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Let $K$ be an extension field of a field $k$. We say $K$ is separably generated over $k$ if $K$ has a transcendence basis $S$ over $k$ such that $K$ is separably algebraic over $k(S)$.

Let $k$ be a perfect field of characteristic $p \neq 0$. Let $K$ be an extension field of $k$. It is well-known that if $K$ is finitely generated over $k$, $K$ is separably generated over $k$. I would like to know examples of $K$ which is not separably generated over $k$ in the following cases if any.

(1) tr.dim $K/k < \infty$

(2) tr.dim $K/k = \infty$

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1 Answer 1

up vote 1 down vote accepted

Edit New proof.

Let $k$ be a perfect field of characteristic $p>0$. Let $I$ be a non-empty set. Fix an algebraic closure $L$ of $k(t_i)_{i\in I}$ and consider $$K=k(t_i^{1/p^{\infty}})_{i\in I}:=\cup_{n\ge 1} k(t_i^{1/p^n})_{i\in I}\subseteq L.$$ We will show that $K$ can't be algebraic separable over a purely transcendental subextension $k(S)$.

Suppose the contrary. The transcendental basis $S$ can be indexed by $I$ and we write $S=\{ s_i\}_{i\in I}$. Fix an index $i_0\in I$. Then $s_{i_0}$ is a fraction in some $t_j$'s. So $s_{i_0}^{1/p}$ exists in $K$ (here we use $k$ is perfect, but this doesn't really matter). But $K$ is separable over $k(S)$. Thus $s_{i_0}^{1/p}$ belongs to the purely trancendental $k(S)=k(s_i)_{i\in I}$. But this is obviously false.

Let $k'/k$ be any field extension. As $K$ is union of finitely generated subextensions $K_i$ and $K\otimes_k k'$ is the union (colimit to be exact) of the reduced $k$-algebras $K_i\otimes_k k'$, $K\otimes_k k'$ itself is also reduced.

So any field extension $K/k$ over a perfect field $k$ is separable.

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Dear QiL, If an extension field $K$ of a field $k$ has a transcendence basis $S$ over a field $k$ such that $K$ is separably algebraic over $k(S)$, we say $K$ is separably generated over $k$. The question asks for examples of $K$ which is not separably generated over a perfect field $k$. –  Makoto Kato Dec 31 '12 at 13:09
    
@MakotoKato: ah I had "geometrically reduced" in mind ! –  user18119 Dec 31 '12 at 16:30
    
Dear QiL, Could you give us a hint for (2)? –  Makoto Kato Jan 1 '13 at 20:44
    
Dear Makoto: if $K$ is algebraic separable over some subextension $K'$, then for any $m\ge 1$, $K'$ contains some $t_m^{1/p^{r_m}}$. Replacing $t_m$ with this root, we have $k(t_1, \dots, t_m, \dots)\subseteq K'\subseteq K$. As $K$ is purely inseparable over $k(t_1,\dots, t_m, \dots)$, this implies that $K=K'$. So the question is whether $K$ its self is purely transcendental over $k$. –  user18119 Jan 1 '13 at 23:06
    
Could you explain why $K'$ contains some $t_m^{1/p^{r_m}}$ for any $m \ge 1$? –  Makoto Kato Jan 2 '13 at 1:05

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