people who, as you just did, acknowledge their previous misconceptions. I think this honesty is worth infinitely more than any knowledge of the strange behaviour of fields. – Georges Elencwajg Jun 30 '10 at 7:35

For (b) recall that a prime ideal is the kernel of a map from $G$ into a field. Such a field must be an extension field of $H$ and the image of $G$ is generated by $H$ and the $p^{r_i}$-th roots of certain elements $a_i$ of $k$ where $p$ is the characteristic. Thus the image of $G$ is a purely inseparable extension of $H$ and so embeds in the perfect closure $H^i$ of $H$. Now see that there can only one such map from $G$ extending the identity on $H$.

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