Problem regarding proving an extension of a field to be separable
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The whole question looks like-
Let, $x^p-x-1$ be a polynomial over a field $F$ of characteristic
$pne 0$ and $alpha$ be a root of it. Prove that $F(alpha)$ is
separable extension over $F$.
I have tried a bit of the problem which goes as follows-
Note that, if $F$ is finite i.e. $FsimeqBbb{Z}_p$ then the result is obvious, since any irreducible polynomial over a finite field cannot have multiple root.
Again, $f’(x)=-1ne0$, hence $f(x)$ has all roots simple.
Now, let $p(x)in F[x]$ is the minimal polynomial of $alpha$ over $F$. Then $p(x)|f(x)implies$ any root of $p(x)$ is a root of $f(x)implies p(x)$ cannot have multiple roots.
So, I get $alpha$ is separable over $F$. From this I cannot get any idea how to show $F(alpha)$ is separable over $F$.
Can anybody complete this proof? Thanks for assistance in advance.
abstract-algebra field-theory extension-field separable-extension
asked Dec 5 at 12:53
Biswarup Saha
4769
add a comment |
up vote
0
down vote
favorite
The whole question looks like-
Let, $x^p-x-1$ be a polynomial over a field $F$ of characteristic
$pne 0$ and $alpha$ be a root of it. Prove that $F(alpha)$ is
separable extension over $F$.
I have tried a bit of the problem which goes as follows-
Note that, if $F$ is finite i.e. $FsimeqBbb{Z}_p$ then the result is obvious, since any irreducible polynomial over a finite field cannot have multiple root.
Again, $f’(x)=-1ne0$, hence $f(x)$ has all roots simple.
Now, let $p(x)in F[x]$ is the minimal polynomial of $alpha$ over $F$. Then $p(x)|f(x)implies$ any root of $p(x)$ is a root of $f(x)implies p(x)$ cannot have multiple roots.
So, I get $alpha$ is separable over $F$. From this I cannot get any idea how to show $F(alpha)$ is separable over $F$.
Can anybody complete this proof? Thanks for assistance in advance.
abstract-algebra field-theory extension-field separable-extension
asked Dec 5 at 12:53
Biswarup Saha
4769
AN extension generated by separable elements is a separable extension.
– Lord Shark the Unknown
Dec 5 at 12:54
And how do you deal with the case $F = mathbb{F}_p(t)$ ? To get the intuition for separability you probably need the concept of fixed field. Distinct roots implies we'll have enough automorphisms (of the splitting field) for $F$ being the fixed field of the Galois group.
– reuns
Dec 5 at 12:58
Lord Shark the Unknown, yes intuitively looks like. But I can't write the proof properly.
– Biswarup Saha
Dec 5 at 13:07
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The whole question looks like-
Let, $x^p-x-1$ be a polynomial over a field $F$ of characteristic
$pne 0$ and $alpha$ be a root of it. Prove that $F(alpha)$ is
separable extension over $F$.
I have tried a bit of the problem which goes as follows-
Note that, if $F$ is finite i.e. $FsimeqBbb{Z}_p$ then the result is obvious, since any irreducible polynomial over a finite field cannot have multiple root.
Again, $f’(x)=-1ne0$, hence $f(x)$ has all roots simple.
Now, let $p(x)in F[x]$ is the minimal polynomial of $alpha$ over $F$. Then $p(x)|f(x)implies$ any root of $p(x)$ is a root of $f(x)implies p(x)$ cannot have multiple roots.
So, I get $alpha$ is separable over $F$. From this I cannot get any idea how to show $F(alpha)$ is separable over $F$.
Can anybody complete this proof? Thanks for assistance in advance.
abstract-algebra field-theory extension-field separable-extension
asked Dec 5 at 12:53
Biswarup Saha
4769
The whole question looks like-
Let, $x^p-x-1$ be a polynomial over a field $F$ of characteristic
$pne 0$ and $alpha$ be a root of it. Prove that $F(alpha)$ is
separable extension over $F$.
