Mixing
Is this a counterexample?
$begingroup$
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 want to prove that every finite extension of $L $ is cyclic (and hence separable).
Isn't this a counterexample ?
Let $p$ be an odd prime, $K= mathbb F_p (t^{2p}) $. Then the polynomial $X^{2p}-t^{2p} $ is irreducible over $K $ with exactly two roots $pm t$ in $overline K $. So there is a $K $-automorphism $f $ of $mathbb F_p (t) $ that sends $t $ to $-t $. This automorphism can be extended to an automorphism of $overline K $.
Now, $L $ (the fixed field defined above) does not contain $t $ (since $f (t)neq t $). So the extension $L(t)/L $ is finite, but not separable as $t $ is not separable.
Is this reasoning correct ? Thank you.
field-theory galois-theory cyclic-groups galois-extensions separable-extension
edited Jan 14 at 19:41
Jyrki Lahtonen
109k13169372
asked Jan 13 at 21:47
FriedrichFriedrich
542112
$endgroup$
add a comment |
$begingroup$
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 want to prove that every finite extension of $L $ is cyclic (and hence separable).
Isn't this a counterexample ?
Let $p$ be an odd prime, $K= mathbb F_p (t^{2p}) $. Then the polynomial $X^{2p}-t^{2p} $ is irreducible over $K $ with exactly two roots $pm t$ in $overline K $. So there is a $K $-automorphism $f $ of $mathbb F_p (t) $ that sends $t $ to $-t $. This automorphism can be extended to an automorphism of $overline K $.
Now, $L $ (the fixed field defined above) does not contain $t $ (since $f (t)neq t $). So the extension $L(t)/L $ is finite, but not separable as $t $ is not separable.
Is this reasoning correct ? Thank you.
field-theory galois-theory cyclic-groups galois-extensions separable-extension
edited Jan 14 at 19:41
Jyrki Lahtonen
109k13169372
asked Jan 13 at 21:47
FriedrichFriedrich
542112
$endgroup$
add a comment |
$begingroup$
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 want to prove that every finite extension of $L $ is cyclic (and hence separable).
Isn't this a counterexample ?
Let $p$ be an odd prime, $K= mathbb F_p (t^{2p}) $. Then the polynomial $X^{2p}-t^{2p} $ is irreducible over $K $ with exactly two roots $pm t$ in $overline K $. So there is a $K $-automorphism $f $ of $mathbb F_p (t) $ that sends $t $ to $-t $. This automorphism can be extended to an automorphism of $overline K $.
Now, $L $ (the fixed field defined above) does not contain $t $ (since $f (t)neq t $). So the extension $L(t)/L $ is finite, but not separable as $t $ is not separable.
Is this reasoning correct ? Thank you.
field-theory galois-theory cyclic-groups galois-extensions separable-extension
edited Jan 14 at 19:41
Jyrki Lahtonen
109k13169372
asked Jan 13 at 21:47
FriedrichFriedrich
542112
$endgroup$
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 want to prove that every finite extension of $L $ is cyclic (and hence separable).
Isn't this a counterexample ?
Let $p$ be an odd prime, $K= mathbb F_p (t^{2p}) $. Then the polynomial $X^{2p}-t^{2p} $ is irreducible over $K $ with exactly two roots $pm t$ in $overline K $. So there is a $K $-automorphism $f $ of $mathbb F_p (t) $ that sends $t $ to $-t $. This automorphism can be extended to an automorphism of $overline K $.
Now, $L $ (the fixed field defined above) does not contain $t $ (since $f (t)neq t $). So the extension $L(t)/L $ is finite, but not separable as $t $ is not separable.
Is this reasoning correct ? Thank you.
field-theory galois-theory cyclic-groups galois-extensions separable-extension
field-theory galois-theory cyclic-groups galois-extensions separable-extension
edited Jan 14 at 19:41
Jyrki Lahtonen
109k13169372
asked Jan 13 at 21:47
FriedrichFriedrich
542112
edited Jan 14 at 19:41
Jyrki Lahtonen
109k13169372
asked Jan 13 at 21:47
FriedrichFriedrich
542112
edited Jan 14 at 19:41
Jyrki Lahtonen
109k13169372
edited Jan 14 at 19:41
Jyrki Lahtonen
109k13169372
edited Jan 14 at 19:41
Jyrki Lahtonen
109k13169372
109k13169372
asked Jan 13 at 21:47
FriedrichFriedrich
542112
asked Jan 13 at 21:47
FriedrichFriedrich
542112
asked Jan 13 at 21:47
FriedrichFriedrich
542112
542112
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
That is not a counterexample.
