For field extension $M/L/K$ with $M/K$ normal, if $\sigma\in\mathrm{Aut}(M/K)$ implies $\sigma(L)\subset M^{\mathrm{Aut}(M/L)}$, must $L/K$ be normal?

Question

Let $M/K$ be a normal extension, $L$ be an intermediate field. Suppose that for every $\sigma\in\mathrm{Aut}(M/K)$, $\sigma(L)\subset M^{\mathrm{Aut}(M/L)}$ ($M^{\mathrm{Aut}(M/L)}$ is the subfield of $M$ fixed by $\mathrm{Aut}(M/L)$). Is it true that for every $\sigma\in\mathrm{Aut}(M/K)$, $\sigma(L)\subset L$?

This question says that $M^{\mathrm{Aut}(M/L)}$ is a purely inseparable extension of $L$, but what can happen if $M^{\mathrm{Aut}(M/L)}$ is strictly larger than $L$?

score 1 · Accepted Answer · answered Mar 10 '23 at 01:37

1

Try $K = \Bbb{F}_p( x^p+y^p,x^py^p),L =\Bbb{F}_p(x^p,y), M=\Bbb{F}_p(x,y)$

answered Mar 10 '23 at 01:37

reuns

79,880

You really saved my day, thanks a lot! – Jianing Song Mar 10 '23 at 08:08

For field extension $M/L/K$ with $M/K$ normal, if $\sigma\in\mathrm{Aut}(M/K)$ implies $\sigma(L)\subset M^{\mathrm{Aut}(M/L)}$, must $L/K$ be normal?

1 Answers1