Prove that a polynomial of the form $x^p−x−a in F[x]$ is either irreducible or splits in $F$.
$begingroup$
Let $F$ be a field with char$F = p$. Prove that a polynomial of the form $x^p−x−a in F[x]$ is either irreducible or splits in $F$.
I've seen a few different ways of proving this with $x^p-a$. I know that $a in F$ being a root of $F(x)$ means $a^p=a$, so $x^p−a=(x−a)^p$. But I don't think this property works for $x^p-x-a$.
abstract-algebra irreducible-polynomials
asked May 2 '18 at 16:01
TimTim
305
$endgroup$
add a comment |
$begingroup$
Let $F$ be a field with char$F = p$. Prove that a polynomial of the form $x^p−x−a in F[x]$ is either irreducible or splits in $F$.
I've seen a few different ways of proving this with $x^p-a$. I know that $a in F$ being a root of $F(x)$ means $a^p=a$, so $x^p−a=(x−a)^p$. But I don't think this property works for $x^p-x-a$.
abstract-algebra irreducible-polynomials
asked May 2 '18 at 16:01
TimTim
305
$endgroup$
$begingroup$
Close to being a duplicate of this.
$endgroup$
– Jyrki Lahtonen
May 3 '18 at 18:52
add a comment |
$begingroup$
Let $F$ be a field with char$F = p$. Prove that a polynomial of the form $x^p−x−a in F[x]$ is either irreducible or splits in $F$.
I've seen a few different ways of proving this with $x^p-a$. I know that $a in F$ being a root of $F(x)$ means $a^p=a$, so $x^p−a=(x−a)^p$. But I don't think this property works for $x^p-x-a$.
abstract-algebra irreducible-polynomials
asked May 2 '18 at 16:01
TimTim
305
$endgroup$
Let $F$ be a field with char$F = p$. Prove that a polynomial of the form $x^p−x−a in F[x]$ is either irreducible or splits in $F$.
I've seen a few different ways of proving this with $x^p-a$. I know that $a in F$ being a root of $F(x)$ means $a^p=a$, so $x^p−a=(x−a)^p$. But I don't think this property works for $x^p-x-a$.
abstract-algebra irreducible-polynomials
abstract-algebra irreducible-polynomials
asked May 2 '18 at 16:01
TimTim
305
asked May 2 '18 at 16:01
TimTim
305
asked May 2 '18 at 16:01
TimTim
305
asked May 2 '18 at 16:01
TimTim
305
asked May 2 '18 at 16:01
TimTim
305
305
$begingroup$
Close to being a duplicate of this.
$endgroup$
– Jyrki Lahtonen
May 3 '18 at 18:52
add a comment |
$begingroup$
Close to being a duplicate of this.
$endgroup$
– Jyrki Lahtonen
May 3 '18 at 18:52
$begingroup$
Close to being a duplicate of this.
$endgroup$
– Jyrki Lahtonen
May 3 '18 at 18:52
$begingroup$
Close to being a duplicate of this.
$endgroup$
– Jyrki Lahtonen
May 3 '18 at 18:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is an Artin-Schreier polynomial. If $alpha$ is a root,
the other roots are $alpha+k$ for $kinBbb F_p$.
If $alphanotin F$ then for some nonzero $kinBbb F_p$,
$alpha+k$ is a conjugate of $alpha$ over $F$. Then $alpha+2k$
is a conjugate of $alpha+k$ and so also a conjugate of $alpha$, etc.
Then all $alpha+mk$ are conjugates of $alpha$, and these are all
the roots. Therefore $x^p-x-a$ is irreducible in this case.
answered May 2 '18 at 16:08
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$begingroup$
So does it split when $a=0$? Because this seems to only work for $a$ nonzero?
$endgroup$
– Tim
May 2 '18 at 17:02
$begingroup$
Of course it splits when $a=0$, but also in other cases when $FneBbb F_p$. For instance, $X^2-X-1$ splits over $Bbb F_4$.
$endgroup$
– Lubin
May 2 '18 at 18:15
add a comment |
Your Answer
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is an Artin-Schreier polynomial. If $alpha$ is a root,
the other roots are $alpha+k$ for $kinBbb F_p$.
If $alphanotin F$ then for some nonzero $kinBbb F_p$,
$alpha+k$ is a conjugate of $alpha$ over $F$. Then $alpha+2k$
is a conjugate of $alpha+k$ and so also a conjugate of $alpha$, etc.
