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Rectified erroneous "F" to "k"
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Georges Elencwajg
Georges Elencwajg
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Dear Zev, as LASER pointed out, Lang's Proposition 6.11 indeed solves your problem. Still, I'd like to add two remarks that might put your problem in perspective.

A) Given a field $K$ and a group of automorphisms $G$ of $K$ we get a fixed field $K^G$ and an extension $K^G \subset K$ . This extension is algebraic if and only if all orbits of $G$ are finite and if this is the case the extension is always separable: each element of $x \in K$ has as minimal polynomial over $K^G$ the product $\prod \limits_{y \in Orb(x)}(T-y)$. [Let me emphasize that i) there is no base field $k$ in this general statement and ii) the group $G$ may very well be infinite with all its orbits finite: every infinite dimensional galois extension gives an example.]

B) The statement is false if you don't assume normality of $k \subset K$ (as I'm sure you guessed!). Here is an example taken from Bourbaki's Algebra V, exercise 3) for §5.

Take a field $F$ of characteristic $p>2$ and define $k=F(x,y)$ where $x,y$ are indeterminates. Let $\theta$ be a zero of the polynomial $P(T)=T^{2p}+xT^p+y$ (in an algebraic closure of $k$, say).Then if we put $K=F(\theta)$$K=k(\theta)$, we have our counter-example: no element in $K$ is purely inseparable over $k$, except if it already is in $k$. And yet $K$ is not separable over k since the minimal polynomial of $\theta$ over $k$ is $P(T)$ , which is obviously not separable.

Dear Zev, as LASER pointed out, Lang's Proposition 6.11 indeed solves your problem. Still, I'd like to add two remarks that might put your problem in perspective.

A) Given a field $K$ and a group of automorphisms $G$ of $K$ we get a fixed field $K^G$ and an extension $K^G \subset K$ . This extension is algebraic if and only if all orbits of $G$ are finite and if this is the case the extension is always separable: each element of $x \in K$ has as minimal polynomial over $K^G$ the product $\prod \limits_{y \in Orb(x)}(T-y)$. [Let me emphasize that i) there is no base field $k$ in this general statement and ii) the group $G$ may very well be infinite with all its orbits finite: every infinite dimensional galois extension gives an example.]

B) The statement is false if you don't assume normality of $k \subset K$ (as I'm sure you guessed!). Here is an example taken from Bourbaki's Algebra V, exercise 3) for §5.

Take a field $F$ of characteristic $p>2$ and define $k=F(x,y)$ where $x,y$ are indeterminates. Let $\theta$ be a zero of the polynomial $P(T)=T^{2p}+xT^p+y$ (in an algebraic closure of $k$, say).Then if we put $K=F(\theta)$, we have our counter-example: no element in $K$ is purely inseparable over $k$, except if it already is in $k$. And yet $K$ is not separable over k since the minimal polynomial of $\theta$ over $k$ is $P(T)$ , which is obviously not separable.

Dear Zev, as LASER pointed out, Lang's Proposition 6.11 indeed solves your problem. Still, I'd like to add two remarks that might put your problem in perspective.

A) Given a field $K$ and a group of automorphisms $G$ of $K$ we get a fixed field $K^G$ and an extension $K^G \subset K$ . This extension is algebraic if and only if all orbits of $G$ are finite and if this is the case the extension is always separable: each element of $x \in K$ has as minimal polynomial over $K^G$ the product $\prod \limits_{y \in Orb(x)}(T-y)$. [Let me emphasize that i) there is no base field $k$ in this general statement and ii) the group $G$ may very well be infinite with all its orbits finite: every infinite dimensional galois extension gives an example.]

B) The statement is false if you don't assume normality of $k \subset K$ (as I'm sure you guessed!). Here is an example taken from Bourbaki's Algebra V, exercise 3) for §5.

Take a field $F$ of characteristic $p>2$ and define $k=F(x,y)$ where $x,y$ are indeterminates. Let $\theta$ be a zero of the polynomial $P(T)=T^{2p}+xT^p+y$ (in an algebraic closure of $k$, say).Then if we put $K=k(\theta)$, we have our counter-example: no element in $K$ is purely inseparable over $k$, except if it already is in $k$. And yet $K$ is not separable over k since the minimal polynomial of $\theta$ over $k$ is $P(T)$ , which is obviously not separable.

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Georges Elencwajg
Georges Elencwajg
  • 47.5k
  • 14
  • 159
  • 241

Dear Zev, as LASER pointed out, Lang's Proposition 6.11 indeed solves your problem. Still, I'd like to add two remarks that might put your problem in perspective.

A) Given a field $K$ and a group of automorphisms $G$ of $K$ we get a fixed field $K^G$ and an extension $K^G \subset K$ . This extension is algebraic if and only if all orbits of $G$ are finite and if this is the case the extension is always separable: each element of $x \in K$ has as minimal polynomial over $K^G$ the product $\prod \limits_{y \in Orb(x)}(T-y)$. [Let me emphasize that i) there is no base field $k$ in this general statement and ii) the group $G$ may very well be infinite with all its orbits finite: every infinite dimensional galois extension gives an example.]

B) The statement is false if you don't assume normality of $k \subset K$ (as I'm sure you guessed!). Here is an example taken from Bourbaki's Algebra V, exercise 3) for §5.

Take a field $F$ of characteristic $p>2$ and define $k=F(x,y)$ where $x,y$ are indeterminates. Let $\theta$ be a zero of the polynomial $P(T)=T^{2p}+xT^p+y$ (in an algebraic closure of $k$, say).Then if we put $K=F(\theta)$, we have our counter-example: no element in $K$ is purely inseparable over $k$, except if it already is in $k$. And yet $K$ is not separable over k since the minimal polynomial of $\theta$ over $k$ is $P(T)$ , which is obviously not separable.