Questions tagged [separable-extension]

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3
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1answer
61 views

Does a finite $[E:K]_s$ imply that $[E_s:K]$ is finite?

Let $E/K$ be an algebraic extension of fields and denote by $E_s$ the separable closure of $K$ in $E$, i.e. the set of all elements of $E$ that are separable over $K$ (which can be shown is a (...
3
votes
1answer
146 views

If $L/K$ normal and $H = \operatorname{Aut}(L/K)$, then $L/L^H$ is separable and $L^H/K$ is purely inseparable.

I need to prove the following: Let $L/K$ be a normal field extension. Denote by $H=\operatorname{Aut}(L/K)$ the Galois group of the extension, and by $L^H$ the fixed field of $H$ in $L$. Prove that ...
3
votes
1answer
173 views

“Intersection” of separable subfields [duplicate]

I have the following question, from Isaacs' Algebra book. Suppose $F\le E$ is a finite-degree normal field extension, and that $K$ and $L$ are intermediate subfields (between $F$ and $E$). If $E$ is ...
0
votes
1answer
147 views

If $K$ is a field extension of $F$ and if $\alpha\in K$ is not separable over $F$, show that $\alpha^{p^m}$ is separable over $F$ for some

If $K$ is a field extension of $F$ and if $\alpha\in K$ is not separable over $F$, show that $\alpha^{p^m}$ is separable over $F$ for some $m\geq 0$, where $p = $char$(F)$. I know that $x^p-\alpha^p=(...
3
votes
1answer
77 views

Is this a counterexample?

Suppose $K $ is a field and $\overline K $ an algebraic closure. Let $f $ be a $K $-automorphism of $\overline K$, let $L$ be the subfield of $\overline K $ fixed by $f $. In this post : (link), they ...
2
votes
1answer
65 views

Let $L:K$ be a Galois extention, show that $L:M$ is a normal.

Assume the field extension $L:K$ is Galois (i.e. finite, normal and separable), with $M$ an intermediate field. Show that $L:M$ and $M:K$ are finite separable field extensions. Attempt: ...
0
votes
0answers
42 views

If $a \in L-k $ satisfies $k(a^n)=L$ (for all $n \geq 1$), then $L/k$ is Galois?

Let $k \subsetneq L$ be a finite separable field extension, and let $a \in L-k$ satisfy: For every $n \geq 1$, $k(a^n)=L$. In other words, all the non-zero powers of the primitive element $a$ are ...