All Questions
6 questions
10
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1
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243
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If $E_\text{sep}/F$ is normal, then must $E/F$ be normal?
This question has been asked in Math.StackExchange (see here) for more than a week and I even put a bounty on it. But still it hasn't been correctly answered (the current answer there was written by ...
1
vote
1
answer
149
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$F=\mathbb{C}(u,v)$ satisfying: For every $a,b \in \mathbb{C}[y],c,d \in \mathbb{C}[x]$: $\mathbb{C}(x,y)=F(ax+b)=F(cy+d)$
Let $u,v \in \mathbb{C}[x,y]$, where $u$ and $v$ are algebraically independent over $\mathbb{C}$ and $F=\mathbb{C}(u,v)$. Of course, $d:=[\mathbb{C}(x,y):F] < \infty$.
Denote the following ...
0
votes
1
answer
473
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A subfield $R \subseteq \mathbb{C}(x,y)$ with 'many' generators $w$, $R(w)=\mathbb{C}(x,y)$
Let $R \subseteq \mathbb{C}(x,y)$ and assume that $R=\mathbb{C}(u,v)$, where $u,v \in \mathbb{C}[x,y]$ are algebraically independent over $\mathbb{C}$.
Here $\mathbb{N}$ includes $0$.
Assume that $R$ ...
1
vote
0
answers
59
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If $E \subseteq F=k(x_1,\ldots,x_r)$, satisfies $E(x_1^{i_1},\ldots,x_r^{i_r})=F$, for every $(i_1,\ldots,i_r) \neq (0,\ldots,0)$, then $[F:E] \leq 2$
For $r \geq 2$, let $A_r=\mathbb{C}[x_1,\ldots,x_r]$,
$F_r=\mathbb{C}(x_1,\ldots,x_r)$ the field of fractions of $A_r$, and $E_r \subseteq F_r$ an arbitrary subfield of $F_r$ with $[F_r:E_r] < \...
1
vote
0
answers
360
views
A composition of a simple extension and a separable extension is simple
Let $K/L/M$ be a tower of finite field extensions with $K/L$ separable and $L/M$ simple (in the sense of being generated by a single element). How does one show that $K/M$ is also simple?
I know that ...
37
votes
4
answers
12k
views
Finite extension of fields with no primitive element
What is an example of a finite field extension which is not generated by a single element?
Background: A finite field extension E of F is generated by a primitive element if and only if there are a ...