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 4.2.1.4, 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

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