I have tried a bit of the problem which goes as follows-
Note that, if $F$ is finite i.e. $FsimeqBbb{Z}_p$ then the result is obvious, since any irreducible polynomial over a finite field cannot have multiple root.
Again, $f’(x)=-1ne0$, hence $f(x)$ has all roots simple.
Now, let $p(x)in F[x]$ is the minimal polynomial of $alpha$ over $F$. Then $p(x)|f(x)implies$ any root of $p(x)$ is a root of $f(x)implies p(x)$ cannot have multiple roots.
So, I get $alpha$ is separable over $F$. From this I cannot get any idea how to show $F(alpha)$ is separable over $F$.
Can anybody complete this proof? Thanks for assistance in advance.
abstract-algebra field-theory extension-field separable-extension
abstract-algebra field-theory extension-field separable-extension
asked Dec 5 at 12:53
Biswarup Saha
4769
asked Dec 5 at 12:53
Biswarup Saha
4769
asked Dec 5 at 12:53
Biswarup Saha
4769
asked Dec 5 at 12:53
Biswarup Saha
4769
asked Dec 5 at 12:53
Biswarup Saha
4769
4769
AN extension generated by separable elements is a separable extension.
– Lord Shark the Unknown
Dec 5 at 12:54
And how do you deal with the case $F = mathbb{F}_p(t)$ ? To get the intuition for separability you probably need the concept of fixed field. Distinct roots implies we'll have enough automorphisms (of the splitting field) for $F$ being the fixed field of the Galois group.
– reuns
Dec 5 at 12:58
Lord Shark the Unknown, yes intuitively looks like. But I can't write the proof properly.
– Biswarup Saha
Dec 5 at 13:07
add a comment |
AN extension generated by separable elements is a separable extension.
– Lord Shark the Unknown
Dec 5 at 12:54
And how do you deal with the case $F = mathbb{F}_p(t)$ ? To get the intuition for separability you probably need the concept of fixed field. Distinct roots implies we'll have enough automorphisms (of the splitting field) for $F$ being the fixed field of the Galois group.
– reuns
Dec 5 at 12:58
Lord Shark the Unknown, yes intuitively looks like. But I can't write the proof properly.
– Biswarup Saha
Dec 5 at 13:07
AN extension generated by separable elements is a separable extension.
– Lord Shark the Unknown
Dec 5 at 12:54
AN extension generated by separable elements is a separable extension.
– Lord Shark the Unknown
Dec 5 at 12:54
And how do you deal with the case $F = mathbb{F}_p(t)$ ? To get the intuition for separability you probably need the concept of fixed field. Distinct roots implies we'll have enough automorphisms (of the splitting field) for $F$ being the fixed field of the Galois group.
– reuns
Dec 5 at 12:58
And how do you deal with the case $F = mathbb{F}_p(t)$ ? To get the intuition for separability you probably need the concept of fixed field. Distinct roots implies we'll have enough automorphisms (of the splitting field) for $F$ being the fixed field of the Galois group.
– reuns
Dec 5 at 12:58
Lord Shark the Unknown, yes intuitively looks like. But I can't write the proof properly.
– Biswarup Saha
Dec 5 at 13:07
Lord Shark the Unknown, yes intuitively looks like. But I can't write the proof properly.
– Biswarup Saha
Dec 5 at 13:07
add a comment |
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AN extension generated by separable elements is a separable extension.
– Lord Shark the Unknown
Dec 5 at 12:54
And how do you deal with the case $F = mathbb{F}_p(t)$ ? To get the intuition for separability you probably need the concept of fixed field. Distinct roots implies we'll have enough automorphisms (of the splitting field) for $F$ being the fixed field of the Galois group.
– reuns
Dec 5 at 12:58
Lord Shark the Unknown, yes intuitively looks like. But I can't write the proof properly.
– Biswarup Saha
Dec 5 at 13:07