Then all $alpha+mk$ are conjugates of $alpha$, and these are all
the roots. Therefore $x^p-x-a$ is irreducible in this case.
answered May 2 '18 at 16:08
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$begingroup$
So does it split when $a=0$? Because this seems to only work for $a$ nonzero?
$endgroup$
– Tim
May 2 '18 at 17:02
$begingroup$
Of course it splits when $a=0$, but also in other cases when $FneBbb F_p$. For instance, $X^2-X-1$ splits over $Bbb F_4$.
$endgroup$
– Lubin
May 2 '18 at 18:15
add a comment |
$begingroup$
This is an Artin-Schreier polynomial. If $alpha$ is a root,
the other roots are $alpha+k$ for $kinBbb F_p$.
If $alphanotin F$ then for some nonzero $kinBbb F_p$,
$alpha+k$ is a conjugate of $alpha$ over $F$. Then $alpha+2k$
is a conjugate of $alpha+k$ and so also a conjugate of $alpha$, etc.
Then all $alpha+mk$ are conjugates of $alpha$, and these are all
the roots. Therefore $x^p-x-a$ is irreducible in this case.
answered May 2 '18 at 16:08
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$begingroup$
So does it split when $a=0$? Because this seems to only work for $a$ nonzero?
$endgroup$
– Tim
May 2 '18 at 17:02
$begingroup$
Of course it splits when $a=0$, but also in other cases when $FneBbb F_p$. For instance, $X^2-X-1$ splits over $Bbb F_4$.
$endgroup$
– Lubin
May 2 '18 at 18:15
add a comment |
$begingroup$
This is an Artin-Schreier polynomial. If $alpha$ is a root,
the other roots are $alpha+k$ for $kinBbb F_p$.
If $alphanotin F$ then for some nonzero $kinBbb F_p$,
$alpha+k$ is a conjugate of $alpha$ over $F$. Then $alpha+2k$
is a conjugate of $alpha+k$ and so also a conjugate of $alpha$, etc.
Then all $alpha+mk$ are conjugates of $alpha$, and these are all
the roots. Therefore $x^p-x-a$ is irreducible in this case.
answered May 2 '18 at 16:08
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
This is an Artin-Schreier polynomial. If $alpha$ is a root,
the other roots are $alpha+k$ for $kinBbb F_p$.
If $alphanotin F$ then for some nonzero $kinBbb F_p$,
$alpha+k$ is a conjugate of $alpha$ over $F$. Then $alpha+2k$
is a conjugate of $alpha+k$ and so also a conjugate of $alpha$, etc.
Then all $alpha+mk$ are conjugates of $alpha$, and these are all
the roots. Therefore $x^p-x-a$ is irreducible in this case.
answered May 2 '18 at 16:08
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered May 2 '18 at 16:08
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered May 2 '18 at 16:08
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered May 2 '18 at 16:08
Lord Shark the UnknownLord Shark the Unknown
104k1160132
104k1160132
$begingroup$
So does it split when $a=0$? Because this seems to only work for $a$ nonzero?
$endgroup$
– Tim
May 2 '18 at 17:02
$begingroup$
Of course it splits when $a=0$, but also in other cases when $FneBbb F_p$. For instance, $X^2-X-1$ splits over $Bbb F_4$.
$endgroup$
– Lubin
May 2 '18 at 18:15
add a comment |
$begingroup$
So does it split when $a=0$? Because this seems to only work for $a$ nonzero?
$endgroup$
– Tim
May 2 '18 at 17:02
$begingroup$
Of course it splits when $a=0$, but also in other cases when $FneBbb F_p$. For instance, $X^2-X-1$ splits over $Bbb F_4$.
$endgroup$
– Lubin
May 2 '18 at 18:15
$begingroup$
So does it split when $a=0$? Because this seems to only work for $a$ nonzero?
$endgroup$
– Tim
May 2 '18 at 17:02
$begingroup$
So does it split when $a=0$? Because this seems to only work for $a$ nonzero?
$endgroup$
– Tim
May 2 '18 at 17:02
$begingroup$
Of course it splits when $a=0$, but also in other cases when $FneBbb F_p$. For instance, $X^2-X-1$ splits over $Bbb F_4$.
$endgroup$
– Lubin
May 2 '18 at 18:15
$begingroup$
Of course it splits when $a=0$, but also in other cases when $FneBbb F_p$. For instance, $X^2-X-1$ splits over $Bbb F_4$.
$endgroup$
– Lubin
May 2 '18 at 18:15
add a comment |
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$begingroup$
Close to being a duplicate of this.
$endgroup$
– Jyrki Lahtonen
May 3 '18 at 18:52