We have
$$
f(t^2)=f(t)^2=(-t)^2=t^2,
$$
so $t^2in L$.
And $t$ is separable over $Bbb{F}_p(t^2)$ because the zeros of $X^2-t^2=(X-t)(X+t)$ are simple. Hence $t$ is also separable over $L$.
Seems to me that confusion was to think of separability being an intrinsic property of an element when, in fact, it says something about an algebraic element over a certain field. While $t$ is not separable over the smaller field $K$, it is separable over the bigger field $L$.
It is not difficult to show that in your setting actually $L=Bbb{F}_p(t^2)$.
answered Jan 14 at 19:38
Jyrki LahtonenJyrki Lahtonen
109k13169372
$endgroup$
add a comment |
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$begingroup$
That is not a counterexample.
We have
$$
f(t^2)=f(t)^2=(-t)^2=t^2,
$$
so $t^2in L$.
And $t$ is separable over $Bbb{F}_p(t^2)$ because the zeros of $X^2-t^2=(X-t)(X+t)$ are simple. Hence $t$ is also separable over $L$.
Seems to me that confusion was to think of separability being an intrinsic property of an element when, in fact, it says something about an algebraic element over a certain field. While $t$ is not separable over the smaller field $K$, it is separable over the bigger field $L$.
It is not difficult to show that in your setting actually $L=Bbb{F}_p(t^2)$.
answered Jan 14 at 19:38
Jyrki LahtonenJyrki Lahtonen
109k13169372
$endgroup$
add a comment |
$begingroup$
That is not a counterexample.
We have
$$
f(t^2)=f(t)^2=(-t)^2=t^2,
$$
so $t^2in L$.
And $t$ is separable over $Bbb{F}_p(t^2)$ because the zeros of $X^2-t^2=(X-t)(X+t)$ are simple. Hence $t$ is also separable over $L$.
Seems to me that confusion was to think of separability being an intrinsic property of an element when, in fact, it says something about an algebraic element over a certain field. While $t$ is not separable over the smaller field $K$, it is separable over the bigger field $L$.
It is not difficult to show that in your setting actually $L=Bbb{F}_p(t^2)$.
answered Jan 14 at 19:38
Jyrki LahtonenJyrki Lahtonen
109k13169372
$endgroup$
add a comment |
$begingroup$
That is not a counterexample.
We have
$$
f(t^2)=f(t)^2=(-t)^2=t^2,
$$
so $t^2in L$.
And $t$ is separable over $Bbb{F}_p(t^2)$ because the zeros of $X^2-t^2=(X-t)(X+t)$ are simple. Hence $t$ is also separable over $L$.
Seems to me that confusion was to think of separability being an intrinsic property of an element when, in fact, it says something about an algebraic element over a certain field. While $t$ is not separable over the smaller field $K$, it is separable over the bigger field $L$.
It is not difficult to show that in your setting actually $L=Bbb{F}_p(t^2)$.
answered Jan 14 at 19:38
Jyrki LahtonenJyrki Lahtonen
109k13169372
$endgroup$
That is not a counterexample.
We have
$$
f(t^2)=f(t)^2=(-t)^2=t^2,
$$
so $t^2in L$.
And $t$ is separable over $Bbb{F}_p(t^2)$ because the zeros of $X^2-t^2=(X-t)(X+t)$ are simple. Hence $t$ is also separable over $L$.
Seems to me that confusion was to think of separability being an intrinsic property of an element when, in fact, it says something about an algebraic element over a certain field. While $t$ is not separable over the smaller field $K$, it is separable over the bigger field $L$.
It is not difficult to show that in your setting actually $L=Bbb{F}_p(t^2)$.
answered Jan 14 at 19:38
Jyrki LahtonenJyrki Lahtonen
109k13169372
answered Jan 14 at 19:38
Jyrki LahtonenJyrki Lahtonen
109k13169372
answered Jan 14 at 19:38
Jyrki LahtonenJyrki Lahtonen
109k13169372
answered Jan 14 at 19:38
Jyrki LahtonenJyrki Lahtonen
109k13169372
109k13169372
add a comment |
add a comment